1 Groups

Definition 1.1 (Group). A Group G is a set with a binary operation m :G×G→ G (m(g, h) := gh) such that

1. ∀a, b, c ∈ G, (ab)c = a(bc)

2. ∃e ∈ G such that ∀a ∈ G, ea = ae = a

3. ∀a ∈ G, ∃a−1 ∈ G such that aa−1 = a−1a = e

If only 1 is satisfied, then G is a semigroup.If only 1 and 2 are satisfied, then G is a monoid.If a structure has ∀a, b ∈ G, ab = ba, then it is abelian.

Definition 1.2 (Group Homomorphism). A map f : G→ H between groups isa homomorphism if f(ab) = f(a)f(b)

If the homomorphism is injective, it is a monomorphism.If the homomorphism is surjective, it is an epimorphism.If the homomorphism is bijective, it is an isomorphism.

Lemma 1.1. Let ϕ : G → H be a group homomorphism. Then ϕ(eG) = eHand ϕ(a−1) = ϕ(a)−1

Proof. ϕ(a) = ϕ(aeG) = ϕ(a)ϕ(eG)ϕ(a)−1ϕ(a) = ϕ(a)−1ϕ(a)ϕ(eG) = eH = eHϕ)eG) = ϕ(eG)The other part is similar.

Definition 1.3 (Kernel of a Homomorphism). The kernel of a homomorphismf : G→ H is the set a ∈ G : f(a) = eH and is denoted ker f

Definition 1.4 (Subgroup). If G is a group and H ⊆ G is itself a group underG’s multiplication, then H is a subgroup of G, denoted H < G

Trivially, ker f < G

Lemma 1.2. A nonempty subset H ⊆ G is a subgroup iff ∀a, b ∈ H, ab−1 ∈ G

Definition 1.5 (hom(G,H)). If G,H are groups, then hom(G,H) is the set ofhomomorphisms from G to H.

Definition 1.6 (Cosets). Let H < G. The the left coset of H containing a isaH = ah : h ∈ H.

Right cosets are similarly defined.

Lemma 1.3. If |H| <∞ then |aH| = |H| = |Ha|

Proof. Let h : H → aH : h 7→ ah. Then ah = ah′ ⇒ h = h′, so f isbijective.

Lemma 1.4. Let H < G. Then for a, b ∈ G, either aH = bH or aH ∩ bH = ∅.Also, aH = bH iff a−1b ∈ H

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Proof. Suppose aH ∩ bH 6= ∅. Then ∃h, h′ ∈ H such that ah = bh′.So a = bh′h−1 thus a ∈ bH. Thus aH ⊂ bH. This argument is symmetric,

thus aH = bH.Then, if aH = bH we have a−1b = hh′−1 ∈ HIf a−1b = h ∈ H, then b = ah so b ∈ aH and aH ∩ bH 6= ∅, so aH = bH.

Definition 1.7 (Order of g). The order of g ∈ G is the smallest positive n suchthat gn = e, or ∞ is no such n exists.

Theorem 1.5 (Lagrange’s Theorem). If G is a finite group, then the order ofany element divides |G|.

Proof. We first check that each element has finite order.Since |G| <∞, there must be m,n > 0 such that gn = gm, where n < mSo gm−n = e.Suppose that g has order n. Then consider e, g, g2, . . . , gn−1 = HThe left cosets of H partition G, and each is size n, so |G| = nk, where k is

the number of left cosets of G.

Definition 1.8 (Direct Product Group). If G,H are groups, then G ×H haselements (g, h) : g ∈ G, h ∈ H and multiplication (g, h)(g′, h′) = (gg′, hh′)

Definition 1.9 (Normal Subgroup). A subgroup H of G is said to be normalif ∀a ∈ G, aH = Ha. We write H E G

Lemma 1.6. N E G iff aNa−1 = H ∀a ∈ G iff aNa−1 ⊆ N ∀a ∈ G.

Proof. If N is normal, then aN = Na, so aNa−1 ⊆ N .Also, ∀n′ ∈ N , ∃n ∈ N such that an = n′a so ana−1 = n′ ∈ N , thus

n′ ∈ aNa−1. Thus aNa−1 = NIf aNa−1 = N , then ∀n ∈ N , ∃n′ ∈ N such that an′a−1 = n so an′ = na

and thus Na ⊂ aN . This argument is symmetric, so N E G.

Definition 1.10 (Index). If H < G then the index of H in G, written [G : H]is the number of left cosets of H. This may be infinite.

Theorem 1.7. If N E G then the set aN : a ∈ G is a group of order [G : N ]with operation (aN)(bN) = (ab)N

Proof. If aN = a′N , bN = b′N then abN = a′b′N . as aN = a′N, a−1a′ ∈ N ,similarly b−1b′ ∈ N .

(ab)−1(a′b′) = b−1a−1a′b′ = b−1nb for some n ∈ N .Then, there is an n′ such that b−1nb′ = b−1b′n′ = n′′ ∈ NThus, abN = a′b′NThis multiplication is associative, because G is a group, and the other group

properties follow similarly.

Definition 1.11 (Quotient Group). The group in the theorem above is calledG/N .

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Note: f : G→ H a homomorphism, then ker f E G.

Theorem 1.8. If f : G→ H is a homomorphism and N E G, N ⊆ ker f then∃!f : G/N → H such that f(aN) = f(a) ∀a ∈ G. Then Im f = Im(f) andker f = ker f/N and f is an isomorphism iff f is surjective and N = ker f .

Proof. See Hungerford.

Corollary 1.9 (First Isomorphism Theorem). If f : G→ H is a group homo-morphism, then G/ ker f ' Im f

Definition 1.12 (HK). If H and K are subsets of a group, then HK = hk :h ∈ H, k ∈ K

Definition 1.13 (Join). If H,K < G then H ∨K is the smallest subgroup ofG containing both H and K.

Whenever the smallest subgroup of G containing some set S is mentioned,it is the subgroup ∩Hi, where Hi < G and S ⊆ Hi.

Lemma 1.10. If N E G,K < G, then

1. N ∩K E K

2. N E N ∨K

3. NK = N ∨K = KN

Proof. 1. If n ∈ N∩K, a ∈ K, then ana−1 ∈ N∩K, so a(N∩K)a−1 ⊂ N∩K,so N ∩K E K

2. N E G and N < N ∨K < G, so N E N ∨K.

3. Since N ∨K is closed under multiplication, NK ⊆ N ∨K. Let a ∈ N ∨K,write a = n1k1 . . . , nrkr

Since N is normal, ∀n ∈ N , we can write kjn = n′kj for some n′ ∈ N .

So a = nk1 . . . kr = nk with n ∈ N, k ∈ K. So a ∈ NK.

Theorem 1.11 (Second Isomorphism Theorem). If K,N < G and N E G thenK/N ∩K ' NK/N

Proof. Define a homomorphism f : K → NK/N by f(k) = kN .

K NK/N

N ∨K

................................................................................................................................................................................................................................................... ............f

............................................................................................................................................................................ ............

ι

.............................................................................................................................................................................

π

ker f = N ∩K. By the first isomorphism theorem, K/N ∩K ' Im f

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Thus, it remains to show that Im f = NK/N .Consider nkN ∈ NK/N . Since N is normal, ∃n′ such that nkN = kn′N =

kN = f(k).

Theorem 1.12 (Third Isomorphism Theorem). Let K E G and H E G, withK < H. Then H/K E G/K and (G/K)/(H/K) E G/H

Proof. Define a homomorphism f : G/K → G/H by f(aK) = aH. f is ahomomorphism, as K ⊂ ker(G→ G/H)

ker f = H/K, and f is surjective by definition, so by the first isomorphismtheorem, the result follows.

Definition 1.14 (Simple Group). A group G is a simple group if it has noproper nontrivial normal subgroups.

Now we will determine solutions to the problem of how can a group bedescribed.

1. Listing the elements and making a table

2. Give the generators for G as a subgroups of Sn, finite groups only.

3. Give it as AutX for some structure X.

4. Build up from simpler groups

5. Give generators for G as a subgroup of GLn (giving a homomorphismf : G→ GLn is the beginning of representation theory)

6. Generators and Relations.

Theorem 1.13 (Cayley). Any finite group G is isomorphic to a subgroup ofSn for some n.

Proof. Let n = |G|, and fix a bijection 1, . . . , n → G such thatG = g1, . . . , gn.Given g ∈ G, ∀i∃j such that ggi = gj . Note that if ggi = ggk then gi = gk.Define σg : 1, . . . , n → 1, . . . , n by σg(i) = j if ggi = gjThis is an injection from a finite set into itself, and so a bijection. We will

define ϕ : G→ Sn : g 7→ σgCheck that σgh = σgσh. ϕ is a homomorphism.Also, if ggi = gi then g = e, so σg 6= 1 for g 6= e. So ϕ is an injection, this

ϕ(G) ' G and is a subgroup of Sn.

Corollary 1.14. Every finite group is isomorphic to a subgroup of GLn(R) forn = |G|.

Proof. By Cayley’s Theorem, it suffices to show that Sn is isomorphic to asubgroup of GLn

Define ϕ : Sn → GLn(R) : σ 7→ A written Aij =

1 if σ(j) = i0 else

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Definition 1.15 (F (X)). Given a set X, we choose a set X−1 which is disjointfrom X and has the same cardinality, with a bijection X → X−1 : x 7→ x−1,and we define an element 1 ∈ X ∪X−1.

A word is a sequence (a1, . . . , ) with ai ∈ X ∪X−1 ∪ 1 such that ∃N suchthat an = 1∀n ≥ N .

A word is reduced if ai = x ⇒ ai+1, ai−1 6= x−1 and if an = 1 then am =1∀m ≥ n.

F (X) is the set of reduced words in X.

Lemma 1.15. F (X) is a group with respect to concatenation and reduction ofreduced words.

Lemma 1.16. If G is a group and f : X → G is a map of sets, then there is aunique homomorphism f : F (X)→ G such that f = f ι, where ι : X → F (X)is the inclusion map. That is, F (X) is free in the category of groups.

Proof. Define f(1) = e ∈ G.If xλ1

1 . . . xλnn is a nonempty reduced word, then f(xλ1

1 . . . xλnn ) = f(x1)λ1 . . . f(xn)λn .

f is a homomorphism.

Corollary 1.17. Every group is a homomorphic image of a free group.

Proof. Let X be a set of generators for G, that is, no proper subgroup containsX.

Let ι : X → G be the inclusion. Then ι : F (X) → G is a group homomor-phism and is surjective. Thus, by the first isomorphism theorem, ι(F (X)) ' G.

That is, G ' F (X)/ ker ι

Definition 1.16 (N(H)). The normal subject generated by a set H is the in-tersection, N(H), of all normal subgroups containing H.

And so, any group can be given by generators X and relations R by 〈X|R〉where G ' F (X)/N(R).

G is finitely generated if X is finite.G is finitely presented if both X and R are finite.

Definition 1.17 (Free Product of Groups). If G,H are groups, then the groupG ∗H consists of all reduced words in G ∪H.

2 Structure of Groups

Recall that a free group F (X) is free on X in the category of groups. Whatabout the category of abelian groups?

If X = x1, . . . , xn then Zn =⊕n

i=1 Z is free on X is the category of abeliangroups.

For X infinite, we want the restricted direct sum, that is, we have onlyfinitely many nonzero terms.

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Definition 2.1 (Basis of an Abelian Group). A basis of an abelian group F isX ⊆ F such that F = 〈X〉 and

∑nixi = 0⇒ ni = 0, for ni ∈ Z.

Lemma 2.1. F has a basis X iff F ' ⊕′x∈XZ, where ⊕′ is the restricted directsum.

Proof. Define a map ϕ⊕′x∈X Z→ F : a 7→∑x∈X,ax 6=0 axx

ϕ is a homomorphism. ϕ is also surjective, as F = 〈X〉. If a ∈ kerϕ then∑axx = 0⇒ ax = 0, so a = 0.Thus, ϕ is an isomorphism.

Lemma 2.2. Assume G ' ⊕ni=1Gi, H E G, H ' ⊕ni=1Hi and Hi E Gi. ThenG/H ' ⊕ni=1Gi/Hi.

Proof. Define πi : Gi → Gi/Hi : gi 7→ giHi.Ππi : G→ ⊕Gi/Hi : (gi) 7→ ⊕giHi.Ππi is a surjective homomorphism with kernel H.

Theorem 2.3. If X,Y are bases of an abelian group F , then |X| = |Y | = n orboth are infinite. |X| is the rank of F .

Proof. We may assume |X| = n. Consider 2F = 2u : u ∈ F < F , andconsider F/2F .

As F has basis X, F ' Zn and 2F ' ⊕2Z.By the lemma, F/2F ' Zn/⊕ 2Z ' ⊕Z/2Z, so |F/2F | = 2n.If |Y | > n, the same argument says |F/2F | > 2n, so |Y | = |X| = n.

Theorem 2.4. If F is a free abelian group of rank n and G < F is nontrivial,then ∃ basis x1, . . . , xn of F and r(1 ≤ r ≤ n) and d1, . . . , dr ∈ N such thatd1| . . . |dr such that G is a free abelian group with basis dixi|1 ≤ i ≤ r.

Proof. The main ideas of the proof are the Euclidean algorithm: given n, d ∈ Z,∃q, r ∈ Z with 0 ≤ r < |d| with n = qd+ r and if x1, . . . , xn is a basis for Zn,then so is x1 +

∑ni=2 aixi, . . . , xn.

We proceed by induction on n.Suppose n = 1. Then F ' Z, G ≤ F .Let d be the smallest positive element of G. Let n ∈ G. Write n = qd + r,

0 ≤ r < d.As r = n− qd ∈ G, and r < d, r = 0. So n = qd ∈ 〈d〉 so G = dZ.Let S be the set of r ∈ Z such that ∃ a basis y1, . . . , yn for F and v ∈ G

with v = ry1 + k2y2 + . . .+ knyn.AsG is nonempty, S is nonempty. Since y2, y1, . . . , yn is a basis, k2, . . . , kn ∈

S.Let d be the smallest positive element of S, which exists if G 6= 0.So there is a basis y1, . . . , yn for F and v ∈ G with v = d1y1 +

∑ni=2 kiyi.

Write ki = d1qi+ri, then v = d1(y1+∑ni=2 qiyi)+

∑riyi and y1+

∑ni=2 qiyi, y2, . . . , yn

is again a basis for F so ri ∈ S. So since ri < d1, ri = 0 ∀i.Let x1 = y1 +

∑ni=2 qiyi, v = d1x1.

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Let H = 〈y2, . . . , yn〉, and consider G ∩H.We know F = 〈x1〉 ⊕H. We will show that G ' 〈v〉 ⊕ (G ∩H)First 〈v〉 ∩ (G ∩H) = 0, since av = ad1x1 =

∑ni=1 kiyi. them −ad1xi +∑n

i=1 kiyi = 0, so −ad1 = ki = 0 for all i, so av = 0.Next let g ∈ G. Write g = ax1 +

∑ni=2 kiyi. Write a = qd1 + r, 0 ≤ r ≤ d1.

So g − qv ∈ G and g − qv = rx1 +∑kiyi so r ∈ S, and as r < d1, r = 0 so

q − qv =∑kiyi ∈ G ∩H.

Now we see 〈v〉⊕G∩H → G : (av, g′) 7→ av+g′ is a surjective homomorphismwith trivial kernel, and so is an isomorphism.

Since H is free abelian of smaller rank, by induction, ∃ a basis x2, . . . , xn forH such that G ∩H = 〈d2x2, . . . , drxr〉 with d2| . . . |dn.

Now x1, . . . , xn is a basis for F and G = 〈d1x1, . . . , dnxn〉Write d2 = qd1 + r 0 ≤ r < d1, then d2x2 + d1x1 ∈ G and d2x2 + d1x2 =

qd1x2 + rx2 + d1x1 = rx2 + d1(x1 + qx2), since x2, x1 + qx2, . . . , xn is a basisfor F , r ∈ S so r = 0. Thus, d1|d2.

Corollary 2.5 (Classification Theorem of Finitely Generated Abelian Groups).Every finitely generated abelian group is isomorphic to a finite direct sum ofcyclic groups in which finite cyclic summands, if any, are of orders m1, . . . ,mt

where m1| . . . |mt.

Proof. If G is generated by n > 0 elements, then there is a surjection π : Zn →G.

If π is an isomorphism, then done. Else, let K = kerπ since K ≤ Zn, thereis a basis x1, . . . , xn for Zn such that K = 〈d1x1, . . . , drxr〉, dr ∈ N, r ≤ n.

Now G ' Zn/K ' ⊕x∈XZ/diZ, with di = 0 if i > r.If di = 0, then Z/diZ ' Z, if di = 1, then Z/diZ ' 0.Let m1, . . . ,mt in order be the number of di > 1.Then G ' Zs ⊕ Z/m1Z⊕ . . .⊕ Z/mtZ.

Corollary 2.6. Z/mZ ' ⊕Z/pnii Z, where m =

∏pnii

Corollary 2.7. Any finitely generated abelian group is isomorphic to Zs ⊕⊕ti=1 Z/qiZ with s ≥ 0 and qi = pni

i for pi prime.

Definition 2.2 (Group Action). The action of a group G on a set X is afunction G×X → X : (g, x) 7→ g · x such that

1. e · x = x for all x ∈ X.

2. (g1g2) · x = g1 · (g2 · x)

We say that G acts on X.

Example: Let G be a group. G acts on G by left multiplication and by rightmultiplication by inverses, and also by conjugation.

Definition 2.3 (Orbits). The orbit of x ∈ X is G · x = g · x : g ∈ G.

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Note, if y ∈ G · x then G · x = G · y.The orbits partition X.

Definition 2.4 (Transitive Action). We say that G acts on X transitively ifG · x = X for any x ∈ X.

Definition 2.5 (Conjugacy Classes). The conjugacy classes of a group G arethe orbits of G acting on itself by conjugation.

Definition 2.6 (Stabilizer Subgroup). The stabilizer of x ∈ X is Gx = g ∈G : g · x = x. This is also called the isotropy subgroup.

Lemma 2.8. |G · x = [G : Gx]

Proof. [G : Gx] is the number of left cosets of Gx.Let f : gGx 7→ g · x. To check that f is well defined, we need to check that

if gGx = hGx then g · x = h · x.gGx = hGx implies g−1h ∈ Gx. Thus, (g−1h) ·x = x so h ·x = (gg−1)h ·x =

g(g−1h) · x = g · xNow we must check that it is a bijection.If g · x = h · x then g−1h · x = x. So g−1h ∈ Gx so gGx = hGx, thus f is

injective.If g · x ∈ G · x then gGx 7→ g · x and so f is surjective.

Definition 2.7 (Center of G). The center of G is C(G) = g ∈ G : gh = hg∀h ∈ G.

Definition 2.8 (Centralizer of x in G). The Centralizer of x in G is CG(x) =g ∈ G : gx = xg

Corollary 2.9. 1. The number of elements in the conjugacy class of x ∈ Gis [G : CG(x)]. If |G| <∞ then the size of a conjugacy class divides |G|.

2. If x1, . . . , xk are distinct conjugacy class representatives, then |G| =∑ki=1[G :

CG(xi)] = |C(G)|+∑xi /∈C(G)[G : CG(xi)] = |C(G)|+

∑x/∈C(G)

[G:CG(x)]|G·x|

Corollary 2.10. If G has order pn then the center of G is nontrivial.

Proof. |G| = |C(G)| +∑xi /∈C(G)[G : CG(xi)], and p divides |G| and p divides

[G : CG(xi)], thus p|C(G).

Definition 2.9. Xg = x : g · x = xXH = ∩h∈HXh, H < G.

Lemma 2.11 (Burnside’s Lemma). Let G be a finite group acting on a finiteset X. Then the number of orbits is 1

|G|∑g∈G |Xg|

Proof.∑g∈G |Xg| = the number of pairs (g, x) : g · x = x =

∑x∈X |G · x| =∑

x∈X|G||G·x| = |G|

∑x∈X

1|G·x|

= |G|∑A an orbit

∑x∈A

1|A| = |G|

∑A an orbit 1, which is |G| times

the number of orbits.

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The above lemma is often referred to as Burnside’s Lemma or “The Lemmawhich is not Burnside’s”, as it was discovered in 1900 by Burnside, 1845 byCauchy, and in 1887 by Frobenius.

Theorem 2.12. Let G act on a set X. Then there is a unique correspondinghomomorphism τ : G → Sym(X) = SX = Aut(X) sending g to τg, whereτg(x) = g · x.

Corollary 2.13. 1. G acts on itself by conjugation. Therefore, conjugationby g is an automorphism.

2. τ : G→ AutG : g 7→ τg such that τg(h) = ghg−1 is a homomorphism withker τ = C(G).

Definition 2.10 (Inner Automorphisms). We define an automorphism to be aninner automorphism if it is in Im(τ). That is, it is the set of all automorphismsinduced by conjugation. We will denote it by Inn(G).

Lemma 2.14. Inn(G) E Aut(G).

Proof. Set τg(h) = ghg−1

Let ϕ ∈ AutG. We must check that ϕτgϕ−1 is again conjugation. Let h ∈ G.ϕ(τg(ϕ−1(h))) = ϕ(gkg−1) where k = ϕ−1(h).This is, then, ϕ(g)ϕ(k)ϕ(g−1) = ϕ(g)hϕ(g)−1 = τϕ(g)(h).

Definition 2.11 (Outer Automorphisms). We define an outer automorphismto be an element of OutG = AutG/InnG.

Definition 2.12 (Semidirect Product). Let G,H be groups and τ : H → AutGa homomorphism.

Then we define a new group, X with set G×H and multiplication (g, h)(g′, h′) =(gτh(g′), hh′). We write X = Goτ H

Theorem 2.15. Goτ H is a group and (G, 1) ' G E Goτ G

Definition 2.13. Let G act on a set S. Then S0 = x ∈ S : hx = x, ∀h ∈ G

Lemma 2.16. If G acts on a finite set S and |G| = pn, p prime, then |S| ≡ |S0|mod p.

Proof. S = S0 ∪G · x1 ∪ . . . ∪G · xr where the xr are such that |G · xi| > 1Now |G ·xi| = [G : Gxi

] so 1 < |G ·xi| divides |G| = pn. So p divides |G ·xi|.Thus, |S| ≡ |S0| mod p.

Theorem 2.17 (Cauchy). If p prime divides |G|, then G has an element oforder p.

Proof. Consider S = (a1, . . . , ap) : a1a2 . . . ap = eLet Zp act on S by k · (a1, . . . , ap) = (ak+1, . . . , ap, a1, . . . , ak)Zp maps S to S as a1 . . . ap = e⇒ a2 . . . ap = a−1

1 .Thus, S0 = (a, . . . , a) : ap = e. |S0| ≥ 1, as (e, . . . , e) ∈ S0.

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By the lemma, |S0| ≡ |S| mod p.|S| = |G|p−1 ≡ 0 mod pAs S0 6= ∅, we have that |S0| = np for n > 0. Thus, ∃a 6= e such that ap = e.

|a| = p as p is prime.

Definition 2.14 (p-groups). A group in which every element has order a powerof p, p prime, is called a p-group.

Theorem 2.18. If |G| <∞, then G is a p-group iff |G| = pn for some n.

Proof. Necessary by Cauchy, Sufficient by Lagrange.

Definition 2.15 (Normalizer). The normalizer of a subgroup H of G is NG(H) =g ∈ G : gHg−1 = H. This is, in fact, a subgroup containing H, and ifNG(H) = G then H E G.

Lemma 2.19. If H is a p-subgroup of a finite group G then [NG(H) : H] ≡[G : H] mod p

Proof. Let S be the set of left cosets of H in G. H acts on S by left translation,h · aH = haH.|S| = [G : H].xH ∈ S0 ⇐⇒ hxH = xH∀h ∈ H ⇐⇒ x−1hxH = H∀h ∈ H ⇐⇒

x−1Hx = H ⇐⇒ x ∈ NG(H)|S0| = [NG(H) : H]By the lemma and the fact that |H| = pn and H acts on S we have |S| ≡ |S0|

mod p.So [G : H] ≡ [NG(H) : H] mod p.

Corollary 2.20. If H is a p-subgroup of G such that p|[G : H] then NG(H) 6=H.

Proof. 0 ≡ [G : H] ≡ [NG(H) : H] ≥ 1.So [NG(H) : H] ≥ 1, thus NG(H) 6= H.

Definition 2.16 (Sylow p-subgroup). If p is prime, then a Sylow p-subgroup ofG is a maximal p subgroup of G. That is, if P ≤ H ≤ G and |H| = pn and Pis a Sylow p-subgroup, then H = P .

Theorem 2.21 (First Sylow Theorem). Let G be a group of order pnm, m ≥ 1and (p,m) = 1. Then G contains a subgroup of order pi for 1 ≤ i ≤ n and eachsubgroup of size pi i < n is normal in a subgroup of size pi+1.

Proof. By induction on i. Cauchy is the base case.Suppose H is a subgroup of G of order pi, i < n. Then [G : H] = |G|

|H| , sop|[G : H]. Thus NG(H) 6= H.

1 < |NG(H)/H| = [NG(H) : H] ≡ [G : H] ≡ 0 mod p, so p divides|NG(H)/H| by Cauchy, there is a subgroup H ′ ≤ NG(H)/H with |H ′| = p.

Let K = g ∈ NG(H) : gH ∈ H ′ ≤ NG(H) ≤ GThen |K| = pi+1 and H E K.

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Corollary 2.22. All Sylow p-subgroups have order pn where |G| = pnm, (p,m) =1

Theorem 2.23 (Second Sylow Theorem). If H is a p-subgroup of a finite groupG and P is any Sylow p-subgroup, then ∃x ∈ G such that H < xPx−1. Inparticular, all Sylow p-subgroups are conjugate.

Note, that if there is exactly one Sylow p-subgroup for some p, then it isnormal.

Theorem 2.24 (Third Sylow Theorem). If G is a finite group and p prime,then the number of Sylow p-subgroups divides |G| and is congruent to 1 mod p.

Corollary 2.25. If |G| = p2q2 with p, q prime and p, p2 6≡ 1 mod q and q, q2 6≡1 mod p then G is abelian.

Definition 2.17 (Even Permutation). A permutation τ in §n is even if it canbe written as a product of an even number of transpositions. Similarly, oddpermutations.

Note: each permutation is either even or odd, not both.

It is based on the effect of τ on ∆ = det

1 x1 . . . xn−1

1

1 x2 . . . xn−12

......

. . ....

1 xn . . . xn−1n

=∏i≤i<j≤n(xi−

xj).Then τ∆ = ±∆

Definition 2.18 (Sign of a Permutation). The sign of τ ∈ Sn is +1 is τ iseven and −1 is τ is odd.

Definition 2.19 (Alternating Group). An = τ ∈ Sn : τ is even .

An is a group of order n!/2. An = kerϕ,ϕ : Sn → ±1, so An E Sn.

Lemma 2.26. Let r 6= s ∈ [n]. Then An = 〈(rsk) : 1 ≤ k ≤ n, k 6= r, k 6= s〉 =H.

Proof. Assume n > 3. Since every element of An is a product of an even numberof transpositions, we must show that (ab)(cd), (ab)(ac) ∈ H

(ab)(cd) = (acb)(acd), (ab)(ac) = (acb). So just need all three cycles in H.This follows by brute force.

Corollary 2.27. If N is a normal subgroup of An and N contains a 3-cycle,then N = An.

Proof. Suppose (rsc) ∈ N .(rsk) = (rs)(ck)(rsc)2(ck)(rs) = aNa−1 ∈ N .So (rsk) ∈ N , for 1 ≤ k ≤ n, k 6= r, s so An ⊆ N . Thus N = An.

11

Theorem 2.28. Let n ≥ 5. Then An is simple.

Theorem 2.29. Let N be a proper normal subgroup of An. By the corollary,N contains no 3-cycles.

Proof. Suppose that N contains π = c1 . . . ck disjoint cycles, and c1 is a cycleof length r ≥ 4. Write c1 = (a1, . . . , ar).

Let δ = (a1a2a3) /∈ N .π = c1τ , τ = c2 . . . ck.So π−1(δπδ−1) ∈ N , as N is normal.But π−1(δπδ−1) = τ−1(a1arar−1 . . . a2)(a1a2a3) = (a1a3ar).Suppose π ∈ N , π = c1 . . . ck = c1c2τ where c1, c2 are 3-cycles.c1c2 = (a1a2a3)(a4a5a6), δ = (a1a2a4).π−1(δπδ−1) ∈ N , and equals (a1a4a2a6a3), which reduces to the last case.Thus, every element of N is a product of at most 1 3-cycle and a bunch of

transpositions.Now, if π = (a1a2a3)τ ∈ N , with τ =product of disjoint transpositions.π2 = (a1a2a3)τ(a1a2a3)τ = (a1a2a3)2 = (a1a3a2).So every element of N is a product of disjoint transpositions.Finally, choose σ ∈ N with σ = (a1a2)(a3a4)τ , δ = (a1a2a3)So σ−1(δσδ−1) ∈ N is equal to (a1a3)(a2a4) = σ′.Since n ≥ 5, ∃b such that b /∈ a1, a2, a3, a4, π = (a1a2b).σ′(πσ′π−1) ∈ N is (a1a3b) ∈ N . Thus, there is no proper nontrivial normal

subgroup of An for n ≥ 5.

