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2 INTRO TO GROUPS Algebra Study Guide

1 Proofs to Know

Prop: Let R be commutative and let M be a left R-module. Suppose M has an R-basis B of cardinality n and anR-basis Eof cardinality m. Then n = m.Proof. By the remark from class (which states that ifM is a free R-module for a commutative ring R with a basis ofn elements, M

= Rn), Rm

= M

= Rn (can use the map : Rn

M given by (r1,

, rn) = n1 risi, where {si}

n1

is a basis for Mits surjective as it generates M and injective since ker = (0)).Let I be a maximal ideal of R, which exists as R is commutative. Then F = R/I is a field. By an exercise,Rn/IRn = R/IR R/IR

n times

= Fn = Fm. By vector space theory, n = m.

Prop: Assume the set A = {v1, , vn} spans the vector space V but no proper subset ofA spans V. Then A is a basisof V.

Proof. Suffices to show vi are LI. If a1v1 + + anvn = 0 and WLOG a1 = 0, write v1 as a linear combination ofthe others. So {v2, , vn} spans V. ()

Prop: Let F be a field and V an F-vector space. Let T S be subsets of V such that T is LI and S spans V. Thenthere exists a basis B such that T B S.Proof. Let = {A : T A S, A LI}. Note = as T A. is a poset under . Let C be a totally orderedsubset of . Let A =

AC

A. Show that A .By Z-L there exists a maximal element of , say B.Claim: B is a basis for V.Proof: It is enough to show that S spanF B. Ifs S such that s / spanF B, then B{s} is linearly independent.Thus B {s} . Since B is maximal, B B {s} = B = B {s}.

Thm: Let V be an F-vector space and let W be a subspace of V. Then V /W is a vector space with dim V = dim W +dim V /W (where if one side is infinite then both are).

Proof. Let dim W = m and dim V = n and let w1, w2, , wm be a basis for W. These are LI and can be built upto a basis w1, w2, , wm, vm+1, , vn of V. The natural surjective projection of V into V /W sends all of the wito 0 and none of the vi to 0 (as this would imply vi

W). Hence, V /W of the projection map is isomorphic to the

subspace of V spanned by the vi, giving that dim V/W = n m.

2 Intro to Groups

Prop: 1 (Z/nZ) = {a Z/nZ : gcd(a, n) = 1}.Proof. gcd(a, n) = 1 = x, y Z such that ax + ny = 1. Then n|ax 1 = ax = 1 so a is a unit. Conversely, ifax = 1 in Z/nZ then n|ax 1 = ax 1 = ny = gcd(a, n) = 1.

Def: Let F be a field. Then the general linear group of F of degree n is defined by GLn(F) := Mn(F) =

{A Mn(F) : det A = 0}.Def: Let G be a group and g G. Then the order of g is the least positive integer n (if it exists) s.t. gn = 1. If no

such n exists, then|g|

=

.Def: Define the dihedral group of order 2n to be the group of rigid motions of the regular n-gon. Then |D2n| = 2n.

Let r be the counterclockwise rotation by 2/n radians and s be the reflection about the x-axis. Note: |r| = nand |s| = 2. Then sri = srj for 0 i < j n 1 (if sri = srj = ri = rj = 1 = rji). ThenD2n =

r, s|rn, s2 = 1, rs = sr1 .

Prop: 2 Let G be a group and x G. Let d = |x| < . Then for all n Z, xn = 1 d|n. (For the forward direction,use the division algorithm.)

Prop: Let G be a group, x G. Then |x| |G|. If G is a finite group, then every element ofG has finite order.Proof. Nothing to show if |G| = . Assume m = |G| < . Consider 1, x , x2, , xm G. Since G has only mdistinct elements, there exists xi = xj where 0 i < j m Then xij = 1 and 0 < j i m.

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2 INTRO TO GROUPS Algebra Study Guide 2.1 Generating Groups

Def: Let T be a set, and let ST = {f : T T : f is a bijection}. This is a group under composition. In the caseT = {1, 2, , n}, denote this group ST by Sn, and its called the symmetric group of degree n. Note: |Sn| = n!.

Def: A k-cycle in Sn is a permutation such that there exists {a1, a2, , ak} {1, 2, , n} such that (a1) = a2,(a2) = a3, , (ak) = a1 and (x) = x for all x {1, , n}\{a1, , ak}. We write this k-cycle as (a1 a2 ak).

Rmrk: Any two disjoint cycles commute, and any permutation can be uniquely written as a product of disjoint cycles.Prop:

Let a, b G and |a| = m < and |b| = n < . Suppose ab = ba and i, j Z, ai

= b

j

ai

= b

j

= 1. Then|ab| = lcm(m, n).Cor: Suppose , are disjoint cycles in Sn. Then || = lcm(||, ||).Thm: Let = c1c2 ct, where this is a product of disjoint cycles. Let ci be an li-cycle. Then || = lcm(l1, l2, , lt).Prop: 3 Let Sn be a k-cycle. Then || = k.Def: Let Sn \ {1}. Suppose is a product of disjoint cycles = p1 pl where p1, , pl are of lengths 1 < k1

k2 kl. We say has cycle type {k1, k2, , kl}.Def: The quaternion group, Q8, is defined by Q8 = {1, 1, i, i,j, j, k, k}.Def: Let S G. The centralizer of S in G is CG(S) := {g S : gx = xg x S}.Def: Let H G. The normalizer of H in G is defined to be NG(H) :=

g G : gH g1 = H.

Rmrk: CG(H) NG(H).Prop: Let G be a group and {H}I be a family of subgroups of G. Then IH G.

2.1 Generating Groups

Prop: Let S G with S = . Then S = {se11 senn : si S, ei = 1i}. IfG is finite, S = {s1 , sn : si S}.E.g.: 4 Q = {1/pm : p prime, m N}.Prop: Sn = {(i j) : 1 i < j n} = {(12), (13), , (1 n)} = {(12), (23), , (n 1 n)} = {(12), ( 1 2 3 n)}.Def: If G = a, G is said to be a cyclic group (the cyclic group generated by a).Prop: | a | = |a|. If |a| = n, then a = 1, a, , an1.Prop: Any subgroup of a cyclic group is cyclic.Thm:

5 Let G = a be a cyclic group with |G| = n < .1. For all i Z, ai = ad where d = gcd(i, n). Thus, ad : d > 0, d|n is the set of all subgroups ofG (these

are all distinct since | ad | = |ad| = n/d).2.

|ai

|= n/gcd(i, n).

3. Every subgroup of G has order dividing n.

Cor: Let G = a be a cyclic group of order n. Then ai is a cyclic generator for G if and only if gcd(i, n) = 1.Cor: Let G = a, |G| = n, d a positive divisor of n. Then G has (d) elements of order d (where is Eulers

function).Cor: Let n be a positive integer. Then

d1d|n

(d) = n.

Prop: Let G = a be a cyclic group. Then if |G| = n then G =Z/nZ. If |G| = then G =Z.Def: 6 The alternating group of order n, denoted by An, is the subgroup ofSn consisting of all even permutations.Prop: If n > 2, An is generated by the set of all 3-cycles.Def: Let G be a group, H G. Then H is called normal, denoted H G if gH g1 = H for all g G.Rmrk: Let G be a finite group and H

G such that 2

|H

|=

|G

|. Then H G.

Proof. Let G \ H. Note |H| = |H| and H H = . Then |H| = |G|/2 = |H| so H H = G. Similarly,H H = G and H H = . Therefore G \ H = H = H, which implies H 1 = H.

Cor: An Sn.Prop: |A4| = 12. A4 has no subgroup of order 6.

Proof. The number of 3-cycles in A4 (or S4) is

43

2 = 8. Suppose H A4 such that |H| = 6 = |A4|/2, whichimplies H A4. So H contains some 3-cycle. WLOG, assume the 3-cycle (1 2 3) H. Then (1 3 2) H (as(132)1 = (1 2 3). Since H is normal in A4, setting = (1 2) implies (123)1 = (142) H, so (124) H.Doing this again with, say, (1 3)(2 4) yields two more 3-cycles. Therefore H contains at least 6 3-cycles. ()

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2 INTRO TO GROUPS Algebra Study Guide 2.2 Cosets

2.2 Cosets

Def: Let H G, g G. Then the set gH = {gh : h H} is called a left coset of H.Prop: Let H G. (1) For all g1, g2 G, |g1H| = |g2H|. (2) For all g1, g2 G, g1H = g2H or g1H g2H = . (3)

G = gGgH.

Proof. (1) The maps f : g1H g2H defined by x g2g1

1 x and f : g2H g1H defined by y g1g1

2 y aremutual inverses. (2) Suppose g1H g2H = . Then x g1H g2H, so x = g1h1 = g2h2. Let g1h3 g1H. Theng1h3 = g1h1h

11 h3 = g2h2h

11 h3 g2H.

Def: If H G, let [G : H], called the index of H in G be defined to be the number of distinct left cosets of H in G.Thm: (Lagrange) Let G be a finite group, H G. Then |G| = |H|[G : H].

Proof. Let m = [G : H], say {g1H, , gmH} is the set of distinct left cosets of H. By (3) above, G = mi=1giH. By(2), giH gjH = for i = j. So |G| = |g1H| + + |gmH| = m|H| by (1).

Cor: 7 Let G be a finite group. Then |a| | |G| for all a G.

2.3 Group Actions

Def: Let G be a group and A a set. An action of G on A is a function G A A ((g, a) g a) such that (1) forall a A, g1, g2 G, (g1g2)a = g1(g2a) and (2) for all a A, 1 a = a.

Def: Let G be a group acting on a set A. For a A, the stabilizer Ga of a is Ga := {g G : ga = a}. Note thatGa G. The kernel of the action is ker = aAGa. The action is called faithful if ker = {1}. The action is freeif Ga = {1} for all a A. For a A, the orbit Oa of a is Oa := {ga : g G}. The action is called transitive ifOa = A for some a A.

Lma:8 Let : G G be a group homomorphism. Then is 1-1 iff ker = 1.

Prop: Let A and B be sets such that |A| = |B|. Then SA = SB.Def: A homomorphism : G SA is called a permutation representation for .Rmrk: G acts on A iff for : G SA the kernel of the action = ker .Rmrk: G acts faithfully on A if and only if : G SA is injective. In this case G = (G) SA.Thm: (Cayleys Theorem) Let G be a group. Then G is isomorphic to a subgroup of SG. In particular, if |G| = n, then

G

= H

Sn.

E.g.:Examples of Group Actions:

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2 INTRO TO GROUPS Algebra Study Guide2.4 Quotient Groups and Homomorphisms

Prop: Let H G, A = {xH : x G}. Then G acting on A by left multiplication gives a group homomorphism : G SA = Sn (where n = [G : H]) given by (g) = g : A A (where g(xH) = gxH) having kernel

xGxHx1.

IfxGxHx1 = 1 then G is isomorphic to a subgroup of Sn, where n = [G : H].Prop:

9 Let G act on A and let a, b A. Then Oa = Ob or Oa Ob = . Thus, A is the disjoint union of the distinctorbits. If |A| < then |A| =

ni=1

|Oai |, where Oa1 , , Oan are the distinct orbits of A.Cor: If G acts transitively on A, then Oa = A for all a A. Therefore, G acts transitively on A iff a, b A there

exists g G such that ga = b.Prop: Let G act on A. Let a A. Then there exists a bijection from {gGa : g G} Oa given by gGa ga.