3 Rings

Definition 3.1 (Ring). A ring is a nonempty set R together with two binaryoperations +, · such that

1. (R,+) is an abelian group.

2. (ab)c = a(bc)

3. a(b+ c) = ab+ ac and (b+ c)a = ba+ ca

If there is an element 1R such that a1R = 1Ra for all a ∈ R then R is a ringwith identity.

If ab = ba for all a, b ∈ R then R is a commutative ring.

Theorem 3.1. Let R be a ring.

1. 0a = a0 = 0∀a ∈ R

2. (−a)b = a(−b) = −(ab)∀a, b ∈ R

3. (−a)(−b) = ab∀a, b ∈ R

4. (na)b = a(nb) = n(ab)∀n ∈ Z∀a, b ∈ R

12

5. (∑ni=1 an)

(∑nj=1 bj

)=∑ni=1

∑nj=1 aibj∀ai, bj ∈ R.

Definition 3.2 (Zero Divisors). A nonzero element, a, in a ring R is a left(respectively right) zero divisor if ∃b 6= 0 ∈ R such that ab = 0 (respectivelyba = 0)

a is a zero divisor if it is a left and right zero divisor.

WARNING: If R has zero divisors, you cannot automatically cancel multi-plication, ie, ab = ac 6⇒ b = c

Definition 3.3 (Unit). An element a ∈ R, R a ring with identity, is a left unit(respectively right) if ∃c ∈ R such that ca = 1R (respectively right if ∃b suchthat ab = 1R).

The element c can be called the left inverse of a and the element b the rightinverse of a.

Note: if b is a left inverse of a and c is a right inverse, then b = c.

Theorem 3.2. The set of units of R form a group under multiplication.

Definition 3.4 (Ring Homomorphism). A ring homomorphism is a functionf : R→ S such that f(a) + f(b) = f(a+ b) and f(a)f(b) = f(ab).

This is a group homomorphism from (R,+) to (S,+), so f(0) = 0. But ifR,S have identity, we cannot necessarily say f(1R) = 1S

We can now define the category of rings. The objects are rings and themorphisms are ring homomorphisms.

We also have the subcategories of rings with identity and commutative rings.

Definition 3.5 (Integral Domain). If R is a commutative ring with identityand no zero divisors, then R is called an integral domain.

Definition 3.6 (Division Ring). A ring R with identity (not 0) where everynonzero element has an inverse is called a division ring.

Definition 3.7 (Field). A commutative division ring is a field.

Theorem 3.3. No zero divisor is a unit.

Proof. ab = 0, b = a−1ab = a−10 = 0

Definition 3.8 (Group Ring). If R is a ring and G is a group, then R(G) =theset of formal R-linear combinations of group elements with coefficients in R.Addition is componentwise and multiplication is distributive and uses the grouplaw.

Group Rings are a part of representation theory.

Definition 3.9 (Real Quaternions). The real quaternions, H = R(Q8)

13

Definition 3.10 (Endomorphism Ring). Let A be an abelian group. ThenEnd(A) = hom(A,A) is a ring with (f + g)(a) = f(a) + g(a) as addition andcomposition of functions as multiplication.

Definition 3.11 (Ideal). A subset I of R is a left (right) ideal if it is an additivesubgroup and x ∈ I for all r ∈ R, x ∈ I.

An ideal I is a left ideal and a right ideal.

Note: If R has 1R and I contains a unit, then I = R.

Theorem 3.4. If R is a ring and I is an ideal, then the additIf R has identity, then so does R/I.If R is commutative, then so is R/I.

Proof. Need to show that multiplication is well defined. Suppose a+ I = a′+ Iand b+ I = b′ + I.

Then a′ = a+i and b′ = b+i′ for i, i′ ∈ I. So a′b′ = ab+ai+bi′+ii′ ∈ ab+IOther parts follow from definition.

Theorem 3.5. If f : R → S is a ring homomorphism, then ker f = r ∈ R :f(r) = 0 is an ideal in R.

Proof. We know that ker f is an additive subgroup.If x ∈ ker f, r ∈ R, then f(rx) = f(r)f(x) = f(r)0S = 0S .

The converse, that if I is an ideal, then π : R → R/I : r 7→ r + I is ahomomorphism with kerπ = I is true.

Theorem 3.6 (First, Second and Third Isomorphism Theorems). 1. If f :R→ S is a ring homomorphism, then R/ ker f ' Im f .

2. If I, J are ideals of R, then R/(I ∩ J) ' (I + J)/J

3. If I ⊆ J are ideals, then J/I is an ideal of R/I and (R/I)/(J/I) ' R/J

Lemma 3.7. I is a left ideal of R iff ∀a, b ∈ I, r ∈ R we have a − b ∈ I andra ∈ I.

Lemma 3.8. If Aii∈I are ideals of R, then ∩Ai is an ideal of R.

Definition 3.12 (Ideal Generated by X). If X ⊂ R then the ideal generated bX is the intersection of all ideals in R containing X. We write 〈X〉.

Definition 3.13 (PID). An integral domain where every ideal is generated byone element is a principal ideal domain, or PID.

Definition 3.14. If A,B ⊆ R then AB = a1b1 . . . arbr : ai ∈ A, bi ∈ BIf A,B are ideals, then AB ⊆ A ∩B

Definition 3.15 (Prime Ideal). An ideal P ⊆ R is prime if AB ⊆ P ⇒ A ⊆ Por B ⊆ P for A,B ideals of R.

14

Theorem 3.9. If P is an ideal in a ring R, P 6= R and ∀a, b ∈ R we haveab ∈ P ⇒ a ∈ P or b ∈ P then P is prime.

If P is prime and R is commutative, then the converse holds.

Proof. Suppose AB ⊆ P , A 6⊆ P for A,B ideals.Let a ∈ A \ PThen ∀b ∈ B, ab ∈ AB ⊆ P , a /∈ P so b ∈ P . Thus, B ⊆ P , so P prime.Let R be a commutative ring. Suppose ab ∈ P , P prime.Consider 〈a〉, 〈b〉. If R is commutative, 〈a〉〈b〉 ⊆ 〈ab〉. As P is prime, 〈a〉 ⊂ P

or 〈b〉 ⊂ P , so a ∈ P, b ∈ P .

Theorem 3.10. If R is a commutative ring with identity, P ⊆ R is prime iffR/P is an integral domain.

Proof. Suppose P is prime, we know that R/P is commutative with identityand 1R 6= 0, so it is enough to show that R/P has no zero divisors.

If (a+P )(b+P ) = 0+P then ab+P = 0+P so ab ∈ P and R commutative,a ∈ P or b ∈ P so a+P = 0 +P or b+P = 0 +P so R/P has no zero divisors.

Conversely, suppose R/P is an integral domain. Since 0 6= 1R + P , we haveP 6= R.

If ab ∈ P then (a+ P )(b+ P ) = ab+ P = 0 + P so a+ P or b+ P is 0 + P .Thus, P is prime.

Definition 3.16 (Maximal Ideal). An ideal M in a ring R is maximal if M 6= Rand if M ⊆ N ⊆ R, N and ideal implies that N = M or N = R.

That is, M is a maximal element in the poset of proper ideals of R withI ≤ J iff I ⊆ J .

Theorem 3.11. If R is a ring with identity, then every ideal I ⊆ R containedin a maximal ideal M .

Proof. Look at subposets P of proper ideals J ⊆ R with I ⊆ J .Let I ≤ I1 ≤ . . . be a chain in P . Let J = ∪j≥1Ij . We need to prove that J

is a proper ideal.Let a, b ∈ J . Then a ∈ Ii, b ∈ Ij for some i, j. We can take i ≤ j, then

a ∈ Ij as well. Thus, a− b ∈ Ij , and ra, ar ∈ Ij . So J is an ideal.J is a proper ideal as 1R is not in J , because 1R /∈ Ij for all j.So J is an upper bound for the chain. By Zorn’s Lemma, we know that P

has a maximal element, which is a maximal ideal containing J .

Theorem 3.12. If R is a commutative ring with identity, then every maximalideal is prime.

Proof. Suppose M is a maximal ideal, ab ∈M but a, b /∈M .Consider M + 〈a〉,M + 〈b〉 the M + 〈a〉 = M + 〈b〉 = R.As 〈a〉〈b〉 ⊆ 〈ab〉 ⊆ M so that R = (M + 〈a〉)(M + 〈b〉) = MM + 〈a〉M +

M〈b〉〈a〉〈b〉 ⊆M , so M = R, contradiction.

15

Theorem 3.13. Let M be an ideal in R, a ring with identity.IF M is maximal, R commutative, then R/M is a field.If R/M is a division ring, then M is maximal.

Proof. R/M is an integral domain. so it is enough to show that all nonzeroelements have inverses.

Consider a+M ∈ R/M , where a /∈M .Consider 〈a〉+M . As M maximal, 〈a〉+M = R.So 1R = ra+m for m ∈M, r ∈ R.So ra− 1R ∈M , thus ra+M = 1R +M . Thus, (r+M)(a+M) = 1R +M .

So R/M is a field.Assume that R/M is a division ring. Then for a /∈ M , ∃r such that (a +

M)(r +M) = 1R +M , and so ar − 1R ∈M .Suppose M not maximal, so ∃N , M ( N ( R.If we choose a ∈ N , then ar − 1R ∈ M ⊆ N so 1R ∈ N , thus N = R,

contradiction, so M is maximal.

In the category of rings, products exist.Let P =

∏Ai as additive groups, with product (ai)(bi) = (aibi), that is,

componentwise.

Theorem 3.14 (Chinese Remainder Theorem). If A1, . . . , An are ideals in R, aring with identity, such that Ai+Aj = R for all i 6= j, Then R/(∩Ai) =

∏R/Ai

Lemma 3.15. If R2 = R (such as when R is a ring with identity) then M amaximal ideal implies that M is a prime ideal.

Proof. Suppose that M is maximal, and that A,B are ideals such that AB ⊆Mbut A 6⊂M and B 6⊂M .

Let a ∈ A \M and b ∈ B \M .Consider M + 〈a〉 = m+ r : m ∈M, r ∈ A. As M is maximal, M + 〈a〉 =

M + 〈b〉 = R.As R = R2 = (M + 〈a〉)(M + 〈b〉) ⊆ M as (m+ r)(m′ + s) = mm′ +ms+

rm′ + rs.So R ⊆M , thus R = M , contradicting M being maximal.

Definition 3.17 (Divides). A nonzero element a of a commutative ring Rdivides b ∈ R if ∃x ∈ R such that ax = b.

If a|b and b|a then a and b are associates. Also, (a) = (b).

Theorem 3.16. If R is an integral domain, then a, b are associates iff a = br,r a unit.

Definition 3.18 (Irreducibles and Primes). Let R be a commutative ring withidentity.

c is irreducible if c is a nonzero nonunit such that c = ab⇒ a or b is a unit.p is prime if p is a nonzero nonunit such that p|ab⇒ p|a or p|b.

Lemma 3.17. R is an integral domain.

16

1. p prime iff 〈p〉 6= 〈0〉 is prime.

2. Every prime p ∈ R is irreducible.

3. c is irreducible iff 〈c〉 is maximal with respect to inclusion among principalideals.

4. If R is a PID then c is prime iff c is irreducible.

5. Every associate of an irreducible (or prime) is irreducible (or prime).

Proof. 1. Suppose p is prime and ab ∈ 〈p〉. Then p|ab so p|a or p|b, thusa ∈ 〈p〉 or b ∈ 〈p〉. So 〈p〉 is prime.

Suppose 〈p〉 is prime and p|ab. Then ab ∈ 〈p〉 so a ∈ 〈p〉 or b ∈ 〈p〉, so p|aor p|b.

2. If p prime is equals ab then p|a or p|b. WLOG, p|a. so a = pc. So p = pcb,so cb = 1.

3. If c irreducible, suppose 〈c〉 ( 〈d〉, c = de for some e ∈ R. As 〈c〉 ( 〈d〉, enot a unit, but c is irreducible, so d is a unit, thus, 〈d〉 = R.

4. Suppose R is a PID. Then irreducible in the PID means maximal ideal,which means prime.

5. If c = du then 〈c〉 = 〈d〉.

Definition 3.19 (Unique Factorization Domain). An integral domain R is aunique factorization domain if every nonzero nonunit can be written a = c1 . . . ckwith ci irreducible and if a = c1 . . . ck = d1 . . . dm with ci, dj irreducible, thenk = m and, up to reordering, ci, di are associates.

Definition 3.20 (Noetherian Ring). We say that a ring R is Noetherian if itsatisfies the ascending chain condition, that is, A1 ⊂ A2 ⊂ . . . is an ascendingchain of ideals then ∃n such that An = Am for all m ≥ n.

Lemma 3.18. If R is a PID then R is Noetherian.

Proof. Let (a1) ⊂ (a2) ⊂ . . . be a chain of ideals. Let A = ∪i≥1(ai).A is an ideal. As R is a PID, A = 〈a〉 for some a ∈ R. As a ∈ A, a ∈ 〈an〉

for some n.Thus a ∈ 〈an〉 ⊆ 〈aj〉 for j ≥ n, and 〈an〉 = A.

Theorem 3.19. If R is a PID then R is a UFD.

Proof. Let S be the set of nonzero nonunits in R with no irreducible factoriza-tion.

Suppose a ∈ S. Consider 〈a〉 as a is not irreducible, ∃x ∈ R irreducible suchthat a = xa1 for some a1 ∈ R.

17

So a1 ∈ S, else a has irreducible factorization. We inductively construct theascending chain 〈a〉 ⊂ 〈a1〉 ⊂ 〈a2〉 ⊂ . . . which contradicts R being Noetherian.So S is empty.

Suppose a = a1 . . . ak = d1 . . . d` with ci, dj irreducible. As R a PID, irre-ducibles are prime.

So c1 prime, so it must divide di for some i. Then rc1 = di, but di irreducibleand c1 not unit so r is a unit.

So c1, di are associates. Inductively, we obtain uniqueness.

4 Modules

Definition 4.1 (Module). Let R be a ring. A Left R-module is an abeliangroup (A,+) together with a map R × A → A written as (r, a) 7→ ra such that∀r, s ∈ R, a, b ∈ A we have

1. r(a+ b) = ra+ rb

2. (r + s)a = ra+ sa

3. (rs)a = r(sa)

4. If 1R exists, then 1Ra = a. Sometimes, this is called a Unitary R-module.

Note, that if ϕ : S → R is a ring homomorphism then every R-module isalso an S-module by sa = ϕ(s)a.

Definition 4.2 (R-module homomorphism). A function f : A→ B where A,Bare R-modules is an R-module homomorphism if f(a + b) = f(a) + f(b) andf(ra) = rf(a).

This defines the category of R-modules.

Definition 4.3 (R-submodule). Let R be a ring and A and R-module. ThenB ⊆ A is an R-submodule if B < A and rb ∈ B for all r ∈ R, b ∈ B.

Note, that if f : A → B is an R-module homomorphism, then ker f is asubmodule of A.

Lemma 4.1. If B is an R-submodule of an R-module A, then the group A/Bis an R module via r(a+B) = ra+B.

Theorem 4.2 (Isomorphism Theorems). As for groups and rings, we have theisomorphism theorems:

1. f : A→ B an R-module homomorphism, then R/ ker f ' Im f a submod-ule of B.

2. B,C are R-submodules of A. Then B/(B ∩ C) ' (B + C)/C

3. C ⊆ B ⊆ A then B/C ⊆ A/C and (A/C)/(B/C) ' A/B

18

Definition 4.4 (Submodule Generated by X). Let A be an R-module andX ⊆ A be a subset. Then the submodule of A generated by X is this intersectionof all submodules containing X.

Definition 4.5 (Products). If Aii∈I are R-modules then the group ⊕Ai isan R-module with action r(ai) = (rai). This is a product in the category ofR-modules.

Definition 4.6 (Exact Sequence). If f : A→ B and g : B → C are R-modulehomomorphisms, then the sequence

A B C................................................................................................................. ............f

................................................................................................................. ............g

Is exact at B if ker g = Im f .A sequence is exact if it exact at each term.

Definition 4.7 (Short Exact Sequence). A short exact sequence is the exactsequence0 A B C 0................................................................................................................. ............ ................................................................................................................. ............................................................................................................................. ............ ................................................................................................................. ............

In any short exact sequence, f is injective and g is surjective.

Lemma 4.3 (Short 5 Lemma). Let R be a ring and

0 A′ B′ C ′ 0

0 A B C 0

................................................................................................................. ............ ................................................................................................................. ............

................................................................................................................. ............ ................................................................................................................. ............................................................................................................................. ............f

................................................................................................................. ............g

................................................................................................................. ............f ′

................................................................................................................. ............g′

.............................................................................................................................

α

.............................................................................................................................

β

.............................................................................................................................

γ

be a commutative diagram with both rows short exact sequences. Then

1. If α, γ injective then β injective

2. If α, γ surjective then β surjective

3. If α, γ bijective then β bijective

Proof. The third part clearly follows from parts 1 and 2.Part 1: Let b ∈ B with β(b) = 0. As β(b) = 0, g′(β(b)) = 0 = γ(g(b))As γ is injective, g(b) = 0. Also, as b ∈ ker g, we have b ∈ Im f , so b = f(a)

for some a ∈ A.β(f(a)) = 0 = f ′(α(a)) as f ′ is injective, α(a) = 0 and as α is injective,

a = 0.Thus, b = f(a) = f(0) = 0, so β is injective.Part 2: Let b′ ∈ B′. Then consider g′(b′) ∈ C ′. Since γ is surjective,

g′(b′) = γ(c) for some c ∈ C.As g surjective, c = g(b) for some b ∈ B. Thus γ(g(b)) = g′(β(b)) = g′(b′).

So g′(β(b)− b′) = 0.Thus β(b)− b′ ∈ ker g′ = Im f ′, so β(b)− b′ = f ′(a′) for some a′ ∈ A′.As α is surjective, a′ = α(a) for some a ∈ A. Let m = b− f(a) ∈ B.Then β(m) = β(b)− β(f(a)) = β(b)− f ′(α(a)) = β(b)− f ′(a′) = b′.

19

Theorem 4.4. Let R be a ring, 0 A1 B A2 0................................................................................................................. ............ ................................................................................................................. ............................................................................................................................. ............ ................................................................................................................. ............ bea short exact sequence of R-modules. Then the following are equivalent:

1. ∃R-mod homomorphism h : A2 → B such that g h = 1A2

2. ∃R-mod homomorphism k : B → A1 such that k f = 1A1

3. The short exact sequence is isomorphic to 0 A1 A1 ⊕A2 A2 0..................................................................................... ............ι1 ..................................................................................... ............

π2................................................................................................................. ............ ................................................................................................................. ............ soB ' A1 ⊕A2

Proof. The first implies the second easily by looking at the diagram:

0 A1 A1 ⊕A2 A2 0

0 A1 B A2 0

................................................................................................................. ............ ................................................................................................................. ............

................................................................................................................. ............ ................................................................................................................. ............................................................................................................................. ............ ................................................................................................................. ............

..................................................................................... ............ ..................................................................................... ............

.............................................................................................................................

1A1

.............................................................................................................................

ϕ

.............................................................................................................................

1A2

If it is an isomorphism (ie, ϕ is an isomorphism), then we take h = ϕι21A2

and k = π1 ϕ.For 1 implies 3, we have a module homomorphism ϕ : A1 ⊕ A2 → B by

ϕ(a1, a2) = f(a1) + h(a2).

Consider 0 A1

A1 ⊕A2

A2 0

0 A1

B

A2 0

................................................................................................................. ............ ................................................................................................................. ............

................................................................................................................. ............ ................................................................................................................. ................................................................................................. ............ι1 ..................................................................................... ............

π2

................................................................................................................. ............f

................................................................................................................. ............g

.............................................................................................................................

1A1

.............................................................................................................................

ϕ

.............................................................................................................................

1A2

If this is commutative, then the Short Five Lemma says we are done. Fora ∈ A1, ϕ(ι1(a)) = ϕ(a, 0) = f(a) + h(0) and f(1A1(a)) = f(a).

For the other square, (a1, a2) ∈ A1 ⊕ A2 and 1A2(π2(a1, a2)) = a2 andg(ϕ(a1, a2)) = g(f(a1) + h(a2)) = a2, so commutative.

If a sequence satisfies the above conditions, we say that the sequence splits:0 A A 0...................................... ............ ...................................... ............ ...................................... ............⊕ ⊕ ⊕

0 C C 0...................................... ............ ...................................... ............ ...................................... ............

From now on, we will assume that rings have identity and that modules areunitary.

Definition 4.8 (Linearly Independent). A subset X of an R-module is linearlyindependent if for distinct x1, . . . , xn ∈ X and ri ∈ R then r1x1 + . . .+ rnxn =0⇒ ri = 0 for all i.

If X is not linearly independent, then it is linearly dependent.

Definition 4.9 (Spans). If Y ⊆ A generates A as an R-module, then we sayY spans A.

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Definition 4.10 (Basis). A set B which is linearly independent and spans Ais called a basis.

Theorem 4.5. Let R be a ring with identity. The following are equivalent onR an R-module.

1. F has a nonempty basis

2. F '∑x∈X R, the direct sum, all but finitely many terms are zero.

3. Let ι : X → F be the inclusion map, and given any R-module A and f :X → A as a map of sets then ∃!f : F → A an R-module homomorphismsuch that f ι = f . That is, F is free on X in the category of R-modules.

Proof. 1⇒ 2: LetX be a basis for F . Define ϕ :∑x∈X → F by ϕ((rx))sumx∈Xrxx.

This is an R-module homomorphism. As X is linearly independent andspanning ϕ is an injective and surjective homomorphism, so it is an isomorphism.

2⇒ 1: Let ex = (0, . . . , 1R, . . . , 0) in the xth coordinate. Then ex : x ∈ Xis a basis for

∑x∈X R.

2⇒ 3: Let F∑x∈X R. Suppose f : X → A is a map of sets, A an R-module,

and ι : X → F : x 7→ ex. Then f((rx)) =∑x∈X rxf(a).

3 ⇒ 2: F is free on X.∑x∈X R is free on X by the above, and so X and∑

x∈X R both free, and so they must be isomorphic, as free objects on a set Xare equivalent.

Corollary 4.6. Every R-module A is a homomorphic image of a free R-moduleF and thus isomorphic to

∑x∈X R/B where B is some submodule of

∑x∈X R/

If A is finitely generated, then we can take F to be as well.

Proof. If (ai)i∈I is a generating set for A, then f : I → A : i 7→ ai gives a mapf :∑x∈X R→ A so A '

∑x∈X R/ ker f .

Note: If F is a free R-module, there can be submodules which are not freeR-modules.

However, if R is a division ring, then every submodule of a free R-module isfree.

Warning: If X,Y are bases for a free module F then it is possible that|X| 6= |Y |

However, if R is a commutative ring or a division ring, then |X| = |Y |necessarily.

Definition 4.11 (Invariant Dimension Property). Let R be a ring with identity.We say that R has the invariant dimension property if, for any free R-moduleF then any two basis for F have the same cardinality, called rankF

Lemma 4.7. Let R be a ring with identity, I a proper ideal, F a free R-modulewith basis X and π : F → F/IF the canonical projection. Then F/IF is a freeR/I-module with basis π(X) and |π(X)| = |X|.

21

Proof. We first show that π(X) spans F/IF . Let u + IF ∈ F/IF . Thenu ∈ F so u =

∑x∈X rxx so u+ IF =

∑x∈X rxx+ IF =

∑(rx + I)(x+ IF ) =∑

(rx + I)π(x). Thus, π(X) spans F/IF .Now we suppose

∑x∈X(rx+ I)π(x) = 0 So

∑rxx+ IF = 0. Then

∑rxx ∈

IF , so∑rxx =

∑sjuj where sj ∈ Iuj ∈ F .

X is a basis, so each uj is a linear combination of elements of X, so∑sjuj =∑

cxx with cx ∈ I.Sp∑rxx =

∑cxx since x is a basis, rx = cx for all x ∈ X.

So rx ∈ I and thus, cx + I = 0 + I. So π(X) is linearly independent,and x 6= x′ ⇒ π(x) 6= π(x′). Thus π(X) is a basis for F/IF over R/I and|π(X)| = |X|.

Lemma 4.8. If f : R → S is a surjective ring homomorphism, and S has theinvariant dimension property, then so does R.

Proof. S ' R/I for some ideal I of R with bases X and Y . Then F/IF is afree S-module with bases π(X) and π(Y ). As S has the invariant dimensionproperty, |X| = |π(X)| = |π(Y )| = |Y |.

Corollary 4.9. Every commutative ring has the invariant dimension property.

Proof. Let M be a maximal ideal. Then R/M is a field, which has the invariantdimension property. Thus R has the invariant dimension property.

We will now assume R is a PID.

Theorem 4.10. Let F be a free R-module where R is a PID. Let G be asubmodule of F . Then G is a free R-module and rankG ≤ rankF .

Proof. For F <∞.Let X = x1, . . . , xn be a basis for F .Let Fi be the sub,module generated by x1, . . . , xi.Let Gi = G ∩ Fi, note, Gi ⊂ Gi+1 and Gi = Gi+1 ∩ Fi, Gn = G.Also, Fi+1/Fi ' R and G1 ⊂ F1 ' Rxi ' R.So G1 is isomorphic to an ideal, thus G1 = 〈c〉 = Rc ' R is c 6= 0.Thus, G1 ' 0 or R, so G1 is free of rank ≤ rankF .Consider 0→ Gi → Gi+1 → Gi+1/Gi → 0. Then Gi+1/Gi ' Gi+1/(Gi+1 ∩

Fi) ' (Gi+1 ∩ Fi)/Fi < Fi+1/FiSo Gi+1/Gi ' 0 or R. By induction, Gi is free of rank less than rankFi = i.If Gi+1/Gi = 0 then Gi+1 = Gi so done.IfGi+1/Gi = R then 0→ Gi → Gi+1 → R→ 0 so rankGi+1 ≤ rankGi+1 ≤

i+ 1, so G = Gn has rank ≤ rankFn = n.

Definition 4.12 (Θa). Let A be a module over an integral domain R. Thenfor a ∈ A, let Θa = r ∈ R : ra = 0

Lemma 4.11. 1. Θa is an ideal in R.

2. At = a ∈ A : Θa 6= 0 is a submodule of A.

22

3. ∀a ∈ A, R/Θa = Ra = ra : r ∈ R

Proof. Part 2: Let a, b ∈ At. Then ∃r, s ∈ R such that ra = sb = 0. Thenrs(a− b) = rsa− rsb = s0− r0 = 0.

And, if a ∈ At with sa = 0, s 6= 0 then s(ra) = (sr)a = 0. So s ∈ Θra 6= 0.So ra ∈ At.

Definition 4.13 (Torsion). At is called the torsion submodule of A. A is calleda torsion module if A = At. A is called torsion free if At = 0.

Theorem 4.12. 1. Free modules over integral domains are torsion free

2. ∃A, a torsion free module, which is not free.

Proof. Part 1: Let X be a basis for some free module F . If f =∑rixi then

rf =∑rrixi = 0. Then rri = 0 for all i, so r = 0.

Part 2: Q as a Z-module.

Theorem 4.13. A finitely generated torsion free module A over a PID R isfree.

Proof. Assume A 6= 0. Let X be a finite set of generators for A.If x ∈ X then rx = 0⇒ r = 0 as A is torsion free.So ∃S ⊂ X nonempty which is maximal with respect to

∑x∈S rxx = 0 ⇒

rx = 0∀x.That is, S is maximal with respect to “the submodule of A generated by S

is free”.Let y ∈ X \S. Then ∃ry, rx and x ∈ S such that ryy+

∑x∈S rxx = 0. That

is, ryy = −∑x∈S rxx ∈ spanS = F . Note, ry 6= 0, as otherwise

∑x∈X rxx = 0

with some rx 6= 0, contradicting linear independence.So ra ∈ F for all a ∈ A.But consider ϕ : A→ A : a 7→ ra. ϕ is an R-module homomorphism. Then

kerϕ = 0 as A is torsion free. ϕ(A) ⊆ F is a submodule, so A ' ϕ(A) whichis a submodule of a free module over of a PID.

Theorem 4.14. Let A be a finitely generated module over a PID R. ThenA ' At ⊕ F with F free.

Proof. Let F = A/At. F is finitely generate if A is.If r(a+At) = 0 then ra ∈ At., so s(ra) = 0 for some s 6= 0 in R.Thus (sr)a = 0. so a ∈ At or sr = 0⇒ r = 0 as R integral domain s 6= 0.So F is torsion free, thus, F is free.0→ At → A→ F → 0 splits, so A ' At ⊕ F .

Definition 4.14 (A(p)). Let A be a torsion module over a PID R. If p ∈ R isprime, then A(p) = a ∈ A : Θa = 〈pi〉 for some i ∈ N = a ∈ A : pia = 0 forsome i ∈ N.