Hence |Oa| = [G : Ga].Cor: G acts on A, where |A| < . Let Oa1 , , Oan be the distinct orbits. Then |A| =

ni=1

[G : Gai ].

E.g.: (The Class Equation) Let G be a finite group. Let G act on itself by conjugation, i.e., G = A, g a = gag1. LetOa1 , , Oan be the distinct orbits of G under this action. Oa1 =

ga1g

1 : g G

is called the conjugacy class

of a1. Then Ga = g G : gag1 = a = CG(a). Therefore, |G| =n

i=1[G : CG(ai)]. Notice that [G : CG(a)] = 1 ifand only if CG(a) = G if and only if a Z(G). Therefore,

|G| = |Z(G)| +

[G : CG(a)],

where a runs through the distinct conjugacy classes and a / Z(G).Prop: Let G be a group with |G| = pn, where p is prime. Then Z(G) = 1 (follows from the fact that [G : CG(a)] = pl,

l 1).

2.4 Quotient Groups and Homomorphisms

Def: A subgroup H of G is normal in G if xHx1 = H for all x G.Rmrk: H is normal in G iff xHx1 H for all x G.Rmrk: Suppose H = S. Then H G iff gsg1 H for all s S and for all g G.Rmrk: If H Z(G) then H G.Prop: Let : G1 G2 be a group homomorphism. Then ker G1.Rmrk: Let H G, A = {gH : g G}. Let G act on A as usual. Recall, GxH = xHx1. Therefore

xG xHx

1 G(this is the largest normal subgroup of G contained in H). In particular, if [G : H] = n then there exists a grouphomomorphism : G Sn = SA where kernel = xGxHx1.

Prop: 10 Let : G1 G2 be a group homomorphism, let K = ker . Then there exists a bijection {xK : x G1} im (xK (x)). In particular, if |G2| < , [G : K] = |im| divides |G2|.

Prop: Let G be a finite group and p the smallest prime divisor of |G|. Suppose there exists a subgroup H of G suchthat [G : H] = p. Then H G. STUDY THIS PROOF

Cor: If [G : H] = 2. Then H G.Cor: If |G| is odd and [G : H] = 3 then H G.Def: A group G

= 1 is called simple if it has no nontrivial normal subgroups.

Prop: Let G be a group and H G. Define an operation on the left cosets {xH : x G} by (xH) (yH) := (xy)H.This operation is well-defined if and only if H G.

Prop: Suppose H G. Then the set of cosets ofH is a group under the operation defined above. This group is denotedG/H and is referred to as a quotient group.

Rmrk: If |G| < then |G/H| = |G|/|H| (Lagrange).Prop:

11 Let : G A be a surjective group homomorphisms. Then there exists a bijective correspondence between{subgroups of G containing ker } {subgroups of A} given by K (K). Furthermore, K G (K) A.

Prop: Let G be a group and H G. Consider the function : G G/H given by (g) = gH. Then is a surjectivegroup homomorphism with ker = H.

Cor: Let H G, : G G/H be the canonical map. Then there exists a bijection between {K : K G, H K}and {A : A G/H} given by K K/H. Furthermore, if H K G, K G K/H G/H.

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2 INTRO TO GROUPS Algebra Study Guide 2.5 Direct Products

Thm: Let : G A be a group homomorphism and let H G with H ker . Then there exists a group homomor-phism : G/H A given by gH (g). Furthermore, im = im and ker = ker /H.

Cor: (1st Isomorphism Theorem) Let : G A be a group homomorphism. Then G/ ker = im.Cor: Let H K G and H G, K G. We already know that K/H G/H. Then (G/H)/(K/H) = G/K via the

map (gH)/(K/H) gK. (We have ker = K/H in the induced map.)Rmrk:

12

Let : G A be a group homomorphism. Let x G with |x| < . Then |(x)| | |x|.Thm: (Cauchys Theorem) Let G be a finite group and p a prime such that p | |G|. Then G has an element of order p.Proof. Case 1. G is abelian. Use induction on |G| = n. If n = p, then G is cyclic of order p. If n > p, choosex G, x = 1. Let H = x. Ifp||H|, p||x|, say |x| = ps. Then |xs| = p. Ifp |H|, then p|[G : H] = |G|/|H|. Since Gis abelian, H G. Therefore, p | |G/H| = [G : H]. As H = 1, |G/H| < |G|. By induction, G/H has an element oforder p, say yH. Therefore p | |y|, and hence |y| = tp, then |yt| = p.Case 2. G arbitrary. IfZ(G) = G, then G is abelian and were done. So assume Z(G) G. Ifp | |Z(G)|, then Z(G)has an element of order p (as Z(G) is abelian). Ifp |Z(G)|, use the class formula: |G| = |Z(G)|+

x/Z(G)[G : CG(x)].

Since p||G|, p |Z(G)|, there exists x / Z(G) such that p [G : CG(x)] = |G|/|CG(x)|. Therefore, p | |CG(x)|. Asx / Z(G), |CG(x)| < |G|. By induction on |G|, CG(x) has an element of order p. Hence, G does.

Lma: Let a, b G, |a| < , |b| < . Suppose (1) a b = {1} and (2) ab = ba. Then |ab| = lcm(|a|, |b|).Lma: Let H1, H2 G. Suppose H1 H2 = 1. Then ab = ba for all a H1, b H2.Cor: Suppose |G| = pq and let a be an element of order p, b be an element of order q, p < q, p, q primes. Suppose

a G and b G. Then G is cyclic.Proof. a b a , b and a b = 1 by Lagrange. So ab = ba and hence |ab| = |a| |b| = pq.

E.g.: Let G be a group, |G| = 6. Then G =Z/6Z or G = S3.Proof. Suppose G is not cyclic. By Cauchy there exists a, b G such that |a| = 2, |b| = 3. Note if ab = ba then|ab| = 6 which implies G is cyclic (). Let H = a, K = b. As [G : K] = 2, K is normal. Also, H K = 1 (asgcd(|H|, |K|) = 1). By the lemma, A = {xH : x G}. |A| = [G : H] = 3. Let G act on SA as usual. This gives agroup homomorphism : G S3 and ker is the largest normal subgroup of G contained in H. As |H| = 2 is notnormal, ker = 1. Therefore is 1-1. As |G| = |S3| = 6, is onto.

2.5 Direct Products

Def: 13 Let G be a group, A, B G (nonempty). Define AB := {ab : a A, b B}.Prop: Let H, K G. Then |HK| |H K| = |H| |K|.Prop: H, K G. Then HK G HK = KH.Cor: If H, K G and either H or K is normal in G, then HK G.Def: Let G be a group, H1, , Ht G. Suppose (1) Hi G for all i, (2) G = H1H2 Ht = {h1h2 ht : hi Hi},

and (3) Hi H1 Hi Ht = 1 for all i. We say that G is the (internal) direct product of H1, , Ht.Prop: Suppose G is the internal direct product of H1, , Ht. Then G = H1 Ht (externally).Prop: Let G be a group of order p2, where p is a prime. Then G = Cp2 or G = Cp Cp. (For proof, suppose not cyclic

and find two distinct subgroups of order p.)Thm:

14 Let H, K G, K G. Then (1) H K H and (2) HK/K= H/H K.

2.6 Fund. Thm. of FG Abelian Groups

Thm:15 Let G be a finitely-generated abelian group. Then there exists unique integers r 0 and p1, , pk primes,

m1, , mk > 0 such that p1 pk and G = Cr Cpm11

Cpmkk

, so G is finite iff r = 0. If |G| < then|G| = pm11 pmkk . Special Case: Suppose |G| = pn (p a prime) is abelian. Then G = Cpm1 Cpmk , where

mi = n.

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2 INTRO TO GROUPS Algebra Study Guide 2.7 Automorphisms

2.7 Automorphisms

Def: Let G be a group. An isomorphism : G G is called an automorphism of G. Let Aut(G) denote the setof automorphisms, which is a group under function composition.

Def: Let G be abelian, m Z. Then the map m : G G given by m(g) = gm is a group homomorphism. Note thatm

l = ml.

Prop: Let G = Cn, n 1. Then Aut(Cn) = {m : Cn Cn : gcd(m, n) = 1}= (Z/nZ).Prop:

16 Cm Cn = Cmn gcd(m, n) = 1.Cor: If m1, , mk are pairwise relatively prime positive integers then Cm1 Cmk = Cm1mk .Def: Let G be a group. Define the exponent of G by e(G) = inf

d 1 : gd = 1g G. If |G| < , then by

Lagrange, e(G) |G|.Notation: Let G be a group, x G. Define x : G G by x(g) = xgx1.Def: The map x is a homomorphism for all x G, and is called a inner automorphism of G. Define the group of

inner automorphisms to be Inn(G) = {x : x G}.Exercise: Inn(G) Aut(G).Note: Aut(G)/ Inn(G) is called the group of automorphisms.Exercise: Inn(G) = G/Z(G) (isomorphism given by x xZ(G)).Def: Let H G. H is called a characteristic subgroup of G, denoted Hchar G if (H) = H for all Aut(G).Rmrk: Hchar G

(H)

H for all

Aut(G).

Rmrk: Hchar G = H G.Rmrk: If H is the unique subgroup of order |H| then Hchar G.Rmrk: Z(G) char G.Rmrk: In general, K H and H G = K G.Rmrk: Kchar H and Hchar G implies Kchar G.Rmrk: Kchar H and H G = K G.Def: Let G be a group, x, y G. The element denoted [x, y] := xyx1y1 is called the commutator of x and y. The

subgroup of G generated by all commutators is called the commutator subgroup (sometimes denoted G or [G, G]).So G =

xyx1y1|x, y G .

Rmrk: G char G.Rmrk:

17 Let G be a group, H G. (1) If H G then H G and G/H is abelian. (2) The converse is also true. IfH G and G/H is abelian then H G.

2.8 Sylows Theorems

Def: Let G be a finite group, p a prime such that p | |G|. Suppose pn | |G| but pn+1 | |G|. A p-subgroup of G is asubgroup ofG of order p for some 0. Let Sylp(G) denote the set of Sylow p-subgroups ofG. Let np = | Sylp(G)|.

Thm: (Sylow I) If p | |G| then G has a subgroup of order p.Proof. Induction on |G| = n. If n = 1, the statement is vacuously true. Then the class equation gives |G| =|Z(G)| +

li=1

[G : CG(gi)] where CG(gi) = G.Case 1. p | |Z(G)|. By Cauchys Theorem there exists a Z(G) where |a| = p. Let K = a G (as a Z(G)).

Consider G = G/K. Then |G|/p < |G|. Also, p1 | |G|. By the induction hypothesis G has a subgroup of orderp1. Such a subgroup has the form H/K, where K H G. Then |H/K| = |H|/|K| = p1, which implies

|K

|= p =

|H

|= p.

Case 2. p |Z(G)|. Then the class equation implies there exists some i for which p [G : CG(gi)]. Then p |G|/|CG(gi)|implies p | |CG(gi)|. But |CG(gi)| < |G|. Since [G : CG(gi)] > 1, CG(gi) has a subgroup of order p (byinduction).

Thm: (Sylow II) If G is a finite group, pn | |G|, pn+1 |G|. Let P Sylp(G), np = | Sylp(G)|. Then1. np = [G : NG(P)] for all P Sylp(G). In particular, Sylp(G) = {P} if and only if P G (i.e., G = NG(P)).2. np 1 mod p3. Any two Sylow p-subgroups of G are conjugate.