Lemma 4.15. Let A be a torsion module over a PID R. Then A(p) is asubmodule of A for each prime p of R.

23

Proof. Let a, b ∈ A(p), then pia = 0 = pjb. WLOG, i ≤ j. Then pj(a− b) = 0so a− b ∈ A(p).

pi(ra) = r(pia) = ra = 0 so ra ∈ A(p).

Theorem 4.16. Let A be a torsion R-module, R a PID. Then A =∑A(p)

where the sum is over all distinct primes p. If A is finitely generated, then allbut finitely many A(p) are zero.

Proof. We first show that a ∈ A ⇒ a1 + . . . + ak where ai ∈ A(pi). We writeΘa = 〈r〉.

r = upn11 . . . pnk

k where pi prime and ni > 0.Let ri = r

pnii

. Consider 〈r1, . . . , rk〉 = 〈d〉 for some d ∈ R.Then d|ri for all i, thus d is a unit. Thus, 〈r1, . . . , rk〉 = 〈1R〉 = R/Thus 1 = s1r1 + . . .+ skrk so a = 1Ra =

∑ki=1 siria

But pnii (siria) = si(pni

i ri), so siria ∈ A(pi).Thus, a ∈ A⇒ a =

∑ai with ai ∈ A(pi), pi prime.

Fix a prime p ∈ R and let A1 be the submodule of A spanned by the A(q),q 6= p.

Let a ∈ A1 ∩A(p). Then pia = 0 and a =∑ai, ai ∈ A(pi), pi 6= p

So ∀i, ∃n such that pnii ai = 0.

Let d =∏ki=1 p

nii , then da = 0. Consider 〈d, pi〉 = 〈c〉.

Now c|d and c|pi so c is a unit. Thus 〈d, pi〉 = 〈1R〉 = R, so ∃s, t ∈ R suchthat sd+ tpi = 1R.

So a = 1Ra = (sd+ tpi)a = s(da) + t(pia) = 0 + 0 = 0 so A1 ∩A(p) = 0.Consider the homomorphism ϕ :

∑A(p) → A : (ai) 7→

∑ai. This is

surjective by the first part and injective by the second part, this ϕ is an isomor-phism.

Lemma 4.17. Let A be a module over a PID R such that pnA = 0, pn−1A 6= 0for some prime p, n > 0. Let a ∈ A have Θa = 〈pn〉, then

1. If A 6= Ra then ∃b ∈ A, b 6= 0 such that Ra ∩Rb = 0

2. ∃ submodule C of A such that A ' C ⊕Ra

Proof. 1. If A 6= Ra then ∃s ∈ A\ (Ra), since pnc ∈ pnA = 0, ∃ a least j > 0such that pjc ∈ Ra(1 ≤ j ≤ n)

So pj−1c /∈ Ra, pjc = r1a for some r1 ∈ R, r1 = rpk with k ≥ 0 p 6 |rSo k ≥ j ≥ 1.

Write b = pj−1c− rpj−1a. We will check that Ra ∩Rb = 0. Note b 6= 0as pj−1c /∈ Ra but rpk−1a ∈ Ra.

Also note that pb = pjc − rpka = r1a − r1a = 0. Suppose sb ∈ Ra forsome s ∈ R, sb 6= 0.

Then pb = 0 and sb 6= 0, so p 6 |s.

24

So pn, s relatively prime, so 〈pn, s〉 = 〈1R〉, so ∃x, y such that ypn + xs =1R.

b = 1Rb = sxb + ypnb = x(sb) ∈ Ra. b = pj−1c − rpk−1a ∈ Ra sopj−1c ∈ Ra contradiction, so Ra ∩Rb = 0.

2. If A = Ra take C = 0. If A 6= Ra then let S be the set of all submodulesb such that Ra ∩B = 0.By part 1, we know that S is nonempty, so we order it by ⊆. We claimthat S has a maximal element.

Check that if Ai is an increasing chain then ∪Ai is a submodule. ApplyZorn’s Lemma. So let C be a maximal element of S. Consider A/C.

So pn(A/C) = 0, thus pn(a+ C) = 0 but pn−1(a+ C) 6= C as pn−1a 6= 0and Ra ∩ C = 0Apply part 1 to A/C so A/C ' R(a + C) or ∃b + C 6= 0 + C such thatR(a+ C) ∩R(b+ C) = 0 + C.

Consider C ′ = R(b + C). Then Ra ∩ C ′ = 0 as Ra ∩ C = 0 andR(a+ C) ∩R(b+ C) = 0 + C so C ′ ∈ S, contradiction. So A = Ra⊕ C.

Theorem 4.18. Let A be a finitely generated module over a PID R. ThenA ' Rk ⊕

⊕`i=1R/(p

nii ), pi prime and ni ∈ N not necessarily distinct.

Proof. We know that A ' F ⊕At, with F free and At torsion.π1 : A→ F and π2 : A→ At. F is generated by π1(generators of A) and At

is generated by π2(generators of A). So F ' Rk. Thus, A ' Rk ⊕At.We know At = ⊕sj=1A(pj), pj distinct primes. So again, each A(pj) is

finitely generated. It remains to check that each A(pj) ' ⊕mk=1R/pnkj . This

proof is by induction on the number of generators of A(pj). If this is one, thenA(pj) ' Rc ' R/Θc ' R/pnj for some n.

We suppose that this is true for all cases with generators, that pnjA(pj) = 0and pn−1

j A(pj) 6= 0.By lemma, ∃a,C such that A(pj) ' Ra⊕C and C has fewer generators. We

can take a to be a generator of A(pj).So C ' ⊕`−1

k=1R/pnkj and Ra ' R/pnj so A(pj) = ⊕`k=1R/p

nkj as required.

Definition 4.15 (Middle Linear Map). Let A,B be R-modules, A a right R-module (denoted AR) and B a left R-module (denoted RB). Consider f : A ⊕B → C a homomorphism of groups such that f(a1 + a2, b) = f(a1, b) + f(a2, b),f(a, b1 + b2) = f(a, b1) + f(a, b2) and f(ar, b) = f(a, rb). Then f is said to bemiddle linear.

We want to find a module, which we will call A⊗RB such that every middlelinear map factors through it.

25

Definition 4.16 (Tensor Product). Let AR and RB be R-modules, let F be thefree abelian group with basis A ⊗ B and let K be the subgroup of F generatedby (a+ a′, b)− (a, b)− (a′, b), (a, b+ b′)− (a, b)− (a, b′), (ar, b)− (a, rb) for eacha, a′ ∈ A, b, b′ ∈ B

The quotient group, F/K is called the tensor product and is written A⊗RB.The element (a, b) +K is written a⊗ b and (0, 0) +K is written as 0.

Warning: A ⊗R B is GENERATED by a ⊗ b : a ∈ A, b ∈ B, a generalelement is

∑ni(ai, bi).

Definition 4.17 (S,R-bimodule). A is an S,R-bimodule if A is a left S-moduleand a right R-module, and s(ar) = (sa)r. We can denote this as SAR

If B is a left R-module and A is an S,R-bimodule, then A ⊗R B is a leftS-module by s(a⊗ b) = (sa)⊗ b.

Let f : A × B → C be a middle linear map of groups. Let π : A × B →A⊗R B : (a, b) 7→ a⊗ b. π is middle linear.

Theorem 4.19. Let AR and RB be R-modules. If g : A × B → C is middlelinear, then ∃!g : A⊗R B → C a homomorphism such that g π = g.A×B

A⊗R B C

.............................................................................................................................

π

............................................................................................................................................................................ ............

g

........................................................................................ ............g

Proof. Let F be free on A × B. So ∃!g1 : F → C homomorphism such thatg1(a, b) = g(a, b). As g is middle linear, and a homomorphism, K ⊆ ker g1.

Thus g1 : F/K → C is well defined. g(a⊗ b) = g(a, b).

In general, if A is a finitely generated abelian group A ' Zr ⊕ At, thenA⊗Z Q ' Qr, that is, tensor product removes torsion.

Lemma 4.20. Let AR,RB be R-modules, R has identity. Then A ⊗R R ' Aas right R-modules, R⊗R B ' B as left R-modules.

Proof. Let f : A×R→ A by (a, r) 7→ ar. f is middle linear, so ∃f : A⊗R R→A : a⊗ r 7→ ar

As R is an (R,R)-bimodule, A ⊗ R is a right R-module and f(a ⊗ rr′) =arr′ = (ar)r′ = f(a⊗ r)r′ so f is an R-module homomorphism.

f is surjective, as f(a⊗ 1R) = a. If f(∑ni(ai⊗ ri)) = 0⇒

∑niairi = 0⇒∑

ai(niri) = 0.Then

∑ni(ai⊗ ri) =

∑ai(niri)⊗ 1R = (

∑ai(niri))⊗ 1R = 0⊗ 1R = 0

Theorem 4.21. 1. Z[x]⊗Z R ' R[x]

2. Z/m⊗ Z/n ' Z/(m,n)

26

3. If AR,RBS , SC are modules, then A⊗R B is a right S-module, B ⊗S C sa left R-module and (A⊗R B)⊗S C ' A⊗R (B ⊗S C).

4. If Aii∈I are right R-modules and B is a left R-module, then (∑Ai)⊗R

B '∑

(Ai ⊗R B)

5. If F,G are free R-modules with bases X and Y respectively, then F ⊗R Gis a free R-module with basis x⊗ y : x ∈ X, y ∈ Y .

6. If f : A → A′ and g : B → B′ are R-module homomorphisms, then∃! group homomorphism A ⊗R B → A′ ⊗R B′ denoted f ⊗ g such thatf ⊗ g(a⊗ b) = f(a)⊗ g(b).

7. Tensor product is right exact. That is, if D is an R-module, R commuta-tive and A

f→ Bg→ C → 0 is exact, then A⊗D f⊗1→ B⊗D g⊗1→ C⊗D → 0

is also.

Definition 4.18 (k-Algebra). Let k be a commutative ring with identity 1k. Ak-algebra A is a ring such that (A,+) is a left k-module and k(ab) = (ka)b =a(kb).

If A is a division ring then we call it a division algebra.

Definition 4.19 (k-algebra Homomorphism). An algebra homomorphism is aring homomorphism that is also a k-module homomorphism.

Definition 4.20 (Subalgebra). A subalgebra is a subring of A that is a k-submodule of A.

We now have a category whose objects are k-algebras and whose morphismsare k-algebra homomorphisms. There is also the category of commutative k-algebras.

The subalgebra generated by X ⊆ A is the intersection of all subalgebrascontaining X.

A is finitely generated if ∃ a finite set X such that 〈X〉 = A.

Theorem 4.22. If A is a finitely generated commutative k-algebra with identity,k a field, then A ' k[x1, . . . , xn]/I for some n ∈ N and some I ⊆ k[x].

Proof. If X ⊂ A then there is a k-algebra homomorphism f : k[X]→ A : x 7→ x.If A is finitely generated with generating set X then f : k[X]→ A is surjec-

tive, as Im f is a subalgebra of A containing X. So A ' k[X]/ ker f ' k[X]/Ias k-algebras.

5 Fields and Galois Theory

We know that the solutions to ax2 + bx + c = 0 are x = −b±√b2−4ac

2a . There isalso a cubic and a quartic formula.

Galois Theory shows that there is no quintic formula.Field Extensions

27

Definition 5.1 (Extension Field). A field F is an extension of a field K iff Kis a subfield of F . In this case, we can regard F as a K-vector space. We define[F : K] = dimK F .

If [F : K] < ∞ we say F is a finite extension. Otherwise we say F is aninfinite extension.

For example, as z ∈ C an be expressed uniquely as z = r + is with r, s ∈ R,then [C : R] = 2.

Proposition 5.1 (Analogue of Lagrange’s Theorem). Suppose that K ⊆ E ⊆ Fare field extensions.

1. If xi : i ∈ I is a basis of E over K and yj : j ∈ J is a basis of F overE, then xiyj : i ∈ I, j ∈ J is a basis of F over K.

2. [F : K] = [F : E][E : K]

Proof. Let z ∈ F , then there exist αj ∈ E, almost all zero, such that z =∑j αjyj . For each j there exists βji, almost all zero, such that αj =

∑i βjixi.

Substituting in, we get z =∑j

∑i βjixiyj .

Thus, xiyj , i ∈ I, j ∈ J generates F over K as a vector space.To see that they are independent, suppose

∑i,j γijxiyj = 0, where γij ∈

K, almost all zero. Since yj : j ∈ J are independent over E, for each j,∑i γijxi = 0. Similarly, since xi : i ∈ I are independent over K, we get

γij = 0 for all i, j.

Question: Is the analogue of Cauchy also true? That is, if we’ve got [F :Q] = 10, we know any K which is an intermediate field must have degree 2 or5. Can we say that there is an E such that [E : Q] = 5?

Definition 5.2 (Algebraic over K). Let F be an extension of K. Then α ∈ Fis algebraic over K iff there exists a nonzero polynomial f(x) ∈ K[x] such thatf(α) = 0. Otherwise, α is transcendental.

The extension F of K is algebraic iff every α ∈ F is algebraic over K.

e.g.√

2 is algebraic over Q since it is a root of x2 − 2 = 0, while π, e aretranscendental over Q.

Proposition 5.2. If F is a finite extension of K, then F is an algebraic ex-tension of K.

Proof. Suppose [F : K] = n. Let α ∈ F .Then the elements 1, α, α2, . . . , αn can’t be distinct and independent over

K. Hence, there exist ai ∈ K not all zero such that a0 + a1α+ . . .+ anαn = 0

so α is algebraic over K.

Remark: As we will soon see, the converse doesn’t hold. For example,Q(√

2)(√

3) . . . is an infinite algebraic extension.Notation. Suppose F is an extension of K and S ⊆ F .

28

K[S] is the subring of F generated by K ∪ S.K(S) is the subfield of F generated by K ∪ S.When S = s1, . . . , st we will just write K[s1, . . . , st] and K(s1, . . . , st).Suppose now that F is an extension of K and α ∈ F is algebraic over K.

Consider the corresponding ring epimorphism ϕ : K[x] → K[α] : f(x) 7→ f(α).kerϕ is nontrivial. Since K[x] is a PID, there exists a nonzero polynomial0 6= p(x) ∈ K[x] such that kerϕ = (p(x))

Furthermore, we can assume p(x) is monic, asK is a field. SinceK[x]/(p(x)) 'K[α] is an integral domain, it follows that p(x) is irreducible. Hence, (p(x)) isactually a maximal ideal. Hence K[x]/(p(x)) is a field, and so K[α] = K(α).

Definition 5.3. p(x) is the irreducible polynomial of α over K, denoted byIrr(α,K, x).

Continuing out analysis, let deg(p(x)) = n. Then 1, α, . . . , αn−1 are lin-early independent over K. If not, then there exist ai ∈ K not all zero, such thata0+a1α+. . .+an−1α

n−1 = 0, then α is a root of f(x) = a0+a1x+. . .+an−1xn−1

so f(x) ∈ (p(x)). This means p(x)|f(x), contradiction.To see that these elements generate K(α), let β ∈ K(α). Then, there exists

a polynomial f(x) ∈ K[x] such that β = f(α). By the division algorithm, thereexist q(x), r(x) with deg r(x) < n such that f(x) = q(x)p(x) + r(x). Thus,β = f(α) = r(α) so β = b0 + b1α+ . . .+ bn−1α

n−1 for some bi ∈ K.Summing up...

Theorem 5.3. Suppose α is algebraic over K and that p(x) = Irr(α,K, x).

1. K(α) = K[α]

2. K(α) ' K[x]/(p(x))

3. [K(α) : K] = deg p(x)

Definition 5.4 (Finitely Generated Extension). Let F be an extension of K.Then F is finitely generated over K iff there exist α1, . . . , αn ∈ F such thatF = K(α1, . . . , αn).

Proposition 5.4. Let F = K(α1, . . . , αn) be a finitely generated extension ofK. If each αi is algebraic over K, then F is a finite algebraic extension.

Proof. It is enough to show that [F : K] < ∞. Let K0 = K and Ki =K(α1, . . . , αi). Consider the tower of extensions K0 ⊆ K1 ⊆ K2 ⊆ . . . ⊆ KN .

Then Ki+1 = Ki(αi+1) and αi+1 is algebraic over Ki. Hence [Ki+1 : Ki] <∞.

Thus, [F : K] = [Kn : Kn−1] . . . [K1 : K0] <∞.

Definition 5.5. Suppose that F is an extension of K and let σ : K → L be anembedding of fields. Then an embedding τ : F → L extends σ iff the followingdiagram commutes:

29

K

F

L

L

........

........

........

........

........

........

........

........

........

........

........

........

.................

............

inc

................................................................................................................. ............σ

................................................................................................................. ............τ

........

........

........

........

........

........

........

........

........

........

........

........

.................

............

idL

If σ = id, then we say that τ is an embedding of F over K.

K

F L

......................

......................

......................

......................

......................

......................

.............................................

inc

.................................................................................................................................................................................

inc

....................................................................................................................................................................................................................................................................... ............τ

Definition 5.6. Suppose that σ : K → L is an embedding of fields. Then wecan extend σ to an embedding of the corresponding polynomial rings σ : K[x]→L[x] :

∑ni=1 aix

i 7→∑ni=1 σ(ai)xi. We denote the image of each f ∈ K[x] by

σf .

Theorem 5.5. Suppose σ : K → L is an isomorphism of fields and that f(x) ∈K[x] is irreducible. Let u, v be roots of f, σf respectively. Then σ extendsuniquely to an isomorphism τ : K(u)→ L(v) such that τ(u) = v.

Proof. Clearly, there is at most one such map. To see that such an isomorphismexists, note that the isomorphism σ : K → L induces an isomorphism K[x] →L[x]. This map induces an isomorphism K[x]/(f)→ L[x]/(σf).

Recalling the proof of Theorem 1, we can define the desired isomorphismK(u)→ K[x]/(f)→ L[x]/(σf)→ L(v)u 7→ x+ (f) 7→ x+ (σf) 7→ v

Theorem 5.6. Let f(x) ∈ K[x] be a polynomial of degree n ≥ 1. Then thereexists an extension F = K(u) satisfying

1. u ∈ F is a root of f

2. [F : K] ≤ n

3. If f is irreducible, then [F : K] = n and F = K(u) is uniquely determinedup to isomorphism.

Proof. Let p(x) be a monic irreducible factor of f . Then identify K with thecanonical subfield of K[x]/(p), we can take F = K[x]/(p) and u = x+ (p).

Corollary 5.7. If f1, . . . , fn ∈ K[x] are nonconstant polynomials, then thereexist an extension E of K in which each fi has a root.

Definition 5.7 (Algebraically Closed). The field F is algebraically closed ifevery nonconstant polynomial f(x) ∈ F [x] has a root. Equivalently, each non-constant f(x) ∈ F [x] splits into a product of not necessarily distinct linearfactors.

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Theorem 5.8. For each field K, there exists an extension L of K such that Lis algebraically closed.

Proof. The theorem is an easy consequence of the following result:Claim - If E is any field, then there exists an extension F such that each

nonconstant f(x) ∈ E[x] has a root in F .Assuming the claim, we can complete the proof as follows:Define inductively a tower of extensions K = E0 ⊂ E1 ⊂ . . . ⊂ En ⊂ . . .

n ∈ N, such that each nonconstant f(x) ∈ En[x] has a root in En+1. Clearly,L = ∪n∈NEn is an algebraically closed extension of K.

Proof of Claim (E. Artin):Let S = xf : f ∈ E[x] is a nonconstant polynomial.Consider I = (f(xf ) : xf ∈ S) of the polynomial ring in infinitely many

variables E[S].Claim: I 6= E[S]. Suppose not, then there exists an equality g1f1(xf1) +

. . .+ gnfn(xfn) = 1 where gi ∈ E[S].

Then, there exist finitely many variables x1, . . . , xN with n ≤ N such thateach gi only involves x1 up to xN . thus, the equality becomes

∑ni=1 fi(x1, . . . , xN )fi(xfi) =

1.Let E′ be an extension of E in which each fi has a root αi. Let αi = 0 for

n < i ≤ N . Then working in E′[S] and substituting αi for xi, we obtain 0 = 1,the classical contradiction.

Thus, I is a proper ideal. By Zorn’s lemma, there exists a maximal ideal Msuch that I ⊆ M ( E[S]. Thus, E[S]/M is a field in which each nonconstantpolynomial f(x) ∈ E[x] has the root xf + M . Identifying E with the obvioussubfield of E[S]/M , we are done.

Definition 5.8 (An Algebraic Closure). An extension E of K is an algebraicclosure if

1. E is an algebraic extension of K

2. E is algebraically closed.

Corollary 5.9. If K is any field, then there exists an algebraic closure Kalg ofK.

Proof. Let E be an algebraically closed extension of K.Define Kalg = α ∈ E : α is algebraic over KClaim 1: Kalg is a field. Let α, β ∈ Kalg. Then K(α, β) is a finite algebraic

extension of K. Since α − β and α/β are in K(α, β), given β 6= 0, it followsthat α− β, α/β are in Kalg.

Claim 2: Kalg is algebraically closed. Let f(x) ∈ Kalg[x] be a nonconstantpolynomial. Let α ∈ E be a root of f . Let f(x) = β0 + β1x + . . . + βnx

n

where βi ∈ Kalg. Then [K(β0, . . . , βn) : K] < ∞. Since α is algebraicover K(β0, . . . , βn), we have [K(β0, . . . , βn, α) : K(β0, . . . , βn)] < ∞, hence[K(β0, . . . , βn, α) : K] <∞ and so α ∈ Kalg.

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Theorem 5.10. If E and E′ are algebraic closures of K then E and E′ areisomorphic over K.

We will use:

Lemma 5.11. Let σ : F → L be an embedding of the field F into the alge-braically closed field L. Suppose α is algebraic over F and f(x) = Irr(α, F, x).Then the number of extensions of σ to an embedding of F (α) into L is equal tothe number of distinct roots of σf in L.

Proof. Let β be any root of σf in L. By Theorem 2, there is a unique embeddingτ : F (α)→ τ such that τ(α) = β.

Conversely, if τ is any embedding extending σ, then τ(α) must be a root ofσf .

Now, we prove Theorem 5.

Proof. Clearly, Theorem 5 is an immediate consequence of the following (slightly)more general statement.

Theorem 5.12 (AC). Suppose E is an algebraic extension of K and that σ :K → L is an embedding into an algebraically closed field L.

1. There exists an extension of σ to an embedding of E into L.

2. If E is algebraically closed and L is algebraic over σ(K) then any suchextension is an isomorphism E and L.

Proof. 1. Let P be the partially ordered set consisting of all pairs (τ, F ) whereK ⊆ F ⊆ E is a subfield and τ : F → L is an extension of σ ordered by(τ, F ) ≤ (τ ′, F ′) iff F ⊆ F ′ and τ ′|F = τ .

Then (σ,K) ∈ P and so P 6= ∅.Suppose that (τi, Fi) : i ∈ I is a linearly ordered subset of P. LetF = ∪i∈IFi and τ = ∪i∈Iτi. Then (τ, F ) ∈ P and (τi, Fi) ≤ (τ, F ) for alli ∈ I.

By Zorn’s Lemma, there exists a maximal element (τ, F ) of PClaim: E = F .

Suppose not, let α ∈ E \ F . Then ∃ an extension of τ to the algebraicextension F (α) of F . But this contradicts the maximilaity of (τ, F ).

2. Suppose that E is algebraically closed and L is algebraic over σ(K). Letτ : E → L extend σ. Then τ(E) is algebraically closed and L is algebraicover τ(E). Hence, L = τ(E).

Convention: From now on, Kalg will denote some fixed algebraic closure ofK.

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Definition 5.9. Suppose K ⊆ E ⊆ Kalg. Then

1. eK(E,Kalg) is the set of embeddings of E into Kalg over K.

2. AutK(Kalg) is the group of automorphisms of the algebraic closure suchthat π(α) = α for all α ∈ K.

Theorem 5.13. Suppose that K ⊆ E ⊆ Kalg.

1. Each σ ∈ eK(E,Kalg) extends to an automorphism τ ∈ AutK(Kalg).

2. If [E : K] <∞ then |eK(E,Kalg)| ≤ [E : K].

Proof. 1. Immediate consequence of Theorem 6.

2. Let E = K(α1, . . . , αn). Consider the tower of extensions K ⊆ K(α1) ⊆. . . ⊆ K(α1, . . . , αn) = E.

LetK0 = K andKi = K(α1, . . . , αi). Suppose inductively that |eK(Ki,Kalg)| ≤

[Ki : K]. Fix some σ ∈ eK(Ki,Kalg). Then the number of ways of extend-

ing σ to Ki+1 = Ki(αi+1) is equal to the number ri+1 of distinct roots ofIrr(αi+1,Ki, x).

Hence, |eK(Ki+1,Kalg)| ≤ |eK(Ki,K

alg)|ri+1 ≤ [Ki : K][Ki+1 : Ki] =[Ki+1 : K].

Remark: If ri = deg Irr(αi,Ki+1, x), for 1 ≤ i ≤ n then |eK(E,Kalg)| = [E :K].

Galois Theory

Definition 5.10 (Galois Group). Let F be a finite extension of K. Then, thecorresponding Galois Group is AutK(F ) = σ ∈ Aut(F ) : σ|K = idK.

Remark: Clearly, we can suppose that K ⊆ F ⊆ Kalg and so each σ ∈AutK(F ) extends to an automorphism τ ∈ AutK(Kalg).

Also clearly, AutK(F ) ⊆ eK(F,Kalg) and so |AutK(F )| ≤ [F : K]. Inparticular, AutK(F ) is a finite group.

Basic idea of Galois Theory: The finite group AutK(F ) encodes lots of usefulinformation about the extension of F over K.

Counterexamples:

1. Consider F = Q(21/3) then 21/3 is the unique root of x3 − 2 = 0 in F .Hence AutQ(F ) = 1.

2. Let K = Fp(t), where t is transcendental over Fp. Let α satisfy αp = tand consider F = K(α). Since xp − t = xp − αp = (x − α)p, α is theunique root of xp − t = 0 in K and so AutK(F ) = 1.

The first says that we need to adjoin all the roots of an irreducible equation,the second is also something to be wary of.

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Definition 5.11 (Fixed Field). Let F be a field and let G ≤ AutF . Then thecorresponding fixed field of G is FG = α ∈ F : σ(α) = α for all σ ∈ G

Definition 5.12 (Galois Extension). Let F be a finite extension of K and letG = AutK(F ). Then F is a Galois extension of K iff FG = K.

Clearly, this is an attempt to capture the idea that G is “large” in the sensethat it moves as many elements as possible. Another way of capturing this ideawould be that |G| = [F : K]. Fortunately, we’ll eventually see that the twonotions coincide.

Example: F = Q(√

2) is a Galois Extension of Q with Galois Group AutQ F 'C2

Proof. First note that we can define an automorphism σ ∈ AutQ F by r+s√

2 7→r − s

√2 for r, s ∈ Q. Clearly σ(α) = α iff α ∈ Q. Since [F : Q] = 2, we must

have AutQ F ' C2.

Example: Consider the irreducible polynomial p(x) = x3 − 2. Let α = 21/3

and let ω be a primitive third root of unity, ie ω 6= 1 and ω3 = 1. So ω2 +ω+1 =0. Then the roots of p(x) are α, ωα, ωα. Let F = Q(α, ωα, ωα). Then F is aGalois extension of Q and AutQ F ' S3.

Proof. We have already seen that G = AutQ F ' S3. To see that FG = Q

consider the tower of extensions Q3⊆ Q(α)

2⊆ Q(α, ω) = F . Thus [F : Q] = 6.

Finally note that if E = FG then Q ⊆ E ⊆ F and 6 = |G| ≤ |AutE F | ≤ [F :E] ≤ [F : Q] = 6, so all are equal.

Remark: This argument shows that if |AutK F | = [F : K] then F is a GaloisExtension of K.

Open Question: Let G be any finite group. Does there exist a Galois exten-sion F of Q such that AutK F ' G.

Remark: We will soon see that if G is any finite group, then there exists afinite extension K of Q and a Galois Extension F of K such that AutK F ' G.

We next try to understand which finite extensions are Galois.|AutK(E)| ≤ |eK(E,Kalg)| ≤ [E : K]. E is Galois if both are equalities.

The first will be done with splitting fields/normal extensions, and the second willbe done with the number of roots of irreducible equations/separable extensions.

Definition 5.13 (Splitting Field). Let fi : i ∈ I be a family of nonconstantpolynomials in K[x]. Then the extension F of K is a splitting field of fi : i ∈ Iiff

1. Each polynomial fi splits into a product of linear factors in F [x]

2. For each i, let Ri be the set of roots of fiin F . Then F = K(∪iRi)

Example: Using our previous notation, E = Q(α, ωα, ωα) is a splitting fieldof x2 − 2 ∈ Q[x].

Remark: Up to isomorphism, the splitting field of fi : i ∈ I is the subfieldof Kalg generated by all roots of the fi in Kalg.