4. Any p-subgroup of G is contained in a Sylow p-subgroup.

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2 INTRO TO GROUPS Algebra Study Guide 2.9 Solvable Groups

2.8.1 Applications

Prop:18 Let |G| = pq where p < q primes and p q 1. Then G is cyclic.

Proof. Let np | |G|/p = q implies np = 1 or q and nq|p implies nq = 1 or p. And np 1 mod p and nq 1 mod q.As p < q, p 1 mod q. Therefore nq = 1. As q 1 mod p, np = 1. Therefore both the Sylow p-subgroup P andthe Sylow q-subgroup Q are normal. Therefore |P Q| = |P| |Q||P Q| = pq. So G = P Q, P Q = 1, P G, Q G,andhence G = P Q = Cp Cq = Cpq.

Exercise: Suppose |G| = p2q where p, q are primes. Prove G is not simple.E.g.:

19 No group of order 144 is simple.Prop: Let G be a finite group. G is cyclic if and only if G has a unique subgroup of order d for every positive divisor d

of |G|. Study Proof !

2.9 Solvable Groups

Def: Let G be a group. A normal series for G is a sequence of subgroups

1 = Hn Hn

1

H0 = G.

Note: G, nontrivial, is simple iff its only normal series is the trivial normal series. The groups Hi/Hi1 are calledthe factor groups of the series. The length of the series is the number of nontrivial factor groups.

Def: A composition series for G is a normal series such that all the factor groups are simple.E.g.:Easy composition series: {1} A5.E.g.: {1} {1, (12)(34)} V4 A4 S4 (A4/V4 = C3, S4/A4 = C2, V4/H= C2).Def:

20 A subgroup H of G is called a maximal normal subgroup if H = G, H G and there are no normalsubgroups properly between H and G (equivalently, H G and G/H is simple).

Prop: Any finite group has a composition series.Thm: (Jordan-Holder Theorem) Suppose G has a composition series. Then any two composition series for G have the

same length, and the set of factor groups are the same up to isomorphism.Def: A refinement of a normal series for G is just another normal series for G obtained by inserting more subgroups

between the terms of the original.Rmrk: If G has a composition series, any normal series has a refinement which is a composition series.Note: A finite abelian group is simple iff G has prime order.Note: The factor groups for any composition series for an abelian group have prime order.Def: A solvable series for a group G is a normal series in which the factor groups are all abelian, i.e., 1 = Hn

H0 = G such that Hi/Hi+1 is abelian. A group is solvable if it has a solvable series.Prop: Suppose |G| < . Then G is solvable iff all the factor groups in the composition series have prime order.Exercise: Prove this!

Def: 21 Define G(n) := G(n1) where G(0) = G (recall G is the commutator subgroup of G). Recall that G char G.Since char is transitive, G(n) char G for all n 0. In particular, G(n1)/G(n) is abelian. The series G(2) G(1) G(0) = G is called the derived series.

Prop: G is solvable iff G(n) = 1 for some n 1 (i.e., the derived series is normal).

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Lma: Let : G A be a surjective group homomorphism. Then (G(n)) = A(n) for all n 0.Cor: Suppose H G. Then (G/H)(n) = G(n)H/H.Thm: Let G be a group, H G. (1) If G is solvable, so is H. (2) If H G then G is solvable iff H and G/H are

solvable.Prop: Any p-group is solvable. (For proof, use induction and note that Z(G) and G/Z(G) is solvable.)

3 Ring Theory

3.1 Intro to Ring Theory

Def:22 Let R be a ring. Let a R. Then a is called a zero-divisor if there exists nonzero b R such that either

ab = 0 or ba = 0. IfR is commutative and has no nonzero zero divisors (nonzero-divisors or NZDs) then R is calledan integral domain (or just a domain).

Def: Let R be a ring with 1. Let a R. An element b R is called a left-inverse for a if ba = 1 and b is aright-inverse for a if ab = 1.

Rmrk: Let a R. Suppose b is a left-inverse for a and c is a right-inverse. Then b = c.E.g.: Let (G, +) be an abelian group. Let Hom(G, G) = {f : G G : f a group homom.}(= End(G)). Define +

and

on Hom(G, G) as follows: given f, g

Hom(G, G), (f + g)(x) = f(x) + g(x) and (f g)(x) = (f

g)(x) for all

x G. Then Hom(G, G) is a ring with 1.E.g.: Let G = Z, and R = Hom(G, G). Let f, g R be given by f(a1, a2, a3, ) = (0, a1, a2, ) and g(a1, a2, a3, ) =

(a2, a3, ). Then gf = 1 but f g((1, 0, 0, )) = (0, 0, ) so f g = 1. Thus, f is left-invertible but not right-invertibleand vice versa for g. (And actually, if n R is given by n((a0, a1, )) = (na0, 0, 0, ), (g + n)f = 1, so f hasinfinitely many left inverses.)

Rmrk:23 A unit in a ring cannot be a zero-divisor.

Def: A division ring is a ring R = {0} with 1 such that every nonzero element is a unit. A field is a commutativedivision ring.

Prop: Let n 2. Then n is prime iffZ/nZ is a field iffZ/nZ is a domain.

3.2 Polynomial Rings

Rmrk: If R is a domain, then R[x] is a domain.Thm: 24 (Division Theorem) Let R be commutative with 1, and x a variable. Let f(x), d(x) R[x], where the

leading coefficient of d(x) is a unit in R. Then there exist unique polynomials q(x), r(x) R[x] such that f(x) =d(x)q(x) + r(x) and deg r < deg d.

Def: A polynomial is called monic if its leading coefficient is a unit. So the division theorem works whenever youdivide by a monic polynomial.

Cor: Let f(x) R[x], r R. Then f(r) = 0 iff (x r)|f(x).

3.3 Ideals and Ring Homomorphisms

Def: A map : R S (R, S rings) is called a ring homomorphism if(a+b) = (a)+(b) and (ab) = (a)(b)for all a, b R.

E.g.: 25 Suppose : R S is a nonzero ring homomorphism. Suppose that R, S have identity and S is a domain.Then (1R) = 1S.

Def: 26 Let R be commutative with 1. A nonempty subset I of R is an ideal if (I, +) (R, +) and for all r R andi I, ri I.

Def: Let R be commutative with 1. Let A R. Define the ideal of R generated by A by (A) := I idealAR

I.

Exercise: (A) = {r1a1 + + rnan : ri R, ai A, n N}.Exercise: Consider the ideal (2, x) Z[x] is not principal.Def: Let R be commutative with identity, I R an ideal. Let R/I denote the set of left cosets of (I, +) in (R, +).

Define +, on R/I by (1) (a + I) + (b + I) = (a + b) + I and (2) (a + I) (b + I) = (a b) + I.Prop:

27 Let : R S be a homomorphism of commutative rings. Then R/ ker = im (as rings).Prop: Suppose I J are ideals ofR. Then J/I is an ideal of R/I.

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3 RING THEORY Algebra Study Guide 3.4 Operations on Ideals

Prop: (R/I)/(J/I) = R/J.Prop: Let Ibe an ideal ofR. Then there exists a bijection, inclusion-preserving correspondence between {J : J an ideal of R

and {ideals of R/I} given by J J/I = (J) and K 1(K), where : R R/I is the canonical surjectivehomomorphism.

3.4 Operations on IdealsDef: Let I, J be ideals ofR. Then we define I+J := {a + b : a I, b J}. This is the smallest ideal ofR containingI J (which is not an ideal). Further, IJ = {ij} i I, jJ =

n

k=1

ikjk : ik I, jk J

, which is the smallest

ideal containing {ij} i I, j J.Thm: (Chinese Remainder Theorem) Let R be commutative with identity, I1, , In ideals of R such that Ii + Ij = R

for all i = j. Then (1) I1 In = I1I2 In and (2) there is an isomorphism of rings : R/(I1 In) R/I1 R/In given by r + I1 In (r + I1, , r + In).

Thm: 28 Let d1, , dn be pairwise relatively prime positive integers and a1, , an any integers. Then there existsx Z such that x ai mod di for i = 1, , n.

Rmrk: If R, S are rings with 1, then (R S)= R S as groups.Def: Let R be commutative with 1. An ideal m of R is called maximal ifm = R and the only ideals containing m are

m and R.Prop: An ideal I of R is maximal iff R/I is a field.Rmrk: Let : R F is a surjective ring homomorphism such that 0, F a field. Then ker is a maximal ideal, as

R/ ker is isomorphic to a field.Rmrk: Let m, n be distinct maximal ideals of a commutative ring R. Then m + n = R.Def: An element r R is called nilpotent if rn = 0 for some n N. R is called reduced if there are no nonzero

nilpotents.

3.5 Zorns Lemma

Def: 29 Let A be a set. A partial order on A is a binary relation with the following properties: for all a,b,c A,(1) a a; (2) a b and b a iff a = b; (3) a b, b c implies a c. Additionally, is a total order on A if for alla, b A, a b or b a.

Def: Let (A, ) be a poset, and B a subset of A. An element x A is an upper bound for B if b x for all b B.An element b B is maximal if for every y B with y b implies y = b.

Thm: (Zorns Lemma) Let (A, ) be a nonempty poset and suppose every totally ordered subset of (A, ) has an upperbound in A. Then A is a maximal element.

Prop: Let R = 0 be commutative with 1 and I = R an ideal. Then there exists a maximal ideal m of R containing I.Def: Let R be commutative. An ideal P of R is called prime if P = R and whenever ab P, then either a P or

b P.Rmrk: P is prime iff R/P = 0 is a domain.Rmrk: Let : R S be a commutative ring homomorphism. Suppose = 0 and S is a domain. Then ker is a prime

ideal ofR.Rmrk: R is a domain iff{0} is prime. R is a field iff{0} is maximal.Rmrk: If P is prime and P IJ then P I or P J.

3.6 Quotient Fields

Def: Given a domain R, the smallest field containing R is called the quotient field (or field of fractions) of R,denoted by Q(R).

Prop: Let R be a domain and suppose R F, F a field. Then there exists an injective field homomorphism f : Q(R) F that fixes R.

3.7 Euclidean Domains

Def: 30 Let R be a domain. R is called a Euclidean domain if there exists a function : R\{0} N0 = {0, 1, 2, }such that for all a, b R with b = 0 there exists q, r R such that a = bq+ r with r = 0 or (r) < (b).

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3 RING THEORY Algebra Study Guide 3.8 Factorization Conditions

Thm: Z[i] is a Euclidean domain with : Z[i] N0 by a + bi a2 + b2.Thm: Z[i] = {a + bi : a, b Z} is a Euclidean Domain with : Z[i] N given by a + bi a2 + b2 = (a + bi)(a bi).

Proof. Note that, for all , Z[i], () = ()(). Suppose = 0. Then

=

= x + yi, for x, y Q.Let a, b Z and u, v Q such that x = a + u and y = b + v for |u|, |v| 1/2. Then observe = (x + yi) =(a + bi) + (u + vi). Let = (u + vi) and note = (a + bi) implies Z[i].

Claim: If we show () < (), were done: = (a + bi) + .Notice that can be extended to a function on Q[i] Q0 by x+yi x2 +y2. Then () = ((u+vi)) = ()(u+vi),

and (u + vi) = u2 + v2 14 + 14 = 12 . Thus, () 12 () < ().Def: An integral domain in which every ideal is principal is a PID.Thm: Euclidean Domains are PIDs.