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Theorem 5.14. If K ⊆ E ⊆ Kalg, then the following are equivalent.

1. For all σ ∈ AutK Kalg, σ[E] = E.

2. For all τ ∈ eK(E,Kalg), τ [E] = E.

3. E is the splitting field of a family of polynomials in K[x]

4. Every irreducible polynomial in K[x] which has a root in E splits intolinear factors in E[x].

Proof. 1⇒ 2: Every τ ∈ eK(E,Kalg) extends to a σ ∈ AutK Kalg.2⇒ 4: Suppose p(x) ∈ K[x] is irreducible and α ∈ E is a root of p(x). Let

β ∈ Kalg be any root of p(x). Then the isomorphism K(α) → K(β) extendsto an embedding τ : E>Kalg. Since τ [E] = E, we see that β ∈ E. Since Econtains the roots of p(x) in Kalg, it follows that p(x) splits into linear factorsin E[x].

4⇒ 3: Clearly, E is the splitting field of Irr(α,K, x) : α ∈ E.3⇒ 2: Let E be the splitting field of fi : i ∈ I. For each i ∈ I, let Ri be

the finite set of roots of fi in E. If τ ∈ eK(E,Kalg). Then τ [Ri] ⊆ Ri and soτ [Ri] = Ri. Since E = K(∪iRi), it follows that τ [E] = E.

2⇒ 1: If σ ∈ AutK(Kalg) then τ = σ|E ∈ eK(E,Kalg).

Definition 5.14 (Normal Extension). The extension E of K is normal iff itsatisfies the conditions of Theorem 8.

Corollary 5.15. Suppose that K ⊆ E ⊆ Kalg and that E is a normal extensionof K. Then if K ⊆ F ⊆ E, then E is also a normal extension of F .

Proof. Take your pick of any condition in Theorem 8.

Remark: Of course, if K = Fp(t) and α = t1/p, then E = K(α) is a normalextension of K.

We can eliminate this problem as follows:

Definition 5.15 (Separable Polynomial). An irreducible polynomials f(x) ∈K[x] is separable if f has no repeated roots in Kalg.

Theorem 5.16. If p(x) ∈ K[x] is irreducible, then the following are equivalent:

1. p(x) is separable

2. The formal derivative p′(x) 6= 0.

Proof. Reading Exercise, Hungerford III.6.10.

Let K = Fp(t) and f(x) = xp − t, then f ′(x) = pxp−1 = 0.

Corollary 5.17. If the characteristic of K is zero, then every irreducible poly-nomial is separable.

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Definition 5.16 (Separable Extension). Suppose that K ⊆ E ⊆ Kalg.

1. The element α ∈ E is separable over K if Irr(α,K, x) is separable

2. The extension E of K is separable if every α ∈ E is separable over K.

Theorem 5.18. Suppose that K ⊆ F ⊆ E ⊆ Kalg.

1. If α ∈ E is separable over K, then α is separable over F .

2. If E is a separable extension of K, then E is also a separable extension ofF .

Proof. 1: Irr(α, F, x)| Irr(α,K, x).2: Immediate from 1.

Theorem 5.19. If E is a finite extension of K, then the following are equiva-lent:

1. E is a separable extension of K.

2. E = K(α1, . . . , αn) where each αi is separable over K

3. |eK(E,Kalg) = [E : K].

Proof. 1⇒ 2: Utterly obvious.2 ⇒ 3: Consider the tower of extensions K ⊆ K(α1) ⊆ K(α1, . . . , αn). Let

K0 = K and Ki = K(α1, . . . , αi). Suppose inductively that |eK(Ki : Kalg)| =[Ki : K]. Let τ ∈ eK(Ki,K

alg), then the number of extensions of τ to Ki+1 =Ki(αi+1) is equal to the number of distinct roots ri+1 of Irr(αi+1,Ki, x). Sinceαi+! is separable over Ki+1, we have ri+1 = deg Irr(αi+1,Ki, x) = [Ki+1,Ki] So|eK(Ki+1,K

alg)| = [Ki : K][Ki=1 : Ki] = [Ki+1 : K].3 ⇒ 1: Suppose that E is not a separable extension of K. Choose α ∈ E

such that α is not separable over K. Let E = K(α1, . . . , αn) where α1 = α, then|eK(K,Kalg)| < [K(α1) : K] and arguing as above, we find |eK(K,Kalg)| < [E :K].

We can now characterize Galois Extensions

Theorem 5.20. If E is a finite extension of K, then the following are equiva-lent:

1. E is a Galois Extension of K.

2. E is a separable normal extension of K.

3. |AutK(E)| = [E : K].

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Proof. 2 ⇒ 3: Since E is normal over K, eK(E,Kalg) = AutK E; and since Eis separable over K, |eK(E,Kalg)| = [E : K].

3 ⇒ 1: Let G = AutK(E) and let F = EG. Then K ⊆ F ⊆ E. Also[E : K] = |G| ≤ |AutF E| ≤ [E : F ] ≤ [E : K], so F = K thus E is a GaloisExtension.

1 ⇒ 2: Let G = AutK(E), Then EG = K. Let α ∈ E and let f(x) =Irr(α,K, x). Let R be the set of roots of f(x) in E. Consider the polynomialg(x) =

∏β∈R(x− β). Since every σ ∈ G permutes the set R it follows that the

coefficients of g(x) are G-invariant. Hence, g(x) ∈ K[x] and so f(x)|g(x). Also,it is clear that g(x)|f(x). Since both are monic, it follows that g(x) = f(x). Inparticular, Irr(α,K, x) is separable and so α is separable over K. Furthermore,Irr(α,K, x) splits into linear factors in E[x]. It follows that E is a normalextension of K.

Theorem 5.21 (The Fundamental Theorem of Galois Theory). Let E be a finiteGalois Extension of K and let G = AutK E be the Galois Group. Then thereexist mutually inverse bijections between Sub(G) = H|H < G and SubK(E) =F : F is a subfield such that K ⊆ F ⊆ E defined by H 7→ EH the fixed field,F 7→ AutF E, the corresponding Galois Group. Furthermore

1. H ⊆ H ′ iff EH ⊇ EH′

2. If H ⊆ H ′ then [H ′ : H] = [EH : EH′]

3. If K ⊆ F ⊆ E, then F is a Galois Extension of K iff AutF E E AutK E.In this case, AutK F ' AutK E/AutF E

We begin the proof of The Fundamental Theorem.

Proof. Suppose F is a subfield with K ⊆ F ⊆ E. Then E is also a Galoisextension of F . If H = AutF E, then EH = F . Thus, F 7→ H = AutF E 7→EH = F . Thus the map F 7→ AutF E is certainly injective. The surjectivity isan immediate consequence of a result of Artin which will be proved followingthis.

This result of Artin suggests the following question: What are the finitesubgroups of Aut C?

Continuing with the proof, let H,H ′ ≤ G = AutK E. Then there exist cor-responding subfields F, F ′ such that H = AutF E and H ′ = AutF ′ E. Supposethat H ⊆ H ′. Then clearly F = EH ⊇ EH′ = F ′.

Conversely, assume F ⊇ F ′. Then AutF E ⊆ AutF ′ E. Thus, (1) holds.For (2), suppose that H ⊆ H ′. Recall that |H| = [E : F ] and |H ′| = [E : F ′].

Thus [H ′ : H] = |H ′|/|H| = [E : F ′]/[E : F ] = [F : F ′].For (3), we suppose that K ⊆ F ⊆ E. Clearly F is a separable extension of

K. Thus, the following are equivalent:

1. F is a Galois extension of K

2. F is a normal extension of K

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3. σ[F ] = F for all σ ∈ AutK Kalg

4. σ[F ] = F for all σ ∈ AutK E

Suppose now that F is a Galois extension of K. Then we can define ahomomorphism AutK E

π→ AutK F : σ 7→ σ|F . Clearly kerπ = AutF E EAutK F . To see that AutK F ' AutK E/AutF F we must show that π issurjective. Let τ ∈ AutK F .

Then, there exists ϕ ∈ AutK Kalg such that ϕ|F = τ . Thus, if σ = ϕ|E ∈AutK E we have σ|F = τ as required.

Finally, suppose that F isn’t a Galois extension of K. Then there existsσ ∈ G = AutK E such that F ′ = σ[F ] 6= F . It is easily checked that AutF ′ E =σAutF Eσ−1. Since F ′ 6= F , AutF ′ E 6= AutF E and so AutF E is not a normalsubgroup of G = AutK E.

Lemma 5.22 (Artin). Let E be any field and let H be a finite subgroup of AutE.Let F = EH . Then E is a finite Galois extension of F and H = AutF E.

We shall require the following (interesting) result.

Theorem 5.23 (Primitive Element Theorem). If E is a finite separable exten-sion of K, then there exists an element α ∈ E such that E = K(α).

Proof. First suppose that K is a finite field. Then E is also a finite field andhence E∗ is cyclic. Let α ∈ E∗ be a generator, then clearly E = K(α).

Hence, we can suppose that K is infinite. Arguing by induction, it is enoughto consider the case when E = K(β, γ). Let [E : K] = n. Since E is a separableextension of K, there exist exactly n distinct embeddings of E into Kalg, sayσ1, . . . , σn. Consider the polynomial f(x) =

∏i 6=j([σiβ+ xσiγ]− [σjβ+ xσjγ]).

Then f(x) isn’t the zero polynomial, and so has only finitely many roots. Choosesome a ∈ K such that f(a) 6= 0. Let α = β + aγ. Then σ1(α), . . . , σn(α) are alldistinct. Thus, there are at least n distinct embeddings of K(α) into Kalg. So[K(α) : K] ≥ n and so K(α) = E.

Corollary 5.24. Let E be a separable algebraic extension of F . Suppose thereexists n ≥ 1 such that [G(α) : F ] ≤ n for all α ∈ E. Then [E : F ] ≤ n.

Proof. Choose α ∈ E such that [F (α) : F ] = mis maximal. Suppose there existsβ ∈ E \ F (α).

Let γ ∈ F (α, β) satisfy F (γ) = F (α, β). Then [F (γ) : F ] > m, contradiction.

Now we can prove Lemma 2 by Artin.

Proof. Let α ∈ E and let σ1, . . . , σr ∈ H be a maximal subset such thatσ1(α), . . . , σr(α) are distinct. Then if τ ∈ H, by maximality, τσ1(α), . . . , τσr(α) =σ1(α), . . . , σr(α).

Consider the polynomial f(x) =∏ri=1(x−σi(α)). Then α is a root of f and

τf = f for all τ ∈ H. Hence, the coefficients of f lie in EH = F .

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Furthermore, f is separable and splits into linear factors in E[x]. ApplyingCorollary 5, we also see that [E : F ] ≤ |H|.

Thus, E is a finite Galois extension of F . Finally, |G| ≤ |AutF E| = [E :F ] ≤ |H|, and so H = AutF E is the Galois group.

The following easy observation is often useful.

Proposition 5.25. Assume that F is a finite separable extension of K. Thenthere exists a finite Galois extension E of K such that K ⊆ F ⊆ E.

Proof. Choose α ∈ F with F = K(α). Let f(x) = Irr(α, F, x) and let α1, . . . , αnbe the roots of f(x) ins F alg. Then E = K(α1, . . . , αn) satisfies our require-ments.

Lemma 5.26. 1. Each positive r ∈ R has a square root in R.

2. If p(x) ∈ R[x] has odd degree, then p(x) has a root in R.

3. Each z ∈ C has a square root in C.

4. There doesn’t exist a field E ⊇ C such that [E : C] = 2.

Proof. (1) and (2) are the intermediate value theorem, (3) obvious.(4): Suppose E exists. Then E = C(α) for any α ∈ E \ C.Set f(x) = x2 + bx + c = Irr(α,C, x). Then the roots of f(x) are z =

−b±√b2−4c

2 , contradiction.

Theorem 5.27. C is algebraically closed.

Proof. It is enough to show that C has no proper algebraic extensions. Supposethat F is a finite algebraic extension of C. Then F is a finite separable extensionof R and hence there exists a finite Galois extension E of R such that R ⊆ C ⊆F ⊆ E.

Let G = AutR E be the corresponding Galois group and let |G| = 2nm wherem is odd.

By Sylow, there is a subgroup H < G with |H| = 2n. Let K = EH . Then[K : R] = [G : H] = m. Suppose m > 1. Then there exists α ∈ K such thatK = R(α) and hence deg Irr(α,R, x) = [K : R] = m, which is odd, so it has aroot, contradicting irreducibility. So m = 1.

Next note that C is a Galois extension of R, so AutR C ' AutR E/AutC E,hence P = AutC E has order 2n−1. We claim that n = 1, so that E = C asrequired.

We suppose not. By Sylow there exists N < P such that [P : N ] = 2. LetK = EN , then [K : C] = [P : N ] = 2, contradiction.

Definition 5.17 (Galois Group of a Polynomial). Let f(x) ∈ K[x]. Then theGalois group of f over K is AutK E where E is the splitting field of f over K.

Remark: We usually work with separable polynomials, so that E is a Galoisextension.

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Lemma 5.28. Let f(x) ∈ K[x] be a polynomial with Galois group G.

1. If f(x) has exactly n distinct roots, in Kalg, then G is isomorphic to asubgroup of Sn.

2. If f(x) is a separable irreducible polynomial of degree n, then G is isomor-phic to a transitive subgroup of Sn.

Example: Consider f(x) = x4−2 ∈ Q[x]. By Eisenstein, f(x) is irreducible.Let α = 4

√2. Then, the roots are ±α,±iα. Letting E be the splitting field of

f over Q, we have E = Q(α, i), so considering Q ⊂ Q(α) ⊂ Q(α, i) we see that[Q(α) : Q] = 4 and [Q(α, i) : Q] = 2, since i /∈ Q(α) ⊆ R.

Hence [E : Q] = 8. Thus, the Galois Group G of f over Q satisfies |G| = 8and G < S4, and |S4| = 4× 3× 2, it follows that G is the Sylow two subgroupof S4, so G ' 〈(1234)〉o (24).

Theorem 5.29. Suppose p is a prime and that f(x) ∈ Q[x] is an irreduciblepolynomial of degree p with exactly 2 nonreal roots in Qalg. Then, the Galoisgroup of f over Q is isomorphic to Sp.

Proof. Let α1, . . . , αp ∈ Qalg be the roots of f(x). then G is isomorphic to asubgroup of Sym(α1, . . . , αp). Since G acts transitively on the roots, p||G|and so G contains a p-cycle. Also, complex conjugation induces a transposition.So, after suitably ordering the roots, we can suppose G contains τ = (12) andσ = (1i2 . . . ip), hence, there exists 1 ≤ k ≤ p − 1 such that σk = (12j3 . . . jp),as p is prime.

Thus, again relabeling the roots, we can suppose G contains τ = (12) andσ = (12 . . . p). So G also contains O(12)O−1 = (23) and so on. Since G contains(12), (23), . . . , (p− 1p), it follows that G = Sp.

Example: consider f(x) = x5 − 4x + 2 ∈ Q[x]. By Eisenstein, f(x) isirreducible. By Calc 1, it is easily checked that f(x) has exactly three realroots. Thus, the Galois group of f over Q is S5.

Example: Consider f(x) = x5 + 3x + 15 ∈ Q[x]. By Eisenstein, f is irre-ducible. Then f(x) has exactly one real root. This time, complex conjugationinduces a product of two transpositions. So if G is the Galois Group of f overQ, then G contains both a five cycle and a product of two transpositions.

Unfortunately, this still leaves at least three candidates: S5, A5, 〈(12345)〉o〈(25)(34)〉 ' D5

We will use the following, but put off the proof until later:

Theorem 5.30. Let f(x) ∈ Z[x] be a monic polynomial and let p be a prime.Let f(x) = f(x)( mod p), the polynomial obtained by reducing the coefficientsmod p. Suppose that f(x) has no multiple roots in Falg

p . Then, there exists abijection α1, . . . , αn → α1, . . . , αn between the roots of f(x) and the rootsof f(x) together with an embedding of the Galois group G of f over Fp into theGalois group G of f over Q which gives an embedding of the actions of thesegroups on the roots.

40

Reducing mod 2, we get f(x) = x5 + x + 1 ∈ F2. Now working in F2[x],we get x5 +x+ 1 = (x2 +x+ 1)(x3 +x2 + 1). These have no linear factors, andthus are irreducible over F2. f(x) has five distinct roots in Falg

2 . Furthermore,the splitting field of f over F2 is F26 . Hence, the corresponding Galois group off is AutF2 F26 ' C6.

Let σ ∈ AutF2 F26 be a generator. Then σ permutes the roots of x2 + x+ 1and x3 + x2 + 1 so is a 2-cycle times a 3-cycle.

Hence, the Galois group G of f over Q contains an element σ which is aproduct of a 2-cycle and a 3-cycle. So σ3 is a transposition, thus G = S5.

Problem: We find an irreducible f(x) ∈ Q[x]of degree seven with Galoisgroup S7.

Solution: We claim that the polynomial a(x) = x5 + x2 + 1 ∈ F2[x] isirreducible. But b(x) = x2 + x + 1 is the only irreducible quadratic, and itdoesn’t divide a(x), which has no linear factors.

Consider f(x) = a(x)b(x) = x7 + x5 + x4 + 1. The splitting field over f isE = F210 . Thus, the Galois Group G of f over F2 is cyclic of order 10. Letσ ∈ G be a generator. Clearly σ is the product of a 5-cycle and a 2-cycle. Letf(x) = x7 + 3x5 + 3x4 + 3 ∈ Q[x].

Let G be the Galois group of f over Q. Then G contains a product σ of a5-cycle and a 2-cycle. Raise it to the fifth power, and we get a 2-cycle, σ5 ∈ G.As G also contains a 7-cycle, G ' S7.

Theorem 5.31. For each n ≥ 2, there exists an irreducible poynomail f(x) ∈Q[x] of degree n with Galois group Sn.

Clearly, we can suppose that n ≥ 4.

Lemma 5.32. SUppose the n ≥ 4 and G ≤ Sn satisfies:

1. G is transitive

2. G contains a (n− 1)-cycle

3. G contains a transposition.

Then G = Sn.

Proof. Let Ω = 1, . . . , n. Let σ ∈ G be an (n− 1)-cylce and let α ∈ Ω be thefixed point of σ. Then Gα acts transitively on Ω \ α, it follows that for allβ ∈ Ω, Gβ acts transitively on Ω \ β.

This means that G acts 2-transitively on Ω; ie, if α 6= β, γ 6= δ, then thereexists g ∈ G such that g(α) = γ and g(β) = δ.

Hence, if τ ∈ G is a transposition, then we can conjugate τ to any othertransposition by a suitable element of G. Since G contains every transposition,G = Sn.

Lemma 5.33. For each prime p and d ≥ 1, there exists an irreducible f(x) ∈Fp[x] of degree d.

41

Proof. Reading Exercise

We will now prove the Theorem.

Proof. Let a(x) ∈ F2[x] be irreducible of degree n−1 and let b(x) = (x+1)a(x) =xn + bn−1x

n−1 + . . .+ b0 ∈ F2[x].Similarly, let c(x) = xn + cn−1x

n−1 + . . . + c0 ∈ F3[x] be chosen so that itfactors as either g(x)h)(x) where g is an irreducible quadratic and h is irreducibleof odd degree or gh1h2 where g is an irreducible quadratic and h1, h2 are distinctirreducible of odd degree.

By the Chinese Remainder Theorem, for each 0 ≤ i ≤ n − 1, there exists0 ≤ `i < G such that `i ≡ bi( mod 2) ≡ ci( mod 3).

Consider f(x) = xn+7`n−1xn−1 + . . .+7. By Eisenstein, f(x) is irreducible.

Hence, the Galois group G is transitive. Reducing mod 2, G contains ann− 1 cycle. Reducing mod 3, we see that G must also contain a transposition.Thus, G = Sn.

Corollary 5.34. If G is any finite group, then there exists a finite extensionK of Q and a Galois extension E of K such that G ' AutK E.

Proof. Let |G| = n. By Cayley, we can suppose that G ≤ Sn. Let f(x) ∈ Q[x]be irreducible of degree n with Galois group Sn and let E be the splitting fieldof f(x).

Then K = EG satisfies our requirements.

We will now begin our discussion of Solvable groups and radical extensions.In this section we will study Galois groups of prescribed type: cyclic, abelianand solvable

Definition 5.18 (Cyclic/Abelian Extension). a finite extension E of K iscyclic/abelian if E is a Galois Extension of K and AutK E is cyclic/abelian.

Let α ∈ Qalg \Q, and E ⊆ Qalg is maximal such that α /∈ E. Then any finiteextension of E is cyclic.

Let K be a field of characteristic p ≥ 0 and let n ≥ 1 satisfy p 6 |n. Then thepolynomial xn − 1 has n distinct roots in Kalg, since xn − 1 and nxn−1 haveno common roots. These roots form a finite multiplicative subgroup of (Kalg)∗

and hence form a cyclic group. A generator of this group is called a primitiventh root of unity.

e.g., working over Q, we have x3 − 1 = (x − 1)(x2 + x + 1). The primitive3rd roots of unity are −1±

√−3

2 .Let α be a primitive nth root of unity. Then if 1 ≤ r ≤ n, then αr is also a

primitive root iff (r, n) = 1. So there are exactly ϕ(n) primiitive roots of unite.Of course, this is also the order of Z∗n. If fact, αr is primitive iff r ∈ Z∗n.

Definition 5.19 (Cyclotomic Extension). With the above hypotheses, the split-ting field of xn − 1 over K is called the cyclotomic extension of order n.

42

Theorem 5.35. With the above hypotheses, let F be the cyclotomic extensionof order n:

1. F = K(ξ), where ξ is any primitive nth root of unity.

2. F is an abelian extension of K of dimension d for some d|ϕ(n). Further-more, if n is prime, then F is a cyclic extension.

3. AutK F is isomorphic to a subgroup of order d in Z∗n.

Remark: It is possible that d < ϕ(n). For example, let ω be a primitive 5th

root of unity. Then [R(ω) : R] = [C : R] = 2.Note that |F∗23 | = 7, hence F23 contains a primitive 7th root of unity. In

particular, x6 + x5 + x4 + x3 + x2 + x+ 1 isn’t irreducible over F2.

Proof. Clearly the irreducible factors of xn − 1 are separable and so F is aGalois extension of K. Let ξ ∈ F be a primitive nth root of unity. ThenF = K(1, ξ, ξ2, . . . , ξn−1) = K(ξ). Also, if σ ∈ AutK F , then σ is uniquelydetermined by its value σ(ξ) = ξr. Clearly ξr is also a primitive nth root andso r ∈ Z∗n. So we obtain an injective homomorphism AutK F → Z∗n : σ 7→ r.Hence AutK F is isomorphic to a subgroup of Z∗n of order d|ϕ(n). So AutK Fis abelian, and cyclic if n is prime, since Zn is now a field.

Finally, by Galois Theory, [F : K] = |AutK F | = d.

Theorem 5.36. Let K be a field of char p ≥ 0 and let n ≥ 1 with p 6 |n.Suppose that K contains a primitive nth root of unity. Let a ∈ K and let α bea root of xn− a. Then K(α) is a cyclic extension of dimension d for some d|n.Furthermore, αd ∈ K. Hence if b = αd ∈ K then Irr(α,K, x) = xd − b.

Proof. Let ζ ∈ K be a primitive nth root of unity. Then α, ζα, . . . , ζn−1α aredistinct roots of xn − α. If follows that K(α) is a Galois extension of K.

Let G = AutK K(α) be the corresponding Galois group. If σ ∈ G, thenσ(α) = ωσα where ωσ is a not necessarily primitive nth root of unity.

Thus, we obtain an injective homomorphism π : G→ ζi : 0 ≤ i ≤ n− 1 'Cn. Thus G is cyclic of order d for some d|n. Let σ ∈ G be a generator, thenσ(α) = ωσα for some primitive dth root of unity ωσ.

Furthermore, σ(αd) = σ(α)d = (ωσα)d = αd. Thus, αd ∈ K(α)G = K.

Remark: This theorem no longer holds if K doesn’t contain a primitive nth

root of unity. eg, consider x3 − 2 ∈ Q[x]. Then Q( 3√

2) isn’t a Galois extensionof Q.

We next work towards

Theorem 5.37. Let K be a field of charK = 0 which contains a primitive nth

root of unity. If E is a cyclic extension of K of dimension n, then there existsα ∈ E such that E = K(α) and α is a root of xn − a ∈ K[x].

Until further notice, we restrict our attention to fields of characteristic zero.

43

Definition 5.20. Let K be a field of characteristic zero and let E be a finiteextension of K with [E : K] = n. Let eK(E,Kalg) = σ1, . . . , σn. For each α ∈E, the corresponding norm is NE

K(α) = σ1(α) . . . σn(α) and the correspondingtrace is trEK(α) = σ1(α) + . . .+ σn(α).

Lemma 5.38. If E is a finite Galois Extension of K and α ∈ E, then

NEK(α) =

∏σ∈AutK E

σ(α)

andtrEK(α) =

∑σ∈AutK E

σ(α),

and both are in K.

Proof. Under these hypotheses, eK(E,Kalg) = AutK E. Clearly NEK(α), trEK(α)

are fixed by each Aut ∈ AutK E. So NEK(α) and trEK(α) lie in K.

Lemma 5.39. Suppose that E is a finite extension of K and α, β ∈ E. ThenNEK(αβ) = NE

K(α)NEK(β) and trEK(α+ β) = trEK(α) + trEK(β).

Futhermore, if γ ∈ K then NEK(γ) = γ[E:K] and trEK(γ) = [E : K]γ.

So what is the kernel of NEK : E∗ → K∗? Well, if [E : K] = n, and ζ ∈ K is

an nth root of unity, then ζ is in the kernel.

Definition 5.21 (Characters). Let G be a group and let K be a field. Then acharacter of G is a homomorphism χ : G→ K∗.

Theorem 5.40 (Artin). If χ1, . . . , χn are distinct characters of a group G in thefield K, then χ1, . . . , χn are linearly independent over K. ie, if a1, . . . , an ∈ Kare not all zero, then the function

∑ni=1 aiχi is not identically zero.

Proof. Suppose not and that χ1, . . . , χn is chosen with n minimal say a1χ1 +. . .+anχn = 0 where a1, . . . , an ∈ k. Then clearly n > 1 and each ai 6= 0. Sinceχ1 6= χ2, there exists a z ∈ G such that χ1(z) 6= χ2(z). But we have for allx ∈ G a1χ1(zx) + . . .+ anχn(zx) = 0 ie, a1χ1(z)χ1 + . . .+ anχn(z)χn = 0.

Multiplying the original formula by χ1(z), we obtain a1χ1(z)χ1 + . . . +anχ1(z)χn = 0. Subtracting this from the first, we obtian a2(χ2(z)−χ1(z))χ2 +. . . + an(χn(z) − χ1(z))χn = 0. Since a2(χ2(z) − χ1(z)) 6= 0, this contradictsthe minimality of n.

Corollary 5.41. If K is any field and σ1, . . . , σn ∈ AutK are distinct, thenσ1, . . . , σn are linearly independent over K.

Proof. We regard each σi as an isomorphism from K∗ to K∗.

Theorem 5.42 (Hilbert’s Theorem 90). Let K be a field of characteristic zeroand let E by a cyclic extension of K of dimension n. Let σ ∈ AutK E bea generator. If α ∈ E then NE

K(α) = 1 iff there exists β ∈ E∗ such thatα = β/σ(β).

44

Proof. Let G = AutK E be the cyclic Galois group.⇐: Suppose there exists β ∈ E∗ such that α = β/σ(β). Then

NEK(α) = NE

K(β)/NEK(σ(β)) = NE

K(β)/NEK(β) = 1

.⇒: Conversely, suppose that NE

K(α) = 1. To make this proof easier to read,we shall use exponential notation as follows: if τ, σ ∈ G and η ∈ E, then wewrite ητ instead of τ(η). Thus ητ+θ = ητηθ = τ(η)θ(η), etcetera.

In particular, for each η ∈ E, NEK(η) = η1+σ+σ2+...+σn−1

. By Artin’sTheorem on Characters, the map on E defined by id + ασ + α1+σσ2 + . . . +α1+σ+...+σn−2

σn−1 is not identically zero.Hence, there exists θ ∈ E such that θ+αθσ+α1+σθσ

2+. . .+α1+...+σn−2

θσn−1 6=

0. Notice that αβσ = αθσ + α1+σθσ2

+ α1+σθσ3

+ . . . + α1+...+σn−1θσ

n

. Sinceα1+σ+...+σn−1

= NEK(α) = 1 and σn = 1 we see that αβσ = β, and so

α = β/σ(β).

We can now prove Theorem 21

Proof. Let K be a field of characteristic zero which contains a primitive nth

root of unity ζ and let E be a cyclic extension of K of dimension n with Galoisgroup G. Let σ ∈ G be a generator.

Since [E : K] = n, we have that NEK(ζ−1) = (ζ−1)n = 1. Hence, by Hilbert’s

Theorem 90, there exists α ∈ K∗ such that ζ−1 = α/σ(α) and so σ(α) = ζα.Since ζ ∈ K, we have σ2(α) = σ(ζα) = ζ2α. Continuing in this fashion, weobtain σi(α) = ζiα for 1 ≤ i ≤ n. Thus α, ζα, . . . , ζn−1α is a set of n distinctelements under the action of G. Hence [K(α) : K] ≥ n and so E = K(α).

Also, σ(αn) = σ(α)n = (ζα)n = αn. Thus a = αn ∈ EG = K. Clearly α isa root of xn − a ∈ K[x].

We continue to work with fields of characteristic zero.

Definition 5.22 (Radical Extension). An extension F of a field K is a radicalextension iff there exists a tower K ⊆ K(u1) ⊆ . . . ⊆ K(u1, . . . , un) = F suchthat for each 1 ≤ i ≤ n, there exists a di ≥ 1 such that udi

i ∈ K(u1, . . . , ui−1).

Example: Let α = 4√

2 and i =√−1 then Q(α, i) is a radical extension of Q

as Q ⊆ Q(α) ⊆ Q(α, i).

Definition 5.23 (Solution by Radicals). Let f(x) ∈ K[x] and let E be thesplitting field of f over K. Then the equation f(x) = 0 is solvable by radicalsiff there exists a radical extension F of K such that K ⊆ E ⊆ F .

Remark: We have no required that F should be a radical Galois extension,but the next lemma says that we can assume it is also Galois.

Recall that if F is a separable extension of K then there exists a Galoisextension E of K such that E ⊆ F ⊆ E. In fact, there exists a minimal onecalled the normal closure.

45

Lemma 5.43. If F is a radical extension of K and N is the normal closure ofF , then N is also a radical extension of K.

Proof. Let F = K(α) and let g(x) = Irr(α,K, x). Then N is the splittingfield of g(x) over K. Let α1, . . . , αr be the distict roots of g(x) in N . LetK ⊂ K(u1) ⊂ . . . ⊂ K(u1, . . . , un) = F = K(α) wirness that F is a radicalextnesion of K. For each 1 ≤ q ≤ r, let σi ∈ AutK N satisfy σi(α) = αi. ThenK ⊂ K(σi(u1)) ⊂ . . .K(σi(u1), . . . , σi(un)) = K(α) witnesses that K(αi) is aradical extension of K. Hence the following tower witnesses that N is a radicalextension of K.

K ⊂ K(u1) ⊂ . . . ⊂ K(u1, . . . , un) = K(α1) ⊂ K(α1, σ2(u1)) ⊂ . . . ⊂ K(α1, σ2(u1), . . . , σ2(un)) = K(α1, α2) ⊂

. . . ⊂ K(α1, . . . , αn) = N

Suppose that K is a field of characteristic zero. We wish to characterizethose f(x) ∈ K[x] which are solvable by radicals. Let E be the splitting fieldof f and suppose that F is a radical Galois extension such that K ⊆ E ⊆ F .Since E is Galois over K, we have AutK E ' AutK F/AutE F . In particular,AutK E is a homomorphic image of AutK F .

Thus, it is enough to understand the structure of AutK F . Let K = K0 ⊂K1 ⊂ . . . ⊂ Kn = F withness that F is a radical extension of K. To simplifymatters, suppose that K contains all relevant roots of unity. Then each Ki ⊂Ki+1 is a cyclic extension. IN particular, consider the special case where K =K0 ⊂ K1 ⊂ K2 = F .

Then AutK0 K1,AutK1 K2 are cyclic, AutK0 K1 ' AutK0 K2/AutK1 K2.Thus AutK0 K2 is an extension of AutK0 K1 by AutK1 K2, that is, 1→ AutK1 K2/AutK0 K2 → AutK0 K1 → 1 where the outer pair are cyclic.

Thus, we must study the smallest class of groups S such that each abeliangroup is in S and such that S is closed under taking extensions and homo-morphic images.

Definition 5.24 (Commutator). Let G be a group. If a, b ∈ G then the corre-sponding commutator is [a, b] = aba−1b−1.

The commutator subgroup of G is G′ = 〈[a, b] : a, b ∈ G〉.

Example: If G is abelian, then G′ = 1.

Theorem 5.44. Let G be a group.

1. G′ E G

2. G/G′ is abelian

3. if N E G, then G/N is abelian iff G′ ≤ N .

46

Proof. 1. If a, b ∈ G, and π ∈ AutG, then π([a, b]) = [π(a), π(b)] Henceπ[G′] = G′. In particular, this is true if π is an inner automorphism,hence, G′ E G.

2. Let a, b ∈ G. Since (ab)(ba)−1 = aba−1b−1 = [a, b] ∈ G′ it follows thatabG′ = baG′, and so G/G′ is abelian.

3. Finally, suppose N E G. If G′ ≤ N , then the above argument shows thatG/N is abelian. Conversely, suppose that G/N is abelian. Let a, b ∈ G.Since abN = baN , it follows that [a, b] = (ab)(ba)−1 ∈ N .

Examples: Clearly, if S is a simple nonabelian group then S′ = S. Hence ifn ≥ 5, then A′n = An.

Definition 5.25 (Perfect). G is perfect if G = G′.

Let n ≥ 5, since Sn/An ' Z2 it follows that S′n ≤ An, since An is perfect, itfollows that S′n = An.

Consider S3. The normal subgroups of S3 are 1, A3, S3. As S3 is not abelian,and S3/A3 ' Z2, S′3 = A3 and S′′3 = A′3 = 1.

The normal subgroups of S4 are 1, A4, S4 and V = 1, (12)(34), (13)(24), (14)(23).Clearly 1 < V < A4 < S4.

As S4/A4 ' Z2, A4 contains S′4. Since [(12), (23)] = (12)(23)(12)−1(23)−1 =(13)(23) = (132) /∈ V . So S′4 = A4.

Since A4/V ' Z3, so A′4 = V , and so V ′ = 1. Thus, S4 is an iteratedextension of abelian groups.

Definition 5.26 (Derived Subgroups). For each i ≥ 0, the ith derived subgroupG(i) of G is defined by G(0) = G and G(i+1) = G(i)′ .

Definition 5.27 (Solvable Group). G is solvable iff there exists n ∈ N suchthat G(n) = 1.

If G is abelian, then G is solvable. S4 is solvable. If n ≥ 5, then S5 is notsolvable.

Theorem 5.45. 1. If G is solvable, then every subgroup of G is solvable,and every homomorphic image of G is solvable.

2. Suppose G is a group and N E G. If N,G/N are solvable, then so is G.

Proof. 1. First suppose that H ≤ G. An easy induction shows that H(i) ≤G(i) for all i ≥ 0. It follows that H is also solvable.

Next suppose that f : G → H is an epimorphism. An easy inductionshows that f [G(i)] = H(i) for all i ≥ 0. It follows that H is also solvable.

2. Let f : G → G/N be the canonical surjection. Then there exists n ≥ 0such that f [G(n)] = (G/N)(n) = 1. Hence G(n) ≤ N . By the first part,G(n) is also solvable, and hence, there exists k ≥ 0 such that G(n+k) =G(n)(k) = 1.

47

Definition 5.28. A subnormal series of G is a chain of subgroups G = G0 >G1 > G2 > . . . > Gn = 1 such that Gi+1 E Gi for all 0 ≤ i ≤ n − 1 and thequotients Gi/Gi+1 are called the factors of the series.

A subnormal series is a composition series if Gi/Gi+1 is simple for all 0 ≤i ≤ n− 1.

A subnormal series is a solvable series if Gi/Gi+1 is abelian.

Examples, let V =< a > ⊕ < b > where < a >'< b >' Zn and letG = Vo < c > where < c >' C2 and cac−1 = b. Then G > V > 〈a〉 > 1. Notethat < a >6 /G.

This is why it is called a subnormal series.Suppose that G is solvable and n is minimal such that G(n) = 1. Then

G = G(0) > G(1) > . . . > G(n) = 1. This is actually a normal series, and asolvable series.

“Clearly”, every finite group has a composition series. To see this, let G befinite, and inductively let Gi+1 be a maximal normal subgroup of Gi. This givesa composition series.

Of course, not every group has a composition series. For example, Z, has nosuch series.

Theorem 5.46. A group G is solvable iff G has a solvable series.

Proof. The above shows ⇒.So, we suppose that G = G0 > G1 > . . . > Gn = 1 is a solveable series.

We claim that G(i) ≤ Gi for 0 ≤ i ≤ n. Clearly this is true when i = 0.Suppose inductively that G(i) ≤ Gi. Since Gi/Gi+1 is abelian, it follows thatG(i)′ ≤ Gi+1, hence, G(i+1) ≤ Gi+1.

Theorem 5.47. A finite group G is solvable iff G has a composition series allof whose factors are cyclic of prime order.

Proof. ⇐, by theorem 26.⇒, we have already seen the finite group G has a composition series G =

G0 > . . . > Gn = 1. Since G is solvable, it follows that each factor Gi/Gi+1 isalso solvable. Since Gi/Gi+1 is simple and abelian, it is Zp for some p prime.

Theorem 5.48. If K is a field of characteristic zero, and f(x) ∈ K[x], thenthe following are equivalent:

1. The equation f(x) = 0 is solvable by radicals.

2. The Galois group of f(x) over K is solvable.

Proof. 1 ⇒ 2: Suppose that f(x) = 0 is solvable by radicals. Let E be thesplitting field of f over K. Then there exists a radical Galois extension F ofK such that K ⊆ E ⊆ F . Since AutK E ' AutK F/AutE F , it is enough toshow that AutK F is solvable. Since F is a radical extension of K, there existsa tower K ⊂ K(u1) ⊂ . . . ⊂ K(u1, . . . , un) = F such that for each 1 ≤ i ≤ n,there exists di ≥ 1 such that udi

i ∈ K(u1, . . . , ui−1) Let d = d1 . . . dn and let ξ

48

be a primitive dth root of unity. Let L = K(ξ) be the corresponding cyclotomicextension and consider the following diagram of extensions.

F

K

F (ξ)

L = K(ξ).....................................................................................................................................................

cyclotomic

.................................................................................................................................................

.....................................................................................................................................................

cyclotomic

................................................................................................................................................................

Then F (ξ) is a Galois extension ofK. Since AutK F ' AutK F (ξ)/AutF F (ξ),it is enough to show that AutK F (ξ) is solvable. Finally, since AutK L 'AutK F (ξ)/AutL F (ξ), it is enough to show that AutL F (ξ) is solvable. Con-sider the tower of extensions L ⊂ L(u1) ⊂ . . . ⊂ L(u1, . . . , un) = F (ξ).

Let L0 = L and Li = L(u1, . . . , ui) for 1 ≤ i ≤ n. For each 0 ≤ i ≤ n− 1, Licontains a primitive dthi+1 root of unity and Li+1 = Li(ui+1) where udi+1

i+1 ∈ Li.It follows that Li+1 is a cyclic extension of Li, that is, Li+1 is a Galois extensionof Li such that AutLi Li+1 is cyclic.

Suppose inductively that AutL Li is solvable. L ⊂ Li ⊂ Li+1. ThenAutL Li ' AutL Li+1/AutLi

Li+1. By hypothesis AutL Li is solvable andAutLi

Li+1 is cylic, hence AutL Li+1 is also solvable. Thus AutL F (ξ) is solv-able.

To prove that 2 implies 1, it is enough to prove the following theorem:

Theorem 5.49. Suppose K is a field of characteristic 0 and F is a finite GaloisExtension of K. If the Galois Group AutK E is solvable, then there exists aradical Galois Extension F of K such that K ⊆ E ⊆ F .

Proof. We argue by induction on n = [E : K], the case n = 1 being trivial.Suppose inductively that the result holds for al such extensions K ′ ⊆ K ′ withk = [E′ : K ′] < n. Since G = AutK E is a finite solvable group, G has acomposition series, all of whose factors are cyclic of prime order. In particular,there exists H E G such that [G : H] = p for some prime p. Let ξ be a primitivepth root of unity and let N = E(ξ) be the corresponding cyclotomic extension.Also let M = K(ξ) and consider the following diagram.

49

E

K

N = E(ξ)

M = K(ξ).....................................................................................................................................................

cyclotomic

.................................................................................................................................................

.....................................................................................................................................................

cyclotomic

................................................................................................................................................................

Clearly M is a radical extension of K. Hence it is enough to find a radicalextension of M which contains N . Next onbserve N = E(ξ) is a Galois extensionof K. Since E is a Galois Extension of K, it follows that σ[E] = E for everyσ ∈ AutM N ≤ AutK N . Hence, we can define a homomorphism θ : AutM N →AutK E = G by σ 7→ σ|E . Since each σ ∈ AutM N satisfies σ(ξ) = ξ, itfollows that θ is an injection. Hence, AutM N is also solvable and |AutM N | ≤|AutK E|.

There are 2 cases to consider.Case A: Suppose that θ[AutM N ] is a proper subgroup of AutK E. Then

[N : M ] = |AutM N | < |AutK E| = n. Hence, by induction hypothesis, thereexists a radical extension F of M containing N .

Case B: Otherwise, θ : AutM N → AutK E = G is an isomorphism. LetJ = θ−1(H). Then J is a normal subgroup of AutM N of index p, and clearlyJ is also solvable. Let L = NJ and consider the following diagram:

M

L

N

AutMN

J

1..........................................................................................................................................................................................................

................................................................................................................. ............ .............................................................................................................................

................................................................................................................. ............ .............................................................................................................................

........................................................................................ ............ ....................................................................................................

.....................................................................................................

.....................................................................................................

.....................................................................................................

Thus, L is a Galois extension of M and AutM L ' AutM N/AutLN =AutM N/J . Thus |AutM L| = p and so L is a cyclic extension of M . Since Mcontains a primitive pth root of unity, there exists u ∈ L such that L = M(u)and u is a root of some polynomial xp − a ∈M [x].

In particular, L is a radical extension of M . Also notice that [N : L] = |J | <[N : M ] = [E : K] = n and that AutLN = J is solvable. Hence by inductinhypothesis, there exists a radical extension F of L which contains N .

Finally, a couple of loose ends...Example: Let f(x) = x3 − 3x+ 1 ∈ Q[x].

50

1. The equation f(x) = 0 is solvable by radicals

2. The splitting field E of f(x) over Q[x] isn’t a radical extension.

Proof. By Hungerford p272, the Galois group of f(x) is G = A3 ' C3. Hence[E : Q] = 3. Suppose that E is a radical extension of Q. Then there exists α ∈ Esuch that E = Q(α) and α is the root of an irreducible polynomial x3−a ∈ Q[x].Since E is normal, E must contain all roots of x3 − a, ie, α, ξα, ξ2α where ξ isa primitive 3rd root of unity. In particular, ξ ∈ E, which is impossible since[Q(ξ) : Q] = 2.

Another loose end, what are the finite subgroups of AutQ C?

Theorem 5.50 (Artin). Suppose K is an algebraically closed field of charac-teristic 0 and 1 6= G is a finite subgroup. Let F = KG. Then [K : F ] = |G| = 2and K = F (i) where i2 = −1.

Proof. By Artin, [K : F ] = |G| and K is a Galois extension of F . Let E = F (i)where i2 = −1. Then K is also a Galois extension of E. Let H = AutE K.Then it is enough to show that H = 1 so that E = K.

Suppose not and let p be a prime such that p||H|. Let C ≤ H be cyclic oforder p and let L = KC . Then, [K : L] = p and K is a cyclic extension of L. If ξis a primitive pth root fo unity, then Irr(ξ, L, x) has degree ≤ p−1. Thus, ξ ∈ L.Hence, K is the splitting field of some irreducible polynomial xp−a ∈ L[x]. Letα = α1, . . . , αp ∈ K be the roots of xp−a. Since K is algebraically closed, thereexists some β ∈ K such that βp = α.

Computing with N = NKL , we find −a = (−1)pα1 . . . αp = (−1)pN(α) =

(−1)pN(βp) = (−1)pN(β)p. Thus, N(β) ∈ L satisfies N(β)p = (−1)p−1a.Since L doesn’t contain a root of xp − a = 0, it follows that p = 2. ThusN(β)2 = −a. But since i ∈ L, we have that a = (−1)(−a) = (iN(β))2, which isa contradiction.

6 Linear Algebra

Let k be a field. Recall from linear algebra that an endomorphism of a vectorspace (free module over a division ring) can be represented by a matrix in oneway for each basis.

Definition 6.1 (Change of Basis). Let ei and fi be two bases of a vectorspace V . Let C be the matrix define by ej =

∑ni=1 cijfi. C is called the change

of basis matrix, and CAC−1 represents the linear operator A in the basis fi.

Definition 6.2 (Similar Matrices). We say that the matrices A and B aresimilar if ∃C such that CAC−1 = B

Similar matrices are representations of the same linear operator in differentbases.

51

Theorem 6.1 (Rational Canonical Form). Every matrix is similar to a block

diagonal matrix with blocks of the form

0 0 . . . −a1

1 0 . . . −a2

.... . . . . .

...0 . . . 1 −am

Proof. Recall, V is a k-module. We fix ϕ : V → V . Give V the structure of ak[x]-module by setting x · v = ϕ(v).

We notice that if v =∑biei then x · v = ϕ(v) = A · v

Remark: If you change basis, then v =∑bjej =

∑j bj

∑i cijfi =

∑i(∑j cijbj)fi

As k[x] is a PID, we have V ' ⊕`k[x]/f`(x)⊕ k[x]r where f1|f2| . . . fk.But k[x] has ∞ dimension over k, as x, x2, x3, . . . are linearly independent

over k, so r = 0Thus, V ' ⊕`k[x]/f`. Lets write f` =

∑m`

i=1 d`ixi and lets assume as k is a

field that all the f`’s are monic. That is, d` = 1.Let m` = deg f`. So 1, x, x2, . . . , xm`−1 generate V` ' k[x]/f`.Now: how does x act on V`? It acts on this basis 1, x, . . . , xm`−1 by

shifting, so x·xk = xk+1, thus x·xm`−1 = xm` =∑m`−1i=0 (−d`i)xi, so x : V` → V`

has matrix

0 0 . . . −d`01 0 . . . −d`1...

. . . . . ....

0 . . . 1 −d`m`−1

.

Definition 6.3 (Algebraically Closed). A field K is algebraically closed if everypolynomial in k has a root.

Theorem 6.2 (Jordan Canonical Form). If k is algebraically closed, then every

matrix is similar to a block diagonal matrix with blocks of the form

λ 1 . . . 00 λ . . . 0...

. . . . . ....

0 0 . . . λ

Proof. Assume that k is algebraically closed. We know that V ' ⊕m`=1k[x]/pn`

` (x)where p`(x) is prime.

The primes in k[x] with k algebraically closed are of the form x−a for a ∈ k.Suppose p`(x) = x− λ`, with λ` ∈ k. Then pn`

` (x) = (x− λ`)n`

We’ll show that V` has a basis such that ϕ` : V` → V`, where ϕ` = ϕ|V`has

a matrix as above.We will show that the companion matrix of pn`

` is similar to a Jordan block.We define B = A−λI for λ the root of p`. We have a k[y]-module structure

by y · v = Bv. We check that V is a cyclic k[y]-module.Then ynv = Bnv = (A−λI)nv = f(A)v = 0 but yn−1v 6= 0. So V ' k[y]/yn

as a k[y]-module. The the rational canonical form of B has ai = 0 unlessi = n− 1, and an−1 = 1.

CAC−1 = CBC−1 + CλIC−1 = CBC−1 + λI

52

So, in this basis, A is the transpose of a Jordan block. We reverse the orderof the basis and obtain the Jordan Form.

How can we compute the JCF?

Definition 6.4 (Eigenvalues and Eigenvectors). λ ∈ k is an eigenvalue of A if∃x 6= 0 such that Ax = λx.

x ∈ V is an eigenvector of A if ∃λ ∈ k such that Ax = λx.

Thus, the λ’s in the JCF are all eigenvalues.If V has a basis of eigenvectors of A, then A is similar to a diagonal matrix.

Theorem 6.3. The Following Are Equivalent:

1. λ is an eigenvalue.

2. ∃x 6= 0 such that Ax = λx

3. (A− λI)x = 0

4. A− λI is not invertible

5. det(A− λI) = 0.

Definition 6.5 (Generalized Eigenspace). Let Emλ = v ∈ V : (A−λI)mv = 0.Emλ is called the generalized eigenspace and v ∈ Emλ is called a generalizedeigenvector.

Note: Emλ is a subspace, if v ∈ Emλ then (A−λI)v ∈ Em−1λ , (A−λI)Em+1

λ ⊆Emλ ⊆ E

m+1λ , and ∃N such that ENλ = Emλ for all m ≥ N .

If λ 6= µ then Emaxλ ∩ Emax

µ = 0, and Emaxλ : λ eigenvalue span V and

are disjoin, so V ' ⊕λEmaxλ . Thus, dimV =

∑dimEmax

λ = n.So the Jordan Canonical Form gives a lower bound on dimEmax

λ since thebounds must add to n, we have that the number of λ in the Jordan CanonicalForm is equal to dimEmax

λ .

Corollary 6.4. All eigenvalues show up in the JCF.

An algorithm for computing the JCF:

1. Compute the eigenvalues λ of A

2. For each λ, compute E1λ, . . . , E

maxλ .

3. If Emaxλ = Enλ 6= En−1

λ choose a basis for Enλ/En−1λ . Claim: (A−λI)mvi :

1 ≤ i ≤ k, 0 ≤ m ≤ n−1 is linearly independent. Look at (A−λI)vi ⊆En−1λ is linearly independent. Complete this to a basis for Enλ/E

n−1λ .

Continue in this fashion until you obtain B = v1, . . . , vk a basis for Vwhere each vi ∈ Emλ for some m and vi ∈ B implies (A − λI)vi ∈ B ifnonzero. This gives the desired basis.

53

Definition 6.6 (Minimal Polynomial). Let I = f ∈ k[x] such that f(A)v = 0for all v ∈ V . Pick a basis v1, . . . , vn for V . I = ∩ni=1Θvi

, Θvi= f ∈ k[x] :

f · vi = 0 with this being the k[x]-module structure for V with ϕ a linearoperator.

Since each Θvi is nonzero and k[x] is an integral domain, I = ∩Θvi isnonzero, and I is an ideal in k[x]. As k[x] is a PID, I = 〈f〉 for some f ∈ I.

We call f the minimal polynomial.

Lemma 6.5. The minimal polynomial of the companion matrix of f is itself.

Proof. Recall, if A is a block of the rational canonical form, then e1, . . . , en is abasis for V . If g(x) =

∑ki=0 cix

i such that g(A) = 0, then g(A)e1 =∑ciA

iei =∑ciei+1 so if k + 1 ≤ n then g = 0. So degree of minimal polynomial of A is

≥ n. note that, by definition, Ane1 =∑ni=0 aiei+1, 0 = −

∑aiA

iei − Ane1 =(An+

∑aiA

i)e1 = 0 so f(x) = xn+∑n−1i=0 aix

i satisfies f(A)v = 0 for all v ∈ V ,so f(A) = 0.

Thus, the minimal polynomial divides f , but has at least the same degree,so it equals f up to multiplication by a constant.

Corollary 6.6. If A is in rational canonical form, then Ai is the companionmatrix of fi, f1| . . . |f` then f` is the minimal polynomial.

Definition 6.7 (Characteristic Polynomial). The characteristic polynomial ofA is det(A− xI).

det(CAC−1 − xI) = det(C(A− xI)C−1) = det(A− xI). So to compute thecharacteristic polynomial, we may take A to be in JCF .

Theorem 6.7 (Cayley-Hamilton). The minimal polynomial of A divides thecharacteristic polynomial of A.

7 Commutative Algebra

1. Rings of Quotients and localizations

Definition 7.1 (Multiplicative Set). A subset S of a ring R is multiplicativeiff 1 ∈ S and a, b ∈ S implies that ab ∈ S.

Definition 7.2 (Ring of Fractions). Let S be a multiplicative subset of the ringR. Then we define the equivalence relation ∼ on R × S by (r, s) ∼ (r′, s′) iffthere exists t ∈ S such that t(rs′ − r′s) = 0.

For each (r, s) ∈ R × S the corresponding ∼ class is denoted by rs . Let

S−1R = r/s : (r, s) ∈ R × S. It is easily checked that S−1R is a ring withoperations r/s+ r′/s′ = (rs′+ r′s)/ss′ and r/s∗ r′/s′ = rr′/ss′, zero is 0/1 andidentity is 1/1.

Furthermore, there exists a canonical homomorphism φS : R→ S−1R : r 7→r/1 and every element of φS [S] is a unit in S−1R.

Remarks:

54

1. If 0 in S then S−1R is the trivial ring.

2. If 0 /∈ S and S doesn’t contain any zero divisors, then ∼ reduces to theexpected relation.

3. ConsiderR = Z6 = 0, 1, 2, 3, 4, 5 and S = 0, 2, 4. Consider the relation≡ on R × S given by (r, s) ≡ (r′, s′) iff rs′ − r′s = 0. Then (0, 1) ≡ (0, 2)and (0, 2) ≡ (3, 1), but (0, 1) 6≡ (3, 1). It is easily checked that S−1R ' Z3

so φS isn’t injective.

Theorem 7.1. Let S be a multiplicative subset of the ring R. Suppose that0 /∈ S and S doesn’t contain any zero divisors. Then φS : R → S−1R isinjective.

Proof. Suppose that φS(r) = r/1 = 0/1. Then there exists s ∈ S such thats(r1− 0) = 0. That is, sr = 1. Since s isn’t a zero divisor, r = 0.

Examples

1. If R is an integral domain and S = R \ 0, then S−1R is a field, calledthe quotient field of R.

2. Suppose R = Z, and S = 3n : n ≥ 0. Then Z[1/3] = z/3n : z ∈ Z, n ≥0.

Convention: if R is an integral domain and S ⊆ R \ 0 is multiplicative,then we identify R with its canonical copy in S−1R.

Definition 7.3 (Extension and Retraction). Let S be a multiplicative subset ofthe ring R.

1. If I ⊆ R is an ideal, then the extension of I in S−1R is S−1I = a/s :a ∈ I, s ∈ S.

2. If J ⊆ S−1R is an ideal, then the contraction of J in R is φ−1S (J).

Remarks: Clearly, S−1I, φ−1S (J) are ideals.

Lemma 7.2. With the above hypotheses, S−1I = S−1R iff S ∩ I 6= ∅.

Proof. If s ∈ S ∩ I, then 11 = 1

ss1 ∈ S

−1I.Conversely suppose that S−1I = S−1R. Then there exist a ∈ I and s ∈ S

such that as = 1

1 . Hence, there exists t ∈ S such that ts− ta = t(s− a) = 0. Sots = ta ∈ S ∩ I.

Lemma 7.3. With the above hypotheses,

1. I ⊆ φ−1S (S−1I)

2. If I = φ−1S (J) for some ideal J ⊆ S−1R, then S−1I = J . Hence every

ideal of S−1R has the form S−1I for some ideal I ⊆ R.

55

3. If P ⊆ R is a prime ideal and S ∩ P = ∅, then S−1P is a prime ideal ofS−1R and φ−1

S (S−1P ) = P .

Proof. 1 is obvious.2: suppose that I = φ−1

S (J). To see that S−1I ⊆ J , let x ∈ S−1I. Thenthere exists r with φS(r) ∈ J and s ∈ S such that x = r/s. Hence x = 1

sr1 =

1sφS(r) ∈ J . Conversely suppose that r

s ∈ J . Then φS(r) = r/1 = s1rs ∈ J

hence r ∈ I and /s ∈ S−1I.3: By Lemma 1, S−1P is an ideal such that S−1P 6= S−1R. Suppose that

rsr′

s′ ∈ S−1P . Then there exists a ∈ P and t ∈ S such that rr′

ss′ = at . Hence,

there exists t′ ∈ S such that t′trr′ − t′ss′a = 0. Thus, tt′rr′ = t′ss′a ∈ P , sincett′ ∈ S and S ∩ P = ∅, we have rr′ ∈ P . Finally, by part 1, P ⊂ φ−1

S (S−1P ).Conversely, suppose that r ∈ φ−1

S (S−1P ). Then φS(r) ∈ S−1P and so thereexists a ∈ P and s ∈ S such that r

1 = as . Hence, there exists t ∈ S such that

tsr = ta ∈ P . Since ts /∈ P , r ∈ P .

Theorem 7.4. With the above hypotheses, we can define a bijectionP ⊆ R : P prime ideal,P∩S = ∅ → Q ⊆ S−1R : Q prime by P 7→ S−1P .

Proof. By the third part of Lemma 2, the map P 7→ S−1P is an injectionbetween these sets. To see it is a surjection, let J be a prime ideal of S−1R.Then S−1(φ−1

S (J)) = J , so it is enough to show that φ−1S (J) is a prime ideal.

Suppose that ab ∈ φ−1S (J). Then φS(a)φS(b) ∈ J and so φS(a) ∈ J or

φS(b) ∈ J . Thus, a ∈ φ−1S (J) or b ∈ φ−1

S (J).

Definition 7.4 (Localization at P ). Suppose that P is a prime ideal of R. ThenR \P is a multiplicative subset of R. The localization of R at P is defined to beRP = (R \ P )−1R.