Proof. Let R be a ED under and let I R be an ideal. IfI = {0}, were done. Assume I = {0}. Choose d I\{0}such that (d) (a) for all a I\ {0}. Claim: I = (d). Pick a I. Then a = dq+ r for some q, r R with r = 0or (r) < (d). Since r = a dq I, we may not have (r) < (d), as (d) (b) for all b I. Thus, r = 0, anda (d).

Rmrk: The domain R = Z[5] is not a PID (the ideal (3, 2 + 5) is not principal).

Rmrk:31 The domain R = Z

1+

192

is a PID that is not a Euclidean Domain.

Prop: Let R be a PID and I = (p) = 0 a prime ideal. Then I is maximal.

3.8 Factorization Conditions

Def: Let R be a domain. An element r R is called irreducible if r is not a unit and whenever r = ab, then oneof a or b is a unit. An element p R \ {0} is called prime if (p) is a prime ideal (i.e. if p|ab then p|a and p|b).

Prop: Let R be a domain. Then any prime element is irreducible.Rmrk: 2 R = Z[5] is irreducible but not prime.Def:

32

Let R be a domain. Then R is a UFD if (1) every nonunit of R is a product of irreducibles and (2) ifa1a2 am = b1 bn, where ai, bj is irreducible, then m = n and after reindexing ai = uibi for ui R.

Def: Let R be a commutative ring. R is said to satisfy the ascending chain condition (ACC) on principal ideals ifgiven any ascending chain of principal ideals

(a1) (a2) there exists n 1 such that for all k n, (ak) = (ak+1) i.e.,

(a1) (a2) (an) = (an+1) = (an+2) = .Rmrk: Let R be a commutative ring. Then R satisfies ACC on principal ideals iff every nonempty set of principal

ideals of R has a maximal element in .Thm: Let R be a domain. Then R is a UFD if and only if R satisfies ACC on principal ideals and every irreducible

element is prime.

Prop: 33 Let R be a PID. Then R is a UFD.Def: Let R be a domain, a, b R. An element d R \ {0} is a greatest common divisor (gcd) of a, b if (1) d|a, d|b

and (2) if e|a, e|b, then e|d.Rmrk: If d1, d2 are gcds of a and b then (d1) = (d2).Prop: Let R be a UFD. Then gcd(a, b) exists for all a, b R.Def: Let R be a domain, a, b R both nonzero. The least common multiple (lcm) of a and b is an element

m R \ {0} such that (1) a|m, b|m, and if a|m and b|n, then m|n.Rmrk: Any two lcms of a and b are associate.Exercise: Let R be a UFD, a, b R \ {0}. Then (a) (b) = (lcm(a, b)).Exercise: Find an example of a domain R and nonzero elements a, b such that gcd(a, b) does not exist. (R = Z[

5],a = 2(1 +

5), b = 6.)

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4 MODULES Algebra Study Guide

Def:34 Let R be a UFD, x a variable. Let f(x) = anx

n + + a0 R[x] \ {0}. The content of f(x) is defined byc(f) = gcd(an, , a1, a0). f(x) is called primitive if c(f) = 1.

Rmrk: c = c(f) implies f = cf1, where f1 is primitive.Rmrk: If d R, d|c(f), then f /d R[x] and c(f/d) = c(f)/d.Rmrk: If f = cg where c R, then c|c(f).Rmrk:

Suppose f = gh, f , g , h R[x] then c(g)c(h)|c(f).Thm: (Gauss Lemma) Let R be a UFD, f = gh R[x] \ {0}. Then c(f) = c(g)c(h).Cor:

35 The product of primitive polynomials is primitive.Cor: Let R be a UFD, F = Q(R). Let f(x) R[x], g(x) R[x] primitive. Suppose g|f in F[x] (i.e. f = gh for

h F[x]). Then g|f in R[x].Cor: Let R be a UFD and F = Q(R), f(x) R[x]. Suppose f(x) = g(x)h(x), where g, h F[x]. Then there exist

a, b F such that ag,bh R[x] and f = (ag)(bh).Cor: Let R be a UFD and F = Q(R). Let f(x) R[x] be primitive. Then f(x) is irreducible in R[x] if and only iff(x)

is irreducible in F[x].Thm: Let R be a UFD. Then R[x] is a UFD.Cor: If k is a field (or even just a UFD), k[x1, , xn] is a UFD for all n.Rmrk:

36 If deg f(x) > 1 and f(x) is irreducible then f(x) has no roots in F.Rmrk: (x2 + 1)2 Q[x] is reducible but has no roots.Rmrk: If deg f = 2 or 3 then f is irreducible if and only if f has no roots in F.Thm: (Rational Root Theorem) Let R be a UFD and f(x) = anxn + a0 R[x]. Suppose F = Q(R) is a root of

f. Then = c/d where c|a0 and d|an, for c, d R.Thm: (Eisensteins Criterion) Let R be a UFD and f(x) = anx

n + + a0 R[x] and suppose there exists an irreducibleelement R such that (1) an; (2) |ai for 0 i n 1, and (3) 2 a0. Then f(x) is irreducible in F[x],where F = Q(R).

4 Modules

4.1 Intro to Module Theory

Def: 37 Let R be a ring. A left R-module is an abelian group (M, +) together with an R-action, i.e., a map fromR

M

M ((r, m)

rm) with the following properties: For all r, s

R and m, n

M,

r(m + n) = rm + rn (r + s)m = rm + sm (rs)m = r(sm) 1m = m.

E.g.: Let R be a ring

{0}, R Let R be commutative. Any ideal I of R is a left (or right) R-module.

R a ring, n

N. Define the set Rn :=

{(x1,

, xn) : xi

R

}. Rn is a left R-module, and is called a free left

R-module of rank n.