If I ⊆ R is an ideal, then we write IP = (R \ P )−1I.

Example: Let R = Z and P = (3), then ZP = z/n : z ∈ Z, (n, 3) = 1.

Theorem 7.5. Suppose that P is a prime ideal of the ring R.

1. There exists a bijection between Q ⊆ R : Q prime and Q ⊆ P → J ⊆RP : J prime given by Q 7→ QP .

2. PP is the unique maximal ideal of RP .

Proof. 1 is just a special case of the previous theorem.For 2, we let J ⊆ RP be any maximal ideal. Then J is prime. Hence, there

exists a prime ideal Q ⊆ P such that QP = J . But then J = QP ⊆ PP soJ = PP .

Definition 7.5 (Local Ring). A ring R is local if R has a unique maximal ideal.

Theorem 7.6. If R is a ring, then TFAE

1. R is a local ring

56

2. The nonunits of R form an ideal.

Proof. 1 ⇒ 2: Let M be the unique maximal ideal of R. Clearly each elementof M is a nonunit. Conversely, suppose that r ∈ R is a nonunit. Then I = (r)is a proper ideal. Hence, I ⊆ M . So r ∈ M . Thus, M consists of the nonunitsof R.

2⇒ 1: Pretty trivial.

Theorem 7.7 (Nakayama’s Lemma). Let R be a local ring and let m be themaximal ideal of R. Suppose that E is a finitely generated R-module. If mE = Ethen E = 0.

Proof. Suppose that E is a counterexample with a minimal number of gener-ators, say, x1, . . . , xn. Since mE = E, there exist m1, . . . ,mn ∈ m such thatxn = m1x1 + . . .+mnxn, so then (1−mn)xn = m1x1 + . . .+mn−1xn−1.

Since 1 −mn /∈ m, it follows that 1 −mn is a unit. But then, x1, . . . , xn−1

generates E, contradicting minimality of n.

2. Integral Ring Extensions

Definition 7.6 (Integral over R). Let S be a ring extension of the ring R and letα ∈ S. Then α is integral over R iff there exists a monic polynomial f(x) ∈ R[x]such that f(α) = 0.

Example: Let R = Z and S = Qalg. Then α = −√−3

2 ∈ S is integral over Zsince α2 + α+ 1 = 0.

Definition 7.7 (Annihilator). Let R be a ring and let M be an R-module.Then, the annihilator of M , Ann(M) = r ∈ R : rm = 0,∀m ∈M.

The module is faithful iff Ann(M) = 0.

Theorem 7.8. If S is an extension ring of R and α ∈ S, then TFAE

1. α is integral over S.

2. The ring R[α] is finitely generated as an R-module

3. There exists a faithful R[α]-module M which is finitely generated as anR-module.

We shall make use of the following lemma:

Lemma 7.9. Let S be a ring and let A = (aij) be an n × n matrix over S.

Suppose that M is an S-module and that b1, . . . , bn ∈M satisfy A

b1...bn

= 0.

Then (detA)bi = 0 for 1 ≤ i ≤ n.

57

Proof. We will sketch a proof. Consider the case when n = 3. We just

check that detAb2 = 0. Consider the matrix B2 =

1 0 00 b2 00 0 1

. Then

det(AB2) = det(A) det(B2) = detAb2. Also, AB2 =

a11 a12b2 a13

a21 a22b2 a23

a31 a32b2 a33

hence det(AB2) =

a11 a11b1 + a12b2 + a13b3 a13

a21 a21b1 + a22b2 + a23b3 a23

a31 a31b1 + a32b2 + a33b3 a33

= 0

Now we will prove the theorem.

Proof. 1⇒ 2: Let g(x) ∈ R[x] be a monic polynomial such that g(α) = 0. Saythat deg g = n. If β ∈ R[α], there exists f(x) ∈ R[x] such that β = f(α).Singe g(x) is monic, there exists q(x), r(x) ∈ R[x] with degree < n such thatf(x) = q(x)g(x) + r(x). Thus β = f(α) = r(α). Hence 1, α, . . . , αn−1 generatesR[α] as an R-module.

2⇒ 3: Just take M = R[α].3⇒ 1: Let M be a faithful R[α]-module which is finitely generated as an R-

module, say by b1, . . . , bn. Since αM ⊆M , there exist aij ∈ R such that αbi =

aijbj . Hence, by the previous lemma, if d =

∣∣∣∣∣∣∣α− a11 −a12 . . . −a1n

......

......

α− an1 −an2 . . . −ann

∣∣∣∣∣∣∣then dbi = 0 for 1 ≤ i ≤ n. Since M is a faithful R[α]-module, it follows

that d = 0. Thus, α is a root of the polynomial det(xI − A) ∈ R[x]. And so αis integral over R.

Definition 7.8 (Integral Extension). Let S be a ring extension of R. Then Sis integral over R iff every α ∈ S is integral over R.

Corollary 7.10. If S is a ring extension of R and S is finitely generated as anR-module, then S is an integral extension of R.

Proof. Let α ∈ S. Then S is a faithful R[α]-module which is finitely generatedas an R-module. Hence, α is integral over R.

Corollary 7.11. If S is a ring extension of R and s1, . . . , sn ∈ S are integralover R, then R[s1, . . . , sn] is a finitely generated R-module and hence is anintegral extension of R.

Proof. We argue by induction that R[s1, . . . , si] is a finitely generated R-module.For the induction step, note that R[s1, . . . , si+1] = R[s1, . . . , si][si+1].Since si+1 is integral over R[s1, . . . , si], then R[s1, . . . , si+1] is finitely gener-

ated as an R[s1, . . . , si]-module. Since R[s1, . . . , si] is finitely generates as an R-module, it follows that R[s1, . . . , si+1] is finitely generates as an R-module.

58

Corollary 7.12. Let S be a ring extension of R and and let R = α ∈ S : α isintegral over R. Then R is a subring of S called the integral closure of R in S.

Proof. Let α, β ∈ R. Then R[α, β] is an integral extension of R. Hence α −β, αβ ∈ R.

Definition 7.9 (Integrally Closed). 1. With the above hypotheses, R is in-tegrally closed in S iff R = R.

2. An integral domain R is integrally closed iff R is integrally closed in itsquotient field K.

Proposition 7.13. If R is a UFD, then R is integrally closed.

Proof. Suppose that a/b is integral over R, where a, b ∈ R and there existsa prime p such that p|b and p 6 |a. There exist a0, . . . , an−1 ∈ R such that(a/b)n + an−1(a/b)n−1 + . . . + a0 = 0, hence an + anba

n−1 + . . . + bna0 = 0.Since p|b, it follows that p|an and so p|a, contradiction.

Example: Z[i] is the integral closure of Z in Q(i).

Proof. Clearly Z[i] ⊆ Z since i is integral over Z. Since Z[i] is a UFD, it isintegrally closed in Q(i). The result follows.

Definition 7.10 (Number Field). A number field L is a finite extension of Q.If L is a number field, then the ring of algebraic integers of L is the integralclosure of Z in L.

Example: Let ω = −1+√−3

2 . Then Z[ω] is the algebraic closure of Z inQ(√−3).

Proof. Just like before, using the fact that Z[ω] is a UFD.

Definition 7.11 (Lies Above). Let S be an extension ring of R and let I be anideal of R. Then the ideal J of S lies above I iff J ∩R = I.

Nonexample: Clearly no ideal of Q lies above the ideal (2) of Z.

Theorem 7.14 (Lying Over Theorem). Let S be an integral extension of Rand let P be a prime ideal of R. Then there exists a prime ideal Q of S whichlies over P .

Proof DelayedZ ⊂ Z[i], then (2) has (1 + i) and (1− i) lying over it.Z ⊂ Z[ω], then (2) ⊂ Z[ω](2).

Proposition 7.15. Let S be an integeral extension of R and let σ : S → A bea ring homomorphism. Then σ[S] is an integral extension of σ[R].

Proof. Let α ∈ S. Then there exists an−1, . . . , a0 ∈ R such that αn+an−1αn−1+

. . . + a0 = 0. Applying σ we obtain σ(α)n + σ(an−1)σαn−1 + . . . + σ(a0) = 0,thus, σ(α) is integral over σ[R].

59

Application: Let R be an integral domain of char 0 with quotient field Kand let E be a finite algebraic extension of K. If α ∈ E in integral over R, thenNEK(α), trEK(α) are also integral over R. In particular, if R is integrally closed,

then NEK(α), trEK(α) ∈ R.

Proof. If σ ∈ eK(E,Kalg), then σ(α) is integral over R. Hence, so are NEK(α) =∏

σ∈eK(E,Kalg) σ(α) and trEK(α) =∑σ∈eK(E,Kalg) σ(α)

Definition 7.12 (Integral Homomorphism). A ring homomorphism f : R→ Sis integral iff S is an integral extension of f [R].

Definition 7.13. Suppose that f : R→ S is a ring homomorphism and T is amultiplicative subset of R. Then, slightly abusing notation by writing T−1S in-stead of f [T ]−1S, we can define a ring homomorphism T−1f : T−1R→ T−1S :r/t 7→ f(r)/f(t).

Furthermore, the following diagram commutes:

R T−1R

S T−1S

........

........

........

........

........

........

........

........

........

........

........

........

.................

............

f

.................................................................................................... ............

.................................................................................................... ............

........

........

........

........

........

........

........

........

........

........

........

........

.................

............

T−1f

Proposition 7.16. Let f : R → S be integral and let T be a multiplicativesubset of R. Then T−1f : T−1R→ T−1S is also integral.

Proof. To simplify notation, we just consider the case where f and T−1f areinclusions.

Let s/t ∈ T−1S. Then there exist a0, . . . , an−1 ∈ R such that sn+an−1sn−1+

. . .+a0 = 0. Gence (s/t)n+an−1/t(s/t)n−1+. . . , a0/tn = 0, hence s/t is integral

over T−1R.

Proposition 7.17. Suppose that T is an integrally closed integral domain andthat T is a multiplicative subset of R such that 0 /∈ T . Then T−1R is alsointegrally closed.

Proof. Let K be the quotient field of R and hence also of T−1R. Supposethat α ∈ K is integral over T−1R. Then there exists a0, . . . , an−1 ∈ R andt0, . . . , tn−1 ∈ T such that αn + an−1/tn−1α

n−1 + . . . + a0/t0 = 0. Let t =t0 . . . tn−1 ∈ T . Then it is clear that tα is integral over R and so tα ∈ R. Butthis means that α ∈ T−1R.

We will now prove the Lying Over Theorem:

Proof. Let S be an integral extension of R and let P be a prime ideal of R.Let T = R \ P ad consider teh ring of quotients T−1R and T−1S. To simplifynotation, we shall suppose that each of the following maps is an influsion.

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R T−1R = RP

S T−1S

........

........

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f

................................................................. ............

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............

T−1f

(If collapsing occurs, it is easily checked that the following arguments remainvalud using the canonical homomorphisms)

By Proposition 3, T−1S is an integral extension of T−1R, and by Theorem3, T−1R is a local ring with maximal ideal mP = T−1P . Also, by Lemma 2.3,mP lies over P .

Claim: mPT−1S 6= T−1S.

Suppose that mPT−1S = T−1S. Then there exist mi ∈ mP and b−1

i ∈ T−1Ssuch that 1 = m1b1 + . . . + mnbn. Let B = RP [b1, . . . , bn]. Since T−1S is anintegral extension of RP , it follows that B is a finitely generated RP -module.By the equation, for each 1 ≤ i ≤ n, we have mPB = B. By Nakayama’sLemma, B = 0, which is a contradiction.

It follows that mPT−1S is contained in a maximal ideal n of T−1S. Since

mP ⊆ n ∩ T−1R, and mP is a maximal ideal of T−1R, it follows that mP =n ∩ T−1R.

Thus:

R T−1R

S T−1S

........

........

........

........

........

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f

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T−1f

P MP

N

.................................................................................................................................................................................................................................................................... ............

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Let Q = n ∩ S. Clearly Q is a prime ideal of S. Finally note that P =mP ∩R = (n ∩ T−1R) ∩R = (n ∩ S) ∩R = Q ∩R. Thus, Q lies above P .

Proposition 7.18. Let S be an integral extension of R and let P be a primeideal of R. Suppose that that Q is a prime ideal of S which lies above P . ThenQ is a maximal ideal iff P is a maximal ideal.

Proof. Suppose that P is a maximal ideal. Then R/P is a field and S/Q is anintegral domain, which is integral over R/P by Prop 2. Since R/S ⊆ S/Q ⊆(R/P )alg, it follows that S/Q is a field. Thus, Q is a maximal ideal. Nextsuppose that Q is a maximal ideal. Then S/Q is a field which is integral overthe integral domain R/P . Suppose that R/P isn’t a field. Then R/P has amaximal ideal m such that 0 6= m 6= R/P . By the Lying Over Theorem, thereexists a prime ideal of S/Q which lies above m, which is impossible.

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3. Integral Galois ExtensionsStanding Hypotheses: Let R be an integral domain of characteristic 0 which

is integrally closed in its quotient field K. Let E be a finite Galois extension of Kand S be the integral closure of R in E. Let G = AutK E be the correspondingGalois Group.

Remark: Clearly, if σ ∈ G, then σ[S] = S.

Proposition 7.19. Suppose that P is a maximal ideal of R and that P andR are prime ideals of S that lie above P . Then there exists σ ∈ G such thatσ[P] = R.

Proof. Suppose that σ[P] 6= R for all σ ∈ G. Then σ[P] 6= τ [R] for allσ, τ ∈ G. By Proposition 5, each σ[P] and τ [R] is a maximal ideal of S.Hence, by the Chinese Remainder Theorem, there exists x ∈ S such that x ≡ 0mod σ[P] for all σ ∈ G and ≡ 1 mod τ [R] for all τ ∈ G.

Since R is integrally closed in K, it follows that NEK(x) =

∏σ∈G σ(x) ∈

S ∩K = R and so NEK(x) ∈P ∩R = P .

However, since x /∈ τ [R] for all τ ∈ G, it follows that τ(x) /∈ R for allτ ∈ G. Since R is a prime ideal, it follows that NE

K(x) =∏τ∈G τ(x) /∈ R and

so NEK(x) /∈ P , contradiction.

Definition 7.14 (Decomposition Group). Fix a maximal ideal P of R and letP be a maximal ideal of S which lies above P , we define the decompositiongroup of P to be GP = σ ∈ G : σ[P] = P

Then, regarding R/P as a subfield of S/P, each σ ∈ GP induces an auto-morphism σ of S/P which fixes R/P pointwise. Thus we obtain a homomor-phism G→ AutR/P (S/P) : σ 7→ σ.

Definition 7.15 (Decomposition Field). The decomposition field of P is Edec =EGP , the fixed field of GP .

Let Sdec be the integral closure of R in Edec. Let D = P ∩Sdec. Then D isa prime ideal of Sdec which lies above P . Hence D is a maximal ideal of Sdec.

Furthermore, by the previous proposition, P is the unique prime ideal of Swhich lies above D .

Proposition 7.20. Edec is the smallest subfield F of E such that K ⊆ F ⊆ Eand P is the unique prime ideal of S which lies above the prime ideal ∩ ofS ∩ F .

Proof. We’ve just seen that Edec is such a subfield. To see that is it hte smallestsuch subfield, let F be any such subfield and let H = AutF E. Let Q = P ∩F .

Since P is the unique prime ideal of S which lies above Q, it follows thatH ≤ GP . Hence F = EH ⊇ Edec = EGP .

Example: Suppose R = Z, E = Q(i). Let P = (2).GThen (1 + i), (1 − i) are the ideals of S = Z[i] which lie above P . Let

P = (1 + i). Then GP = 1 and so Edec = E1 = Q(i).

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Let R = Z and E = Q(ω), where ω2 + ω + 1 = 0. Again, let Z = (2). ThenP = 2Z[ω] is the unique prime ideal of S = Z[ω] which lies above P . ThenGP = G and Edec = EG = Q.

Proposition 7.21. Under the canonical injection, R/P → Sdec/D , we haveR/P = Sdec/D .

Proof. Let σ ∈ G \GP . Then clearly σ−1P 6= P.Define Dσ = σ−1P ∩ Sdec.Since P is the unique prime ideal of S lying above D, it follwos that Dσ 6= D.Clearly Dσ is a prime ideal of Sdec. Since Dσ lies above the maximal ideal

P of R, then by Proposition 5, we know that Dσ is a maximal ideal of Sdec.Now fix some x ∈ Sdec. We seek an element α ∈ R such that x ≡ α mod D.By the Chinese Remainder Theorem, there exists y ∈ Sdec such that y ≡ x

mod D, y ≡ 1 mod Dσ for all σ ∈ G \GP .In particular, we have that y ≡ 1 mod σ−1P for all σ ∈ G \GP and hence,

σ(y) ≡ 1 mod P for all σ ∈ G \GP .In summary, we have that y ≡ x mod P and σ(y) ≡ 1 mod P for all

σ ∈ G \G√.

Now consider α = NEdec

K (y) ∈ R, since R is integrally closed in K. LeteK(Edec,Kalg) = τ1, . . . , τm so that NEdec

K (y) = τ1(y) . . . τm(y).Then for each 1 ≤ i ≤ m, there exists σi ∈ G such that τi = σi|Edec . Notice

that if σ ∈ GP , then σ|Edec = id.Hence, we can suppose that σ1 = 1 and σi ∈ G \GP for 2 ≤ i ≤ m.It follows that α = NEdec

K (y) ≡ x mod P. Since α, x ∈ Sdec, it follows thatα ≡ x mod D, as required.

Let S = S/P and R = R/P . For each α ∈ S, let α ∈ S be the correspondingelement. Then, for each σ ∈ GP , the corresponding σ ∈ AutR S satisfies σ(α) =σ(α).

Finally, for each f ∈ S[x], let f ∈ S[x] be the corresponding polynomial.More Hypotheses: From now on, we also suppose that R is a finite field.

(This isn’t really essential, but it allows us to ignore separability issues.)

Theorem 7.22. 1. S is a finite Galois extension of R with [S : R] ≤ [E : K]

2. The homomorphism GP → AutR S : σ 7→ σ is surjective.

Proof. 1. It is enough to show that [S : R] ≤ [E : K]. Let α ∈ S and considera corresponding α ∈ S. Let f(x) = Irr(α,K, x). Then deg f ≤ [E : K].Since the roots of f are integral over R, it follows that the coefficients off are also integral over R.

Since R is integrally closed in K, it follows that f(X) ∈ R[x]. Clearlyf(α) = 0, thus f ∈ R[x] is a polynomial of degree ≤ [E : K] such thatf(α) = 0. Hence [S : R] ≤ [E : K].

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2. Since GP = AutEdec E and Sdec/D = R/P , we can suppose that K = Edec

and G = GP . Let α ∈ S satisfy S = R(α). Let f(x) = Irr(α,K, x). Thenwe have already seen that f(x) ∈ R[x] so we can form f(x) ∈ R[x]. Nownotice that any τ ∈ AutR S is uniquely determined by its effect on α andthat α must also be a root of f(x). Let β = τ(α).

Since the roots of f(x) are integral over R, we see that f(x) =∏mi=1(x−

αi), α1 = α, splits into linear factos in S[x]. Applying the canonicalhomomorphism, we obtain that f(x) =

∏mi=1(x−αi) has a corresponding

splitting in S[x]. In particular, there exists 1 ≤ i ≤ m such that αi = β.Let σ ∈ G = GP satisfy σ(α) = αi. Then σ(α) = αi = β and so σ = τ .

Theorem 7.23. With the above hypotheses, suppose further that there exists amonic irreducible f(x) ∈ R[x] such that E is the splitting field of f over K andf(x) has no multiple roots in R

alg. Then:

1. The homomorphism GP → AutR S is an isomorphism.

2. S is the splitting field of f over R.

Proof. 1. Let α1, . . . , αn ⊆ S be the roots of f(x) and let α1, . . . , αn ⊆ Sbe their reductions mod P. For any σ ∈ GP , we have σ(αI) = σ(αi)1 ≤ i ≤ n. hence, if σ = 1, then σ = 1.

2. Clearly S contains the splitting field E = R[α1, . . . , αn] of f over R.Because GP → AutR S surjectively, we see that AutE S = 1, and soE = S.

Warning: Consider the irreducible monic f(x) = x2 + 3 ∈ Z[x]. Reducingmod 2, we have that f(x) = x2 + 1 = (x+ 1)2 ∈ F2[x], so the splitting field of fover F2 is F2. Let P = (2) and let S be the integral closure of Z in the splittingfield E = Q(

√−3). Recall that S = Z[ω] where ω = −1+

√−3

2 . Also, P = 2S isthe unique prime ideal of S lying above P , thus S = S/P = F4.

Extended ExampleConsider the irreducible polynomial f(x) = x4 + 5x2 + 10x− 15 ∈ Z[x]. Let

E be the splitting field and G e the Galois group. It is easily checked that f(x)has exactly two real roots. It follows that G 6≤ A4.

Reducing modulo 3, we obtain the following decomposition into irreduciblesx4 + 2x2 + x = x(x3 + 3x+ 1) ∈ F3[x]. Thus, G contains a 3 cycle. Since G is atransitive subgroup of S4, it contains every 3 cycle, and so A4 ≤ G, so G = S4.

Let S be the ring of algebraic integers in E. Consider the ideal P = (3) ofZ and let P be a maximal ideal of S which lies above P . Since S = S/P ' F33 .

We have that GP ' C3. Hence [G : GP ] = 8. Thus, there are exactly 8maximal ideals of S which lie over P and [Edec : Q] = 8.

Question: How many maximal ideals of Sdec lie above P?

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Answer: Suppose that D′ is any maximal ideal of Sdec which lies above P .By the Lying Over Theorem, there exists a maximal ideal R of S which liesover D′, and clearly R lies over P .

Thus, we must compute the size of Sdec ∩ R : R is a maximal idela of Slying over P.

Let R be any such ideal of S. Then either GR = GP or GR ∩GP = 1.Let R = σP for some σ ∈ G. Then GR = GP iff σ ∈ NG(GP). Hence the

number of such ideals R is [NG(GP), GP ] = [S3 : A3] = 2, so there are two suchideals, P and P ′.

Then D = P ∩ Sdec 6= P ′ ∩ Sdec = D′, since P is the unique maximal idealof S which lies above D = P ∩ Sdec.

Now suppose that R satisfies GR ∩GP = 1, then σR : σ ∈ GP consists ofthree ideals, each lying over R ∩ Sdec. Hence the number of maximal ideals ofSdec lying above P is 2 + (8− 2)/3 = 4.

If we reduce modulo 2 instead, we obtain x4 +x2 +1 = (x2 +x+1)2 ∈ F2[x],so it is unclear what is S/P where P lies above (2). We also don’t know whatGP is.

4. Notherian Rings and Modules

Definition 7.16 (Notherian Module). Let R be a ring and let M be an R-module. Then M is Noetherian iff M satisfies the ascending chain condition(ACC) on submodules. IE, if M1 ⊆M2 ⊆ . . . is an increasing chain of submod-ules, then there is an n such that M` = Mn for all ` ≥ n.

Definition 7.17 (Notherian Ring). The ring R is Notherian iff R is a NotherianR-module.

Theorem 7.24. If R is a ring and M is an R-module, then the following areequivalent:

1. M is Notherian.

2. M satisfies the maximum condition for submodules: ie, whenever S is anonempty set of submodules of M , then S contains a maximal elementunder inclusion.

3. Every submodule of M is finitely generated.

Proof. (1)⇒ (2): Let S be a nonempty collection of submodules of M . SupposeS doesn’t contain a maximal element. Then we can inductively choose Mn ∈ Ssuch that M0 ( M1 ( . . ., contradiction.

(2) ⇒ (3): Let N be a submodule of M . Let S = A : A is a finitelygenerated submodule of N. Then S has a maximal element A. We claim thatA = N . Otherwise, there exists b ∈ N \A and so A ( A+Rb ∈ S, contradiction.

(3)⇒ (1) suppose that M0 ⊆M1 ⊆ . . . is a chain of submodules of M . ThenN = ∪nMn is a submodule of M and so is finitely generated. Then N is finitelygenerated by, say, a1, . . . , at. There exists n such that a1, . . . , at ∈Mn. ClearlyM` = Mn for all ` ≥ n.

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Corollary 7.25. A ring R is Notherian iff every ideal is finitely generated.

Example: In particular, every PID is Noetherian.Question: Suppose that E is a field such that [E : Q] < ∞. Let R be the

ring of algebraic integers in E. Is R Notherian?

Proposition 7.26. 1. If M is a Notherian R-mdoule, then every submodule ofM and every quotient of M is also Noetherian.

2. Suppose that M is an R-module and N is a submodule. If N and M/Nare Notherian, then M is also Notherian.

Proof. 1. Obvious.2. Claim: Suppose that E ⊆ F are submodules of M . If E ∩ N = F ∩ N

and (E +N)/N = (F +N)/N then E = F .To prove the claim, we suppose that x ∈ F . Then there exists y ∈ E and

u, v ∈ N such that x+ u = y+ v. Hence x− y = v− u ∈ F ∩N = E ∩N . Thusx = (x− y) + y ∈ E.

Now, we let M0 ⊆ M1 ⊆ . . . be an increasing chain of submodules of M .Consider teh associated chains M0∩N ⊆ . . . and (M0 +N)/N ⊆ . . . of submod-ules of N,M/N respectively. Since N and M/N are Notherian, there exists nsuch that for all ` ≥ n, M` ∩ N = Mn ∩ N and (M` + N)/N = (Mn + N)/N .Hence M` = Mn for all ` ≥ n.

Corollary 7.27. If M1, . . . ,Mn are Notherian R-modules, then M1⊕ . . .⊕Mn

is also Notherian.Suppose that M is an R-module, and M1, . . . ,Mn are R-modules such that

M = M1 + . . .+Mn. If M1, . . . ,Mn are Notherian, then M is Notherian.

Proof. Arguing by induction, we can suppose that n = 2. Then M1 and (M1 ⊕M2)/M1 'M2 are Notherian, and hence M1 ⊕M2 is.

We can define a surjective homomorphism from M1⊕ . . .⊕Mn to M1 + . . .+Mn = M by (m1, . . . ,mn) 7→ m1 + . . .+mn. The result follows.

Proposition 7.28. If R is a Notherian ring and M is a finitely generatedR-module, then M is Notherian,

Proof. Let x1, . . . , xn generate M . Then define a surjective homomorphismR ⊕ . . . ⊕ R → M by (r1, . . . , rn) 7→ r1x1 + . . . + rnxn. Since R ⊕ . . . ⊕ R isNotherian, the result follows.

Theorem 7.29. Let K be a field such that [K : Q] < ∞ and let R be the ringof algebraic integers in K. Then R is Notherian.

Proof. Note that it is enough to show that R is finitely generated as a Z-module.For then, by Prop 10, R is a Notherian Z-module and hence is also a NotherianR-module.

Choose γ ∈ K such that K = Q(γ). Let Irr(γ,Q, x) = xn + qn−1xn−1 +

. . . + q0, the qi ∈ Q. Clearing denominators, a0, . . . , an−1 ∈ Z such that tγn +an−1γ

n−1 + . . .+ a0 = 0. Thus, α = tγ ∈ R and K = Q(α).

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Warning: It doesn’t follwo that R = Z[α]. For example, let K = Q(√−3).

Then Z[√−3] ( R.

Note that [K : Q] = [Q(α) : Q] = n. Let α = α1 . . . , αn be the distinctconjugates of α in the normal closure E of K. Let ∆ =

∏i<j(αi − αj). Then

d = ∆2 ∈ Z.Claim: R ⊂ d−1Z[α].Before proving the claim, we complete the proof of ht ethorem. Note that

Z[α] is a finitely generated free Z-module, and hence, so is d−1Z[α]. Since R isa Z-submodule, it follows that R is also a finitely generated free Z-module.

We will now prove the claim: Let β ∈ R. Then there exist b0, . . . , bn−1 ∈ Qsuch that β = b0 + b1α + . . . + bn−1α

n−1. It follows that the (not necessarilydistinct) conjugates of β in E are βj = b0 +b1αj + . . .+bn−1α

n−1j for 1 ≤ j ≤ n.

Note that

∣∣∣∣∣∣∣1 α1 . . . αn−1

1...1 αn . . . αn−1

n

∣∣∣∣∣∣∣ =∏i<j(αi − αj) 6= 0, hence we can use

Cramer’s Rule to solve the matrix equation

1 α1 . . . αn−11

...1 αn . . . αn−1

n

b0

...bn−1

=

β1

...βn

.

We obtain bi = 1∆ (γi1β1 + . . . γinγn) 0 ≤ i ≤ n − 1 where each γij is a

polynomial with integer coefficients in αsr : 1 ≤ r ≤ n, 0 ≤ s ≤ n − 1. Inparticular, γij is an algebraic integer of E. Since each βj is an algebraic integer,we decude that dbi = ∆(γi1β1 + . . .+ γinβn) is also an algebraic integer. Sincedbi ∈ Q, if follows that dbi ∈ Z. Thus β = b0 + . . .+ bn−1α

n−1 ∈ d−1Z[α].

Theorem 7.30 (Hilbert Basis Theorem). If R is a Notherian ring, then thepolynomial ring R[x] is also Notherian.

Corollary 7.31. If R is Notherian, then R[x1, . . . , xn] is Notherian.

Corollary 7.32. Let R be a Notherian ring and let S = R[a1, . . . , an] be anextension ring which is finitely generated as a ring over R. Then S is alsoNotherian.

Proof. We can define a surjective homomorphism f : R[x1, . . . , xn]→ R[a1, . . . , an].By the homework, the result follows.

We will now prove the Hilbert Basis Theorem:

Proof. Let I be an ideal of R[x]. For each n ≥ 0, let Jn be the set of all leadingcoefficients an of polynomails of degree n such that

(*) anxn + . . .+ a0 ∈ I.Clearly, Jn is an ideal of R. Furthermore, if (∗) holds, then anx

n+1 + . . .+a0x = x(anxn + . . . + a0) ∈ I, and so J0 ⊆ J1 ⊆ . . .. As R is Notherian, then

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there exists t such that J` = Jt for all ` ≥ t. For each 0 ≤ ` ≤ t, let a`1, . . . , a`s`

be generators of J` and let f`i ∈ I be a polynomial of degree ` with leadingcoefficients a`i. We claim that X = f`i : 0 ≤ ` ≤ t, 1 ≤ i ≤ s` generates I.We shall prove by induction of d = deg f that if f ∈ I then f ∈ (X). First,suppose that d = 0 and that f = r ∈ I ∩R = J0.

Then clearly, r ∈ (a01, . . . , a0s0) ⊆ (X).Now suppose that d > 0 and that the result holds for all 0 ≤ k < d. Let

f ∈ I be a polynomial of degree d with leading coefficient r.First suppose that d ≤ gt. Then r ∈ Jd and so r = r1ad1 + . . . rsd

adsdfor

some ri ∈ R. Thus the polynomial r1fd1 + . . . rsdfdsd

∈ (X) ⊆ is a polynomialof degree d with leading coefficient r. Hence f −

∑sd

i=1 rifdi ∈ I has degreeat most d − 1 and so by induction hypothesis lies in (X). It fllows that f =(f −

∑sd

i=1 rifdi) +

∑sd

i=1 rifdi∈ (X).

Finally, suppose that d > t. Then r ∈ Jd = Jt and so r = r1at1 + . . .+rstatst

sor some ri ∈ R. Thus the polynomial r1xd−tft1 + . . . + rst

ftst∈ (X) ⊆ I has

leading coefficient r and degree d. Arguing as in the previous case, we seeinductively that f ∈ (X).

An application...

Definition 7.18 (Affine n-space). Let k be any field and n ≥ 1. kn is calledaffine n-space over K.

Let k[~x] = k[x1, . . . , xn].

Definition 7.19 (Algebraic Set). A subset V ⊂ kn is algebraic iff there is asubset S = pi(~x) : i ∈ I ⊆ k[~x] such that V = V (S) = ~a ∈ kn : pi(~a) = 0 forall i ∈ I.

Example: We can regard SL2(k) as an algebraic subset of k4 defined byidentifying it with (a, b, c, d) : ad − bc = 1, which is defined by the singlepolynomial p(~x) = x1x4 − x2x3 − 1.

Suppose that V = V (S) is an algebraic set and that I is the ideal of k[~x]generated by S. Clearly, if f(~x) ∈ I then f(~a) = 0 for all ~a ∈ V = V (S),and so V (S) ⊆ V (I). Also since I ⊇ S, it is clear that V (I) ⊆ V (S). ThusV (S) = V (I).

Thus we have a “natural” surjective map from ideals of k[~x] to algebraicsubsets of kn taking I to V (I). The map is NOT injective, however, as V (x)and V (x2) are both 0.

We can also define a “natural” map from the algebraic subsets of kn to idealsof k[~x] by V 7→ J(V ) where J(V ) is the ideal of all polynomials vanishing on V .

Definition 7.20 (Radical of an Ideal). If I is an ideal of the ring R, then theradical of I is the ideal Rad I = r ∈ R : rn ∈ I for some n ≥ 1.

The ideal I is a radical ideal iff I = Rad I.

Example: If V ⊂ kn is an algebraic set, then J(V ) is a radical ideal.Hence, we actually “natural” map from algebraic subsets of kn to radical

ideals of k[~x] by V 7→ J(V ).

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This map must be injective, because V = V (J(V )). By definition, V ⊆V (J(V )). Let I be an ideal such that V = V (I). Then I ⊆ J(V ) and soV = V (I) ⊇ V (J(V )), hence V = V (J(V )).

Remark: If k is an arbitrary field, then the map need not be surjective.(x2+1) ⊂ R[x]. Howeer, the correspondence is a bijection when k is algebraicallyclosed.

This is the content of Hilbert’s Nullstellensatz.

Theorem 7.33. Let k be any field and n ≥ 1.

1. If V ⊆ kn is an algebraic set, then there exists a finite subset of polynomialsS ⊂ k[x1, . . . , xn] such that V = V (S).

2. If V1 ⊇ V2 ⊇ . . . ⊇ Vn ⊇ . . . is a descending sequence of algebraic subsetsof kn, then there exists t such that V` = Vt for all ` ≥ t.

Proof. 1. Let I be an ideal such that V = V (I) since k[x1, . . . , xn] is Notherian,I is finitely generated, say, by S. Clearly V = V (S).

2. Consider J(V1) ⊆ . . .. Then there is a t such that J(Vt) = J(V`) for all` ≥ t. Hence Vt = V (J(Vt)) = V (J(V`)) = V` for all ` ≥ t.

Remark: The following variant of the first part of the theorem is often useful:If S is a subset of k[x1, . . . , xn], then there exists a finite subset S0 such thatV (S) = V (S0).

Definition 7.21 (Variety of Representations). Let G be a finitely generatedgroup with fixed generating set T = t1, . . . , tr.

Let R = W (y1, . . . , yr) : W is a word such that W (t1, . . . , tr) = 1 be thecorresponding set of relations.

Then, the corresponding variety of representations of (G,T ) in SL2(C) isthe algebraic subset of C4r defined by VG,T = (M1, . . . ,Mr) ∈ SL2(C)r :W (M1, . . . ,Mr) = 1 for all W ∈ R

Remark: To see that this is an algebraic set, we must check that the conditionW (M1, . . . ,Mr) = 1 corresponds to a set of polynomial equations.

Let Rep(G,SL2(C)) be the set of all group homomorphisms π : G →SL2(C). Then we can define a bijection Rep(G,SL2(C)) → VG,T by π 7→(π(t1), . . . , π(tr)).

Of course R is infinite, and since most groups are not finitely presented, weapparently need infintiely many words to define VG,T , except...by the HilbertBasis Theorem

Theorem 7.34. With the above hypotheses, there exists a finite R0 ⊆ R suchthat VG,T = (M1, . . . ,Mr) ∈ SL2(C)r : W (M1, . . . ,Mr) = 1 for w ∈ R0.

This is not a contradiction, though the least it can tell us is that most finitelygenerated groups do not have faithful linear representations.

69

Definition 7.22 (Residually Free). A group G is residually free iff for all 1 6=g ∈ G, there exists a homomorphism π : G → F into a free group F such thatπ(g) 6= 1.

Some examples/remarks:Clearly, every residually free group is torsion-free, and every free group is

residually free. If G1, . . . , Gn are residually free, then G1 ⊕ G2 ⊕ . . . ⊕ Gn isresidually free.

Proof. Let 1 6= g = (g1, . . . , gn) ∈ G1 × . . . × Gn, then there exists i such thatgi 6= 1, hence there is a homomorphism G1 × . . . × Gn → Gi → F such that(π pi)(g) = π(gi) 6= 1.

Theorem 7.35. Suppose that G1φ1→ G2

φ2→ . . . → Gn → . . . is a seqeunce ofsurjective homomorphisms between finitely generated residually free groups, thenthere exists t such that φe is an isomorphism for all ` ≥ t.

Basis Facts about Free Groups:

1. The matrices(

1 20 1

),

(1 02 1

)free generate a free group on two gen-

erators.

2. The commutator subgroup of the free group F2 on 2 generators is a freegroup on infinitely many generators. Hence, if G is a finitely generatedresidually free group and 1 6= g ∈ G, then there exists π : G → F2 suchthat π(g) 6= 1.

π : G→ F2 =⟨(

1 20 1

),

(1 02 1

)⟩≤ SL2(C).

Remark: There exists a finitely generated residually free group which isn’tfinitely presented.

Proof. Recall that F2 × F2 is residually free. Hence if G ≤ F2 × F2, then G isalso residually free.

By Google, there exists a finitely generated subgrtoup G ≤ F2×F2 which isnot finitely presented.

We will now prove Theorem 15.

Proof. Let T1 = t1, . . . , tr be a generating set for G1; and for n ≥ 1, letTn = t(n)

1 , . . . , t(n)r be the image of T1 under the surjective homomorphism

G1 → . . . → Gn. Then Tn is a set of generators for Gn. For each n ≥ 1, letVn = VGn,Tn

⊆ (SL2(C)r) ⊆ C4r.Since φn[Tn] = Tn+1, for each word w if w(t(n)

1 , . . . , t(n)r ) = 1, then w(t(n+1)

1 , . . . , t(n+1)r ) =

1.Thus, the corresponding sets of relations satisfy Rn ⊆ Rn+1; and so the

algebraic sets satisfy Vn ⊇ Vn+1. Hence V1 ⊇ . . . Vn ⊇ . . .. By Theorem 13,

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there exists n such that V` = Vn for all ` ≥ n. hence it is enough to prove thefollowing:

If φk : Gk → Gk+1 isn’t an ismorphism, then Vk ) Vk+1.Since φk : G → Gk+1 is not an ismorphism, there exists a word such that

g = w(t(k)1 , . . . , t

(k)r ) 6= 1 but φk(g) = w(t(k+1)

1 , . . . , t(k+1)r ) = 1. In particular,

w ∈ Rk+1. Since Gk is residually free, there exists a homomrophism π : Gk →F2 such that π(g) 6= 1. We can suppose that F2 ≤ SL2(Z) ≤ SL2(C). Thus(π(t(k)

1 ), . . . , π(t(k)r )) ∈ Vk. However, since w(π(t(k)

1 ), . . . , π(t(k)r )) = π(w(t(k)

1 , . . . , t(k)r )) =

π(g) 6= 1 and w ∈ Rk+1, we see that (π(t(k)1 ), . . . , π(t(k)

r )) /∈ Vk+1. And so theclaim is proved.

5. Transcendence Bases

Definition 7.23 (Algebraically Dependent). Let F be an extension field of Kand let S ⊆ F . Then S is algebraically dependent over k iff there exist distincts1, . . . , sn ∈ S and 0 6= p(~x) ∈ k[x1, . . . , xn] such that p(s1, . . . , sn) = 0.

Otherwise, S is algebraically independent.

Examples:

1. √

2 is algebraically dependent over Q.

2. π is algebraically independent over Q.

3. π,√π are algebraically dependent over Q. To see this, let p(x1, x2) =

x1 − x22 ∈ Q[x1, x2]. Then p(π,

√π) = 0.

4. Let F = k(y1, . . . , yn) be the field of rational functions in the variablesy1, . . . , yn. Then y1, . . . , yn is algebraically independent over k.

Definition 7.24 (Transcendence Basis). A subset S ⊆ F is a transcendencebasis of F over K iff S is a maximal independent subset.

Remark: Transcendence Bases always exist. Why? Because if Si : i ∈ I isan increasing chain of independent sets, then ∪iSi is also independent, becauseof the finitary nature of dependence. Now apply Zorn’s Lemma.

Example: Let F = k(y1, . . . , yn) be a field of rational functions. Theny1, . . . , yn is a transcendence basis.

Theorem 7.36. Let F be an extension of K and S ⊆ F be algebraically inde-pendent over K. Then if u ∈ F \K(S), then the following are equivalent:

1. S ∪ u is algebraically independent.

2. u is transcendental over K(S).

Proof. 2 ⇒ 1: Assume u is transcendental over K(S). SUppose there existdistinct s1, . . . , sn−1 ∈ S and f(~x) ∈ k[x1, . . . , xn] such that f(s1, . . . , sn−1, u) =0. Then u is a root of the polynomial f(s1, . . . , sn−1, xn) ∈ K(S)[xn].

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Express f = hrxrn + hr−1x

r−1n + . . . + h0 where hi ∈ k[x1, . . . , xn−1]. Since

u is transcendental over K(S), we have that hi(s1, . . . , sn−1) = 0 for 0 ≤ i ≤ r.Since S is algebraically independent over K, it follows that hi = 0 for 0 ≤ i ≤ r.Thus f = 0. hence S ∪ u is algebraically independent over K.

1 ⇒ 2: Assume that S ∪ u is algebraically independent over K. Supposef(x) =

∑ni=0 aix

i ∈ k(S)[x] is such that f(u) = 0. Then there exists a finitesubset s1, . . . , sr ⊆ S such that ai ∈ K(s1, . . . , sr) for 0 ≤ i ≤ n. Letfi, gi ∈ k[x1, . . . , xr] be such that ai = fi(s1, . . . , sr)/gi(s1, . . . , sr).

Define g = g0g1 . . . gn and for 0 ≤ i ≤ n, let f i = gfi/gi = fig0 . . . gi−1gi+1 . . . gn ∈k[x1, . . . , xr]. Notice that for 0 ≤ i ≤ n, ai = f i(s1, . . . , sr)/g(s1, . . . , sr), andso f(x) = g(s1, . . . , sr)−1

∑ni=0 f i(s1, . . . , sr)xi.

Let h(x1, . . . , xr, x) =∑ni=1 f (x1, . . . , xr)xi. Since f(u) = 0 and g(s1, . . . , sr) 6=

0, it follows that h(s1, . . . , sr, u) = 0. Since S∪u is algebraically independentover K, it follows that h = 0. And so, f i = 0 for 0 ≤ i ≤ n. Hence, ai = 0 for0 ≤ i ≤ n, and so f = 0.

Hence, u is transcendental over K(S).

Corollary 7.37. Let F be an extension of K and S ⊆ F be algebraically inde-pendent over K. Then TFAE:

1. S is a transcendence basis for F over K.

2. F is algebraic over K(S).

An analogy?Vector Spaces over K

1. k-linear span 〈S〉 of S.

2. Basis = Independent Generating Set

Field extension F over K

1. K-algebraic closure of S, ie, K(S)alg ∩ F

2. Transcendence Basis = Independent Generating Set

Question: Suppose S, T ⊆ F are transcendence bases over K. Does |S| =|T |?

Notation: From now on, we fix some extension F of K and write algK(S) =K(S)alg ∩ F .

Theorem 7.38. The closure operator algK(S) satisfies the following properties:

1. If S ⊆ F , then S ⊆ algK(S).

2. S, T ⊆ F and S ⊆ T then algK(S) ⊆ algK(T ).

3. If S ⊆ F then algK(algK(S)) = algK(S)

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4. If S ⊆ F and a, b ∈ F satisfy b ∈ algK(S ∪ a) \ algK(S), then a ∈algK(S ∪ b) \ algK(S).

Proof. Homework

The above result says that (F, algK) is a matroid.Remark: The most important axiom is clearly 4, which is called the exchange

property. The other axioms 1,2,3 are satisfied by every natural closure operator.e.g., let G be a group, and for each S ⊆ G, let cl(S) be the subgroup

generated by S.

Definition 7.25. Let (X, cl) be a matroid.

1. The subset I ⊂ X is dependent iff there exists x ∈ I such that x ∈ cl(I \x). Otherwise, I is independent.

2. The subset B ⊆ X is a basis iff B is an independent subset of X such thatcl(B) = X.

Question: Does every matroid have a basis?

Definition 7.26 (Dependent Sets and Bases). Let (X, cl) be a matroid.

1. The subset I ⊆ X is dependent iff there is x ∈ I such that x ∈ cl(I \ x).Otherwise independent.

2. A basis is an independent set I such thatcl(I) = X.

Warning: There exist matroids without bases.e.g. consider the closure operation cl on N defined by cl(S) = S if |S| <∞,

cl(S) = N if |S| =∞.Remark: Of course, in the matroid (F, algK), matroid independence is ex-

actly the same as algebraic independence.

Lemma 7.39 (5.7). Suppose that I ⊆ X is independent and x ∈ X \ I. ThenTFAE:

1. I ∪ x is independent.

2. x /∈ cl(I).

Proof. (1)⇒ (2): By definition.(2) ⇒ (1): Suppose that I ∪ x is not independent. We claim that x ∈

cl(I). If not, there exists y ∈ I such that y ∈ cl(I \ y ∪ u). Since I isindependent, y /∈ cl(I \ y). Thus y ∈ cl((I \ y) ∪ x) \ cl(I \ y). Hencex ∈ cl((Y \ y) ∪ y) = cl(I) as required.

As an immediate consequence, we obtain:

Lemma 7.40. If I ⊆ X is independent then TFAE:

73

1. I is a basis

2. cl(I) = X.

Theorem 7.41. If B is a fintie basis of X, then every basis C satisfies |C| =|B|.

Proof. The result is clear ifB = ∅. Hence, we can suppose thatB = b1, . . . , bn.Let C be any basis of X.

Claim: There exists c1 ∈ C such that c1, b2, . . . , bn is a basis of X. Sincecl(b2, . . . , bn) 6= X = cl(C).

Thus there exists c1 ∈ C such that c1 /∈ cl(b2, . . . , bn). In particular,c1, b2, . . . , bn is independent. Also, c1 ∈ cl(b1, . . . , bn) \ cl(b2, . . . , bn) andso b1 ∈ cl(c1, b2, . . . , bn). It follows that cl(c1, b2, . . . , bn) = X.

Similarly, there exists c2 ∈ C such that c1, c2, b3, . . . , bn is a basis. Con-tinuing in this fashion, we eventually obtain c1, . . . , cn ⊆ C which is a basis.Thus, C = c1, . . . , cn.

Definition 7.27 (Finitary). The matroid (X, cl) is finitary iff whenever x ∈cl(S), there exists a finite S0 ⊆ S such that x ∈ cl(S0).

Example: (F, algK) is a finitary matroid.

Lemma 7.42. If (X, cl) is a finitary matroid, then X has a basis.

Lemma 7.43 (5.11). If (X, d) is a finitary matroid, then X has a basis.

Proof. We can apply Zorn to the poset of independent subsets of X.

Theorem 7.44. If (X, cl) is a finitary matroid, then any two bases have thesame cardinality.

Proof. Let B,C be bases of X. We can suppose B is infinite. It follows that Cis also infinite. For each b ∈ B, there exists a finite subset Cb ⊆ C such thatb ∈ cl(Cb).

Hence, cl(∪b∈BCb) = X and so C = ∪b∈BCb. Hence |C| ≤ |B|ℵ0 = |B|.Similarly, |B| ≤ |C| and so |B| = |C|.

Definition 7.28 (Transcendence Degree). If F is an extension field of K thenthe transcendence degree is tr dimF/K = |S| where S is any transcendence basisof F over K. If K is the prime subfield of F , we write tr dimF .

Theorem 7.45. Suppose F,E are algebraically closed fields. Then TFAE:

1. F ' E

2. charF = charE and tr dimE = tr dimF .

74

Proof. (1)⇒ (2): Obvious.(2) ⇒ (1): We can suppose that E and F are both extensions of the same

prime field K. Let tr dimE = tr dimF = λ, and let X = xα : α < λ andY = yα : α < λ be transcendence bases of E,F . Then we can define anisomorphism ϕ : k(X)→ k(Y ) : xα 7→ yα.

This extens to an isomorphism ϕ : F = K(X)alg → K(Y )alg = E.

Corollary 7.46. Fix some characteristic p ≥ 0.

1. There are exactly ℵ0 countable algebraically closed fields of char p up toisomorphism.

2. If κ is any uncountable cardinal, there exists a unique algebraically closedfield of char p and cardinality κ up to isomorphism.

Proof. 1. For each 0 ≤ n ≤ ℵ0, there exists a unique algebraically closed fieldof characteristic p and transcendence degree n up to isomorphism.

2. It is easily checked if F is any uncountable field, then tr dimF = |F |. Theresult follows.

6. The Hilbert NullstellensatzLetK be a field and let F be a field extension. If I is an ideal ofK[x1, . . . , xn],

then VF (I) = ~a ∈ Fn : p(a) = 0 for all p ∈ I.In this section, we prove the following:

Theorem 7.47 (Hilbert Nullstellensatz). Suppose F is an algebraically closedextension of K. If I is a proper ideal of K[x1, . . . , xn], then VF (I) 6= ∅.

As we will see, the above theorem is equivalent to:

Theorem 7.48. Let K be a field and let K[a1, . . . , an] be a finitely generatedring extension. If K[a1, . . . , an] is a field, then K[a1, . . . , an] is algebraic overK.

We will prove the equivalence:

Proof. Proof that the above implies Nullstellensatz: Let I be a proper ideal ofK[x1, . . . , xn]. Then there exists a maximal M such that M ⊇ I. Consider thefield k[x1, . . . , xn]/M = k[a1, . . . , an] where ai = xi +M .

By assumption, k[~a] is algebraic over K. Since F is algebraically closed,there exists an embedding ϕ : k[~a]→ F over F . Then clearly (ϕ(a1), . . . , ϕ(an))is a zero of I in Fn. Thus VF (I) 6= ∅.

The other implication: Let k[a1, . . . , an] be a field which is finitely gen-eratoed over K as a ring extension. Consider the ring homomorphism ϕ :k[x1, . . . , xn]→ k[a1, . . . , an] by xi 7→ ai.

Then kerϕ = M is a maximal ideal of K[x1, . . . , xn]. By the Nullstel-lensatz, M has a zero (ξ1, . . . , ξn) ∈ (Kalg)n. Consider teh homomorphism

75

ψ : K[x1, . . . , xn] → Kalg, xi 7→ ξi. Then M ⊆ kerψ, and so kerψ = M . ThusK[a1, . . . , an] ' K[ξ1, . . . , ξn]. Since each ξi ∈ Kalg, it follows that K[~a] is analgebraic extension of K.

We will next prove theorem 22.

Lemma 7.49. Let R ⊆ S ⊆ T be rings such that R is Notherian and T =R[t1, . . . , tn] is finitely generated as a ring over R. If T is finitely generated asan S module, then S is also finitely generated as a ring over R.

Proof. :et w1, . . . , wm be a finite system of generators of the S-module T ,which includes t1, . . . , tn. Then for each 1 ≤ i, j ≤ m there exist aij` ∈ S for1 ≤ ` ≤ m such that wiwk =

∑m`=1 a

ik` w`.

Consider the ring S′ = R[aik` : 1 ≤ i, k, ` ≤ m]. Then T = S′w1 + . . . +S′wm.

Since every product of powers of t1, . . . , tn lies in the right hand side. SinceS′ is a finitely generated ring extension of a Notherian ring, it follows that S′ isalso Notherian. Since T is a finitely generated S′-module, it follows that T is aNotherian S′-module. Since S′ ⊆ S ⊆ T , it follows that S is a finitely generatedS′-module. Since S′ is finitely generated as a ring extension of R, it follows thatS is also finitely generated as a ring extension of R.

Lemma 7.50. Let E = K(z1, . . . , zt) be the field of rational functions in thevariables z1, . . . , zt where t ≥ 1. Then E is NOT finitely generated as a ringover K.

Proof. Suppose that X = f1/g1, . . . , fs/gs satisfies E = K[X], where fi, gi ∈k[z1, . . . , zn]. Let p be any irreducible polynomial which doesn’t divide any ofteh gi. Then clearly 1

p /∈ K[X], contradiction.

We will now prove Theorem 22.

Proof. Let K[a1, . . . , an] be a field which is finitely generated over K as a ring.Suppose that k[~a] isn’t algebraic over K. Let z1, . . . , zt be a transcendencebasis of K[~a] over K where t ≥ 1, and let S = K(z1, . . . , zt). Then K[~a] is afinitely generated algebraic extension of S.

And so, K[~a] is finitely generated as an S-module. Hence, applying Lemma8, we see that S is finitely generated as a ring over K. This contradicts Lemma9.

(The above method of proof of the Nullstellensatz is due to Zariski andArtin)

Theorem 7.51. Let F be an algebraically closed extension of K and let I bean ideal of K[x1, . . . , xn]. If f ∈ K[x1, . . . , xn], then TFAE:

1. f(~a) = 0 for all ~a ∈ VF (I).

2. f ∈ Rad I

76

Proof. 2⇒ 1 : Obvious.1 ⇒ 2 : [Rabinowitz’s Trick] We can suppose that f 6= 0. Consider the

ideal J of K[x1, . . . , xn, y] generated by I ∪ 1− yf. Then clearly VF (J) = ∅,and so J = K[x1, . . . , xn, y], hence, there exist polynomials gi ∈ k[x1, . . . , xn, y]and hi ∈ I such that g0(1 − yf) +

∑i gihi = 1. Substituting 1

f for y in thisequation, we have a1/f

n1h1 + . . . + ar/fnrhr where ai ∈ K[x1, . . . , xn] and

ni ≥ 0. Clearing the denominators, we obtain fm = b1h1 + . . .+ brhr for somem ≥ 0, bi ∈ k[x1, . . . , xn]. Thus, if m ≥ 1, then f ∈ Rad I. If m = 0, then1 ∈ I, and so I = K[~x], and the result is trivial.

In particular, suppose that k is algebraically closed and reconsider the in-jective map from algebraic subsets of kn to radical ideals of k[x1, . . . , xn] byV 7→ J(V ). We have seen that this is injective.

Claim: The above map is also surjective.

Proof. If J is a radical ideal of k[~x], then V (J) 7→ J .

8 Category Theory

Definition 8.1 (Category). A category is a class C of objects together with

1. A class of disjoin sets, hom(A,B), one for each pair of objects A,B ∈ C

2. For each triple, A,B,C ∈ C we have a function from hom(B,C) ×hom(A,B) to hom(A,C), called composition of morphisms, which is

(a) Associative, that is, given f : A → B, g : B → C, h : C → D, wehave (h g) f = h (g f)

(b) For each object B ∈ C , there is a morphism called 1B : B → B suchthat for all f : A→ B and g : B → C, 1B f = f and g 1B = g.

Definition 8.2 (Equivalence). In a category C , a morphism f : A → B isan equivalence if there exists a morphism g : B → A such that f g = 1B,g f = 1A.

If there is an equivalence from A to B, we say that A and B are equivalent.

Definition 8.3 (Products). Let C e a category and Ai : i ∈ I a family ofobjects of C .

A product for the family is an object P of C together with maps πi : P → Aisuch that if B ∈ C has maps fi : B → Ai, then there is a unique map g : B → Psuch that fi = πi g. That is, fi factors through πi.

A1 P A2

B.....................................................................................................................................................................

...............

f1

........................................................................................................................................................................ ............

f2

.............................................................................................................................

g

............................................................................................................................. π1................................................................................................................. ............

π2

77

Lemma 8.1. If (P, πi) and (Q, ψi) are both products of the family Ai :i ∈ I then P and Q are equivalent.

Proof. Since P is a product, ∃!g : Q → P such that ψi = πi g and since Q isa product, ∃!f : P → Q such that πi = ψi f .

Then πi = πi g f for all i ∈ I.So g f : P → P must be the unique option making this commute, so

g f = 1P . Similarly, f g = 1Q.

Products are defined using a universal property.

Definition 8.4 (Universal Object). An object I in C is universal (or initial)if ∀C ∈ C there is a unique morphism f : I → C.

Definition 8.5 (Couniversal Object). An object T in C is couniversal (orterminal) if ∀C ∈ C there is a unique morphism g : C → T .

Lemma 8.2. If I and I ′ are universal for C then I and I ′ are equivalent.Similarly, two couniversal objects must be equivalent.

Proof. Since I is universal, there is a unique morphism f : I → I ′.Since I ′ is universal, there is a unique morphism g : I ′ → I.As I is universal, g f = h = 1I , as we know that the identity morphism

exists.Similarly, f g = 1I′ .

Products are a special case. Given Ai : i ∈ I objects of some categoryC , define the category CA whose objects are the objects B of C with mapsfi : B → Ai for each i.

A morphism (B, fi) → (C, gi) in CA is a morphism h : B → C of C

such that fi = gi h for all i. That is, it satisfiesA1 C A2

B.....................................................................................................................................................................

...............

f1

........................................................................................................................................................................ ............

f2

.............................................................................................................................

h

............................................................................................................................. g1................................................................................................................. ............

g2

The product is then the couniversal object in CA.