Let R = Z. Any abelian group (G, +) is a Z-module via the action of adding elements up m or m times. Let R be a ring and n N. Let S = Mn(R). Then S acts on Rn via matrix multiplication. Let : R S be a ring homomorphism such that (1R) = 1S. Then S is a left R module via the action

r s = (r)s. If R S and 1R = 1S, then S is a left R-module via r s = rs.

Let R be commutative and I and ideal of R. There is a canonical surjective ring homomorphism : R R/Igiven by (r) = r + I. Thus, R/I is an R-module.

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4 MODULES Algebra Study Guide 4.1 Intro to Module Theory

Def: Let F be a field. An F-vector space is a left F-module.Def: Let R be a ring and let M be a left R-module. A left R-submodule of M is a subgroup N ofM such that N is

a left R-module via the action of R on M restricted to N.Prop: Let M be a left R-module and N a nonempty subset of M. Then N is a left R-submodule of M if for all r R

and for all x, y N, x + ry N.E.g.:

Let R be a ring. Then N = (r1, r2, r3) R3 : r1 + 2r2 r3 = 0 is a left R-submodule of R3.E.g.: Let R be a commutative ring and M a left R-module. Let r R and N = {m M : rm = 0}. Then N is a leftR-module.

Rmrk:38 Let : R S be a ring homomorphism with (1R) = 1S. Then any left S-module M is also a left R-module

in a natural way: r R, m M, r m = (r) = m.E.g.: : Z Q. Then Q2 is a Z-module in a natural way.E.g.: Let E be a field and F a subfield. Then : F E is a field homomorphism via the inclusion map. Then E is an

F-vector space and F is an F-subspace of E.Def: Let M, N be R-modules. A function f : M N is an R-module homomorphism (or an R-linear map) if

f(x + y) = f(x) + f(y) for all x, y M. f(rx) = rf(x) for all r R, x M.

Rmrk:

Given an R-linear map f from M to N, ker(f) := {m M : f(m) = 0} and im(f) := {f(m) : m M}. ker(f)and im(f) are submodules of M and N, respectively.Def: Let M be an R-module and {Mi}iI a collection of submodules of M. Define

iIMi :=

mi1 + + min : n N, mij Mij , for j = 1, , n

.

TheniI

Mi is the smallest submodule of M containingiI

Mi. In the case I = {1, , n}, we write M1 + M2 + + Mn =

iI

Mi.

Def: If S is a nonempty subset of M, then RS :=sS

Rs.

Def: Let M be an R-module. The annihilator of M is defined by

AnnR(M) := {r R : rm = 0 m M}.Note: AnnR(M) is a two-sided ideal of R.

Rmrk: Let R be commutative and M an R-module and I an ideal ofR with I AnnR(M). Then M is an R/I module.Def: 39 Let {M1, , Mn} be a collection of R-modules. The external direct product is defined to be

M1 M2 Mn := {(m1, , mn) : mi Mi}.The general case: Let {Mi}iI be a collection of R-modules. Then

iIMi := {(mi)iI : mi Mi, mi = 0 for all but finitely many i}.

Def: Let {Mi}iI be a set ofR-modules. We define the external direct product to beiI

Mi := {(mi) : mi Mii}.

Thus,iI

Mi iI

Mi, with equality if |I| < .Prop: Let M be an R-module and {Mi}iI a collection of submodules of M. Then M =

iI

Mi if and only if (1)

M =iI

Mi and (2) for all i I, Mi

j=iMj

= (0).

Prop: Let {Mi}iI be submodules of M. If M = iIMi (internal) then M= iIMi (external).Def: Let M be an R-module. A subset S of M is called an R-basis for M if

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5 VECTOR SPACES Algebra Study Guide 4.2 Module Isomorphism Theorems

1. M = RS (i.e., S generates M)

2. S is R-linear independent (i.e., if r1s1 + + rnsn = 0 (with ri R, si S and distinct) then ri = 0 for all i).If M has an R-basis, then M is called a free R-module.

Rmrk: If M has a basis of n elements, then M= Rn.Rmrk: If R is not commutative, then a free module may have bases of different cardinalities. If R is commutative,

then any two bases for a given free module have the same cardinality. This cardinality is called the rank of the freemodule.

Rmrk: Suppose R is commutative. Let I be an ideal. Then I is free if and only if I = (0) or I = Rs = (s), where s is anon-zero divisor.

Rmrk: Let R be commutative and I an ideal of R. Then R/I is free if and only if I = R or I = (0).

4.2 Module Isomorphism Theorems

1. Let : M N be an R-module homomorphism. Then M/ ker = (M).2. Let A, B be submodules of M. Then (A + B)/B = A/(A B).3. Let L

M

N be R-modules. Then

(N/L)/(M/L) = N/M.

4. Let N M be R-modules. Then there exists a one-to-one correspondence between the set of R-submodules ofM containing N and the R-submodules of M/N (i.e., Q Q/N).

5 Vector Spaces

Let R = F a field. Then a set V is an F-module if and only if V is an F-vector space40.Def: If V is an F-vector space and S V, then F S = spanF(S). If spanF(S) = V, we say S spans V.Prop: Let F be a field and V and F-vector space. Let T S be subsets of V such that (1) T is linearly independent

and (2) S spans V. Then there exists a basis with T S.Cor: Let V be an F-vector space for a field F. Then

1. V has a basis.

2. Every linearly independent subset ofV is contained in a basis.

3. Every spanning set for V contains a basis.

Thm: (Replacement Theorem) Let V be an F-vector space. Let S be a spanning set for V and T a linearly independentsubset of V. For every (distinct) t1, , tn T there exists s1, , sn S (distinct) such that (S\ {s1, , sn}) {t1, , tn} spans V.

Cor: Let V be a finitely-generated F-vector space. Let S be a finite spanning set for V and T a linearly independentsubset of V. Then |T| |S|.

Cor: Let V be a finitely-generated F-vector