Definition 8.6 (Covariant Functor). Let C and D be two categories.A covariant functor T : C → D is two functions, T : ob C → ob D and T :

mor C → mor D : hom(C,C ′) 7→ hom(T (C), T (C ′)) such that T (1C) = 1T (C)

for all C ∈ C and T (f g) = T (f) T (g)

Example: Let G be the category of groups. Fix g ∈ G .Then hom(G,−) : G → Sets : H 7→ hom(G,H) is a covariant functor. If

H → H ′ is a homomorphism, then h : hom(G,H)→ hom(G,H ′) : ϕ 7→ h ϕ

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Definition 8.7 (Contravariant Functor). Let C and D be two categories.A covariant functor T : C → D is two functions, T : ob C → ob D and T :

mor C → mor D : hom(C,C ′) 7→ hom(T (C ′), T (C)) such that T (1C) = 1T (C)

for all C ∈ C and T (f g) = T (g) T (f)

Definition 8.8 (Opposite Category). Let C be a category. The opposite cate-gory has objects ob C and morphisms hom(Aop, Bop) = hom(B,A)

There is a contravariant functor from C → C op. And given a contravariantfunctor C → D , we can define a covariant functor C op → D .

Definition 8.9 (Natural Transformation). Let C ,D be categories and let S, T :C → D be covariant functors.

A natural transformation α : S → T is a function that assigns to each objectof C a morphism αC : S(C) → T (C) of D such that ∀f : C → C ′ of C thefollowing diagram commutes.

S(C ′) T (C ′)

S(C) T (C)............................................................................................. ............αC

........................................................................................ ............αC′

.............................................................................................................................

S(f)

.............................................................................................................................

T (f)

Definition 8.10 (Free Object). If C is a category whose objects are all setsand every morphism is function, then an object F ∈ C is free on a set X if∃!ι : X → F a set inclusion such that ∀f : X → A of sets, for A ∈ ob C ,∃!f : F → A such that the following diagram commutes.

X

F A

........

........

........

........

........

........

........

........

........

........

........

........

.................

............

ι

.................................................................................................................................................................................

f

................................................................................................................. ............f

9 Representation Theory

Recall that if G is a finite group with |G| = pn for some prime p, then Z(G) 6= 1.Hence, there exists some t ≥ 1 such that Zt(G) = G.

Theorem 9.1. If G is a finite p-group, then G is nilpotent.

In this section, we will prove:

Theorem 9.2 (Burnside). If G is a finite group and |G| = paqb for some primesp, q, then G is solvable.

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Example: Consider the dihedral group D5. Then |D5| = 10 and Z(D5) = 1.Thus, D5 is not nilpotent.

Example: Consider A5 = 22 ∗ 3 ∗ 5.Basic Strategy: Show that G isn’t simple, assuming G isn’t cyclic of order

p. Then the result follows by induction.

Definition 9.1 (k-Representation). Let G be a finite group and let k be a field.Then a k-representation of G is a homomorphism π : G → GL(V ) : g 7→ πgwhere 0 6= V is a finite dimensional vector space over k.

The degree of π is deg π = dimk V .

Remark: We usually just write representation.Examples:

1. A representation of degree 1 is just a homomorphism π : G→ k∗.

2. Suppose that G acts on the finite set X. Let kX = ⊕x∈Xkex. Then thecorrsponding permutation representation is the homomorphism π : G →GL(kX) defined by πg(ex) = eg·x.

Definition 9.2 (G-invariant Subspace). Let π : G → GL(V ) be a representa-tion. Then the subspace W ≤ V is G-invariant iff πg[W ] = W for all g ∈ G.

Examples: 0, V are G-invariant subspaces - the trivial invariant subspaces.

Definition 9.3 (Subrepresentation). If W is a nonzero G-invariant subspace,then the associated subrepresentation is the homomorphism π|W : G→ GL(W )by g 7→ πg|W .

Definition 9.4 (Irreducible). The representation π : G→ GL(V ) is irreducibleiff there are no nontrivial G invariant subspaces.

Example: If deg π = 1, then π is irreducible.Question: Is every representation a direct sum of irreducible representations?

Theorem 9.3 (Maschke). Suppose that p = char k 6 ||G|.If π : G → GL(V ) is a k-representation, and W ≤ V is a nontrivial G-

invariant subspace, then there exists a G-invariant subspace U < V such thatV = W ⊕ U .

Proof. Let S < V be any subspace such that V = W ⊕ S and let θ : V =W ⊕ S →W , that is, w + s 7→ w, be the associated projection.

Then θ(w) = w for all w ∈ W and θ[V ] ⊆ W . Furthermore, if ψ ∈homk(V, V ) satisfies these two conditions, then there exists a corrsponding de-composition V = W ⊕ (1− ψ)[V ] where (1− ψ)[V ] = v − ψ(v) : v ∈ V .

Claim: Suppose that πg(ψ(v)) = ψ(πg(v)) for all g ∈ G and v ∈ V . Then(1− ψ)[V ] is G-invariant.

Proof of Claim: For all g ∈ G and v ∈ V , we ave πg(v − ψ(v)) = πg(v) −πg(ψ(v)) = πg(v) − ψ(πg(v)) = (1 − ψ)[V ]. As πg is an invertible linear trans-formation, we must have that πg((1 − ψ)[V ]) = (1 − ψ)[V ], and so the claimholds.

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Thus, it is enough to find a G-invariant projection.Let θ : V → W be our original projection. Define ψ : V → V by ψ(v) =

1|G|∑g∈G πgθπ

−1g (v). This is the only point where we use the fact that p =

char k 6 ||G|.If h ∈ G, then πhψπ

−1h = 1

|G|∑g∈G πhπgθπ

−1g π−1

h = 1|G|∑g∈G πhgθπ

−1hg =

1|G|∑a∈G πaθπ

−1a = ψ and so πhψ = ψπh. Thus, it is enough to check that ψ

satisfies conditions (i) and (ii).(i): If w ∈ W and g ∈ G then πg(θ(π−1

g (w))) = πg(π−1g (w)) = w since

π−1g [W ] = W .

Hence ψ(w) = 1|G|∑g∈G πgθπ

−1g (w) = w

(ii): Finally, if v ∈ V is arbitrary and g ∈ G, then πG(θ(π−1g (v))) ∈W , since

θ[V ] ⊆W . Thus ψ(v) ∈W .

Corollary 9.4. If char k 6 ||G| then every k-representation of G decomposes intoa direct sum of irreducible representations.

Here if πi : G→ GL(Vi), 1 ≤ i ≤ d, are representations then their direct sumis the representation φ = π1⊕. . .⊕πd = G→ GL(V1⊕. . .⊕Vd), φg(v1+. . .+vd) =π1g(v1) + . . .+ πdg(vd) where vi ∈ Vi and g ∈ G.

Even more explicitly...if we choose a basis Bi of Vi, then each φg has theform of a block diagonal matrix, with respect to B = ∪iBi.

Convention: From now on, we shall only consider representations over C. Inparticular, every representation will be a direct sum of irreducible representa-tions.

A reminder of some linear algebra.Let Mn(C) be the ring of n × n matrices over C. For each A ∈ Mn(C),

the characteristic polynomial is det(Ix − A) = |Ix − A|. The roots of thecharacteristic polynomial are called characteristic roots or eigenvalues. Thetrace is the sum of the diagonal entries which is the sum of the eigenvaluescounted with multiplicities.

Some basic results:

1. Suppose that A ∈ Mn(C) and B ∈ Mn(C) is invertible, then |Ix −BAB−1| = |B(Ix − A)B−1| = |Ix − A|. If particular, it makes senseto speak of the characteristic polynomial and trace of a linear transforma-tion f : V → V .

2. (Cayley-Hamilton) If A ∈Mn(C) and p(x) = |Ix−A|, then p(A) = 0. Inparticular, the minimal polynomial divides the characteristic polynomial.

Theorem 9.5. 1. Suppose that A,B ∈ Mn(C) satisfy AB = BA and letα1, . . . , αn, β1, . . . , βn be the characteristic roots of A,B. Then, afterrenumbering the βi if necessary, the characteristic roots of AB are α1β1, . . . , αnβn.

2. If p(x) ∈ C[x] , then the characteristic roots of p(A) are p(α1), . . . , p(αn)

3. If A is invertible, then the characteristic roots of A−1 are α−11 , . . . , α−1

n .

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Definition 9.5 (Character). If π : G → GL(V ) is a representation, then thecorresponding character is the function χπ : G→ C by χπ(g) = tr(πg).

The character χπ is irreducible iff π is irreducible.

Lemma 9.6. If π : G→ GL(V ) is a representation, then

1. χπ(1) = dimC V = deg π

2. χπ(g) = χπ(hgh−1) for all g, h ∈ G.

Proof. 1. Obvious.

2. Fix some basis B of V and let Mg be the corresponding matrix of πg.Then

χπ(hgh−1) = tr(MhMgM−1h ) = tr(Mg) = χπ(g).

In particular, each character is a class function, ie, it is constant on everyconjugacy class of G.

Definition 9.6 (C`(G)). C`(G) = f ∈ GC is a class function .

Lemma 9.7. [(a)]

1. C`(G) is a vector space over C.

2. dimC C`(G) =the number of conjugacy classes.

Proof. (a) is trivial.Clearly, the characteristic functions on the conjugacy classes of G form a

basis.

Remark, eventually we will show that the distinct irreducible characters forma basis of C`(G).

Lemma 9.8. Let π : G→ GL(V ) be a representation.

1. If g ∈ G then χπ(g) is a sum of roots of unity. In particular, χπ(g) is analgebraic integer.

2. If g ∈ G then χπ(g) = χπ(g)

3. If g ∈ G, then |χπ(g)| ≤ deg π. Furthermore |χπ(g)| = deg π iff πg = λ idVfor some λ ∈ C.

Proof. Fix some basis B of V and let Mg be the matrix of πg. Let d = dimC V =deg π.

1. Let g ∈ G have order n. Then Mng = I. Hence, by Theorem 4, every

characteristic root of Mg is an nth root of unity. The result follows.

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2. Let λ1, . . . , λd be the characteristic roots of Mg. Then the characteristicroots of M−1

g are λ−11 , . . . , λ−1

d . Since the λ+i are roots of unity, λ−1i = λi.

The result follows.

3. Again, let λ1, . . . , λd be the characteristic roots of Mg. Since each λi isa root of unity, we have |λi| = 1. HEnce |χπ(g)| = |λ1 + . . . + λn| ≤|λ1|+ . . .+ |λd| = deg π.

Furthermore, if ξ, η ∈ C∗, then |ξ+ η| ≤ |ξ|+ |η|, and |ξ+ η| = |ξ|+ |η| iffη = rξ for some r ∈ R+.

Thus, |χπ(g)| = deg π iff λ1 = . . . = λd = λ ∈ C. Thus Mg satisfies twopolynomial equations, xn − 1 = 0 and (x− λ)d = 0

Hence, Mg satisfies the gcd of these polynomials, which must be x − λ,since xn − 1 has no repeated roots.

Definition 9.7 (Intertwines). Suppose that π : G → GL(V ) and ρ : G →GL(W ) are representations, then the linear map f : V →W intertwines π andρ iff for every g ∈ G, the following diagram commutes:

V

V

W

W

........

........

........

........

........

........

........

........

........

........

........

........

.................

............

πg

........

........

........

........

........

........

........

........

........

........

........

........

.................

............

ρg

................................................................................................................. ............f

................................................................................................................. ............f

ie, f is a homomorphism between two G-actions. Let homG(π, ρ) be thevector space of intertwiners between π and ρ.

Examples: Suppose π : G → GL(V ) is a representation and that W ≤ Vis a G-invariant subspace. Let f : V → W be a G-invariant projection. Thenf ∈ homG(π, π|W ).

We always have 0 ∈ homG(π, ρ).We always have that λ idV ∈ homG(π, π) for every λ ∈ C.

Definition 9.8 (Equivalence). The represenations π and ρ are equivalent orisomorphic if there exists an invertible intertwiner between π and ρ.

Theorem 9.9 (Schur’s Lemma). Suppose that π : G → GL(V ) and ρ : G →GL(W ) are irreducible representations.

1. if π and ρ aren’t equivalent, then homC(π, ρ) = 0. That is, the zero mapf ≡ 0 is the only intertwiner between π and ρ.

2. If π and ρ are equivalent, then dim homC(π, ρ) = 1, ie, if wlog π = ρ, thenthe only interetwiners are λ idV for λ ∈ C.

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Proof. [(a)]

Suppose dimC homG(π, ρ) 6= 0 and let f : V → W be a nonzero inter-twiner. Clearly ker f is a G-invariant subspace of V . Since ker f 6= V andπ is irreducible, ker f = 0, and so f is injective.

Similarly, Im f is a G-invariant subspace of W . Since Im f 6= 0 and ρ isirreduicble, it follows that Im f = W . Thus, f is invertible and ρ, π areequivalent.

1.2. Since π, ρ are equivalent, we can suppose that π = ρ. It is now enough toshow that if f ∈ homG(π, π), then f = λ idV for some λ ∈ C. Let λ be aneigenvalue of f . Then the corresponding eigenspace U 6= 0 and is clearlyG-invariant. Since π is irreducible, it follows that U = V and the resultfollows.

Corollary 9.10. Suppose that π : G → GL(V ) and ρ : G → GL(W ) areirreducible representations and let h : V → W be any linear map. We defineh = 1

|G|∑g∈G ρ

−1g hπg. Then

1. If π and ρ are inequivalent, then h = 0.

2. if π and ρ are equal, then h = tr(h)dimC V

idV .

Proof. First note that h intertwines π and ρ. If π, ρ are inequivalent, then h ≡ 0.Now suppose that π, ρ are equal. Then h = λ idV for some λ ∈ C. To

evaluation λ, notice that λ dimC V = tr(h) = tr(h), and so λ = tr(g)dimC(V ) .

Definition 9.9. We define a scalar product on GC by 〈ϕ,ψ〉 = 1|G|∑g∈G ϕ(g)ψ(g).

Theorem 9.11. [(i)]

1. If χπ is an irreducble character, then 〈χπ|χπ〉 = 1

2. If χπ, χρ are irreducible characters and π, ρ aren’t equivalent, then 〈χπ|χρ〉 =0.

Proof Delayed.Remark: Thus the distinct irreducible characters form a orthonormal set

in C`(G). WE shall soon see that they actually form an orthonormal basis.Already we see that there are only finitely many non-isomorphic irreduciblerepresentations, and this number is bounded above by the number of conjugacyclasses.

Definition 9.10. We defined a scalar product on GC by 〈φ, ψ〉 = 1|G|∑g∈G φ(g)ψ(g).

Theorem 9.12. If π and ρ are irreducible representations, then 〈χπ, χρ〉 = 1if π, ρ are equivalent and otherwise 0.

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Proof Postponed.

Corollary 9.13. Let π : G→ GL(V ) be a representation and V = W1⊕. . .⊕Wk

be a decomposition into irreducible representations.If ρ : G→ GL(W ) is an irreducible representation, then the rumber of π|Wi

which are isomorphic to ρ is given by 〈χπ, χρ〉.

Proof. Let πi = π|Wi. Then χπ = χπ1 + . . .+ χπk

.Hence 〈χπ, χρ〉 = 〈χπ1 + . . . + χπk

, χρ〉 = 〈χπ1 , χρ〉 + . . . + 〈χπk, χρ〉 = the

number of i such that πi is isomorphic to ρ.

Corollary 9.14. Suppose that π : G → GL(V ) and ρ : G → GL(W ) arerepresentations.

1. π and ρ are isomorphic iff χπ = χρ.

2. π is irreducible iff 〈χπ, χπ〉 = 1.

Proof. a) Immediate from previous corollary.b) If π = `1π1+. . .+`sπs is a decomposition into irreducible representations,

then 〈χπ, χπ〉 = `21 + . . .+ `2s.

From now on, let π1, . . . , πh be the distinct irreducible representations of G,and let ni = deg πi.

Definition 9.11. Consider the action of G on itself by left multiplication andlet ρ : G→ GL(CG) be the corresponding permutation representation, known asthe regular representation. Thus CG = ⊕g∈GCeg, and for each g ∈ g we haveρg(et) = egt.

Lemma 9.15. If ρ is the regular representatation of G, then χρ(g) = |G| ifg = 1 and 0 if g 6= 1.

Proof. If g = 1, then χρ(1) = dimC CG = |G|.If g 6= 1, then ρg(et) = egt 6= et, and so all the diagonal elements of the

corresponding permutation matrix are zero.

Corollary 9.16. Every irreducible representation πi is conmtained in the reg-ular representation ρ with multiplicity ni = deg πi.

Proof. The multiplicity of πi in ρ is given by

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〈χρ, χπi〉 =

=1|G|

∑g∈G

χρ(g)χπi(g)

=1|G|

∑g∈G

χρ(g)χπi(g−1)

=1|G|

χρ(1)χπi(1)

=1|G||G|deg πi

= n

Corollary 9.17. |G| = n21 + . . .+ n2

h.

Proof. |G| = dimC CG =∑hi=1 ni deg πi = n2

1 + . . .+ n2h.

Next we will reluctantly turn to the proof of theorem 7.15

Proof. Let π : G → GL(V ) and ρ : G → GL(W ) be (not necessarily distinct)irreducible representations. Fixing bases of V and W , let Ag = (aij(g)) andBg = (bk`(g)) be the corresponding matrices of πg, ρg. (If π = ρ, we chooseAg = Bg).

Consider any linear map h : V →W and let C = (c`i) be the correspondingm × n matrix, where dimC V = n and dimC W = m. Let D = (djk) be thematrix corrsponding to h = 1

|G|∑g∈G ρ

−1g hπg.

Then djk = 1|G|∑

i`g∈G

bk`(g−1)c`iaij(g).

First suppose that π, ρ are not equivalent. Then h is the zero map, and soeach dkj = 0. Regard the RHS of the above as a linear form in the variables c`i.

Since the form vanishes identically, each coefficient is 0. Hence, Claim 1: Ifπ 6= ρ, then for all k, `, i, j, we have 1

|G|∑g∈G bk`(g

−1)aij(g) = 0.It follows that 〈χπ, χρ〉 = 1

|G|∑g∈G χπ(g)χρ(g−1) = 1

|G|∑g∈G (

∑i aii(g)) (

∑k bkk(g)) =∑

i,k1|G|∑aii(g)bkk(g) = 0.

ext suppose that π = ρ. Then we have that h = tr(g)/ dimC V id. Arguingas in the previous case, we see that if k 6= j, then dkj = 0, and so for all `, i, wehave 1

|G|∑g∈G ak`(g

−1)aij(g) = 0.In particular, Claim 2: If k 6= j, then 1

|G|∑g∈G akk(g−1)ajj(g) = 0.

Now suppose that k = j and choose h such that c`i = 1 if ` = i = k and 0otherwise.

Then tr(h) = 1. Hence the above formulae give Claim 3: for each 1 ≤k ≤ n = dimC V , 1

|G|∑akk(g−1)akk(g) = 1

n . Applying Claim 2 and Claim

86

3, we get 〈χπ, χπ〉 = 1|G|∑χπ(g)χπ(g−1) =

∑i,j

1|G|∑g∈G aii(g)akk(g−1) =∑

k1|G|∑g∈G akk(g)akk(g−1) = n 1

n = 1.This concludes the proof.

Theorem 9.18. The characters χπ1 , . . . , χπhare an orthonormal basis for

C`(G).

Corollary 9.19. h = the number of conjugacy classes of G.

Proof. The characteristic functions of the conjugacy classes of G form a basisof C`(G).

Corollary 9.20. If G is abelian, then every irreducible representation of G hasdegree 1.

Proof. Since G is abelian, h = |G|. As G = n21 + . . .+ n2

|G|, ni = 1 for all i.

Notation: Irreducible representations π1, . . . , πh, irreducible characters χ1, . . . , χh,conjugacy classes C1, . . . , Ch, and Fixed representative gi ∈ Ci.

For each group we have the character table of G.

C1 . . . Cj . . . Ch

χ1

......

...χi . . . . . . χi(gj) . . . . . ....

...

χh...

For A4, we get

1 (12)(34) (123) (132)χ1 1 1 1 1χ2 1 1 ω ω2

χ3 1 1 ω2 ωχ4 3 −1 0 0

The first three are the representations of C3, a homomorphic image of A4,but now we must determine π4.

Consider the action of A4 on X = 1, 2, 3, 4 and let ϕ : A4 → GL(CX)be the corresponding permutation representation. Then χϕ(g) = fix(g), thenumber of fixed points of g, which is χ1 + χ4. Note the following A4-invariantdecomposition.

So CX = Cvx ⊕ V0 where vx = e1 + e2 + e3 + e4 and V0 =∑i aiei|a1 + a2 +

a3 + a4 = 0. Thus χ4 is the character of ϕ|V0 .Remark: Suppose that G acts 2-transitively on X and π : G → GL(CX) is

the corresponding permutation representation. Then χπ − 1 is an irreduciblecharacter.

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Proof. As above, we have a G-invariant decomposition CX = Cvx⊕V0, so χπ−1is the character of π|V0 . Hence, it’s enough to show that 〈χπ, χπ〉 = 2.

Claim:∑g∈G fix(g)2 = 2|G|.

Assuming the claim, 〈χπ, χπ〉 = 1|G|∑g∈G χπ(g)χπ(g) = 1

|G|∑g∈G fix(g)2 =

2.To prove the claim, we count the number of elements of Ω = 〈a, b, g〉 ∈

X ×X ×G|g(a) = a, g(b) = b in two ways.(1) For each g ∈ G, the number of elements 〈a, b, g〉 ∈ Ω is fix(g)2, thus

|Ω| =∑g∈G fix(g)2.

(2) For each a ∈ X, the numebr of elmeents 〈a, b, g〉 ∈ Ω is given by∑b∈X |Ga,b| = |Ga|+

∑b∈X\a |Ga,b| = |Ga|+[Ga : Ga,b]|Ga,b| by 2-transitivity,

and this is |Ga| + |Ga| = 2|Ga|. Thus, |Ω| =∑a∈X 2|Ga| = 2[G : Ga]|Ga| =

2|G|.

To explain why the columns of a character table are orthogonal, recall that1|G|∑g∈G χi(g)χj(g) = δij .

Thus 1|G|∑hk=1 |Ck|χi(gk)χj(gk) = δij . Or

∑hk=1

√|Ck||G| χi(gk)

√|Ck||G| χj(gk) =

δij

In other words, the rows of the matrix with ij term√|Ck||G| χi(gl) form an

orthonormal basis of Ch. In other words, the matrix is unitary. If follows that

the columns are also orthonomal, thus∑hi=1

√|Ck||G| χi(gk)

√|C`||G| χi(g`) = δk`.

And so,√|Ck||C`||G|

∑hi=1 χi(gk)χi(g`) = δk`.

Hence∑hi=1 χi(gk)χi(g`) = |G|

|Ck|δkl.In summary we have proved:

Theorem 9.21 (The Orthogonality Relations). [(a)]

1. 1|G|∑g∈G χi(g)χj(g−1) = δij.

2.∑hi=1 χi(gk)χi(g−1

` ) = |G||Ck|δk`.

In order to prove Burnside’s Theorem, we only require:

Theorem 9.22 (Folklore). For each 1 ≤ i, k ≤ h, the number |Ck|deg πi

χi(gk) isan algebraic integer.

Proof Postponed.Before proving Burnside, we prove

Theorem 9.23. For each 1 ≤≤ h, the degree deg πi divides |G|.

Proof. By the first orthogonality relation, we have∑hi=1 |Ck|χi(gk)χi(g−1

k ) =|G|. And so,

∑hi=1

|Ck|deg πi

χi(gk)χi(g−1k ) = |G|

deg πi, which is a sum of products of

algebraic integers, and so is an algebraic integer. Hence, |G|deg πi

is a rationalalgebraic integer.

88

Burnside’s Theorem is an ”easy” consequence of the following slightly tech-nical result which explains most of the zeros in our above character tables.

Theorem 9.24. If (|Ck|,deg πi) = 1, then either

1. χi(gk) = 0, or

2. χi(g) = deg πiω for some root of unity ω, in which case πi(g) = ωI is ascalar matrix.

Proof Postponed.Now we can prove Burnside.

Proof. Suppose that G is a counterexample of minimal order.Then G must be a simple nonabelian group of order |G| = paqb for some

primes p 6= q and a, b ≥ 1.Let Q be a Sylow q-subgroup of G and let 1 6= g ∈ Z(Q). Since Q ≤ CG(g)

and |gG| = [G : CG(c)], it follows that |gG| = pc for some c ≥ 1. Let g ∈ Ck.Let π1 = 1, π2, . . . , πh be the irreducible representations of G. Consider some

2 ≤ ` ≤ h. If p|deg π`, then we can write deg π` = pd`. On the other hand, ifp 6 | deg π`, then (|Ck|,deg π`) = 1. Since G is simple and nonabelian, we musthavde that χ`(g) = 0 for each such `.

Appealing to the second orthogonality condition, we see that 0 =∑h`=1 χ`(g)χ`(1) =

1 + σ `≥2p| deg π`

χ`(g) deg π` = 1 + p∑d`χ`(g)

Thus, −1/p =∑d`χ`(g) is an algebraic integer.

We still have three things to prove that we have delayed.First we prove Theorem 12.

Proof. Suppose πi(gk) isn’t a scalar matrix. Then, let ck = |Ck| and ni = deg πi.By Theorem 10, |Ck|

deg πiχi(gk) is an algebraic integer. Let α = χi(gk)/ deg πi.

Claim: α is an algebraic integer.Proof of claim: Since (Ck, ni) = 1, there exist a, b ∈ Z such that ack+bni = 1.

Hence α = ackα+ bniα = a |Ck|deg πi

χi(gk) + bχi(gk) is an algebraic integer.Recall that χi(gk) = ωm1 + . . .+ωmni for a suitably chosent root ω of unity.

Also, since πi(gk) isn’t a scalar matrix, Lemma 3c implies that |χi(gk)| < deg πiand so |α| < 1. Let α = α1, . . . , αr be the distinct conjugates of α over Q. Thenαs = ωm1

s + . . .+ ωmnis for some root of unity ωs. It follows that each |αs| < 1.

Let E = Q(α) and consider NEQ (α) = α1 . . . αr. Since α is an algebraic

integer, it follows that NEQ (α) ∈ Z. Since |NE

Q (α)| < 1, it follows that NEQ (α) =

0. And hence, α = χi(gk)/ deg πi = 0.

Next, we will prove Theorem 8.We will make use of the following east observation:

Lemma 9.25. If χ is an irreducible character of G, then so is χ.

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Proof. Let g 7→Mg be the matrix representation corresponding to χ.Then χ corresponds to g 7→ (M−1

g )t. Since 〈χ, χ〉 = 〈χ, χ〉 = 1, it followsthat χ is an irreducible character.

And now the proof of the theorem.

Proof. We suppose that f ∈ C`(G) satisfies 〈χi, f〉 = 0 for 1 ≤ i ≤ h. Wemust show that f = 0. Let ρ : G→ GL(CG) be the regular representation andconsider the map ρf =

∑g∈G f(g)ρg.

Notice that ρf (e1) =∑g∈G f(g)ρg(e1) =

∑g∈G f(g)eg. Hence, it is enough

to show that ρf = 0.As ρ = ⊕π deg πiπi summed over the irreducible representations, we know

that ρf = ⊕π deg π∑g∈G f(g)πg = ⊕π deg ππf and so it is enough to show that

πf =∑g∈G f(g)πg = 0.

To see this, first note that if t ∈ G, then πtπfπ−1t =

∑g ∈ Gf(g)πtπgπ−1

t =∑g∈G f(tft−1)πtgt−1 = πf . Hence, by Schur’s Lemma, it follows that πf = λI

for some λ ∈ C. Finally, we note that

deg πλ =∑g∈G

f(g)χπ(g) = |G|〈f, χπ〉 = 0

Thus, λ = 0.

And finally, we will now prove Theorem 10.

Proof. Identifying each basis vector eg ∈ CG with the corresponding elementg ∈ G, we obtain a natural noncommutative ring structure on CG. Furthermore,each representation θ : G → GL(V ) extends linearly to a ring homomrophismθ : CG → End(V ). For each conjugacy class Ck opf G, define ci =

∑g∈Ci

g ∈CG.

Then it is easily checked that c1, . . . , ch is a basis of Z(CG) (the center).Here there exist `ijk ∈ C such that cicj =

∑hk=1 `

ijk ck.

Since cicj =∑a∑b =

∑ab where a ∈ Ci, b ∈ Cj . We see that each

`ijk ∈ Z.Now let π : G → GL(V ) be an irreducible representation. Then apply π to

the definition of multiplication, adn get

π(ci)π(cj) =h∑k=1

`ijk π(ck)

By Schur’s Lemma, since each ci is in the center of the group ring, it followsthat π(ci) = λiI for some λi ∈ C. Taking traces, we see that deg πλi =∑g∈Ci

χπ(g) = |Ci|χπ(gi). Hence, λi = |Ci|deg πχπ(gi).

Hency, by substituting into the above formula, we obtain

|Ci|deg π

χπ(gi)|Cj |

deg πχπ(gj) =

h∑k=1

`ijk|Ck|deg π

χπ(gk)

90

Hence, the ring R = Z[ |Ci|deg πχπ(gi)|1 ≤ i ≤ h] is finitely generated as a Z-

module, and hence is an integral extension of Z. In particular, each |Ci|deg πχπ(gi)

is an algebraic integer.

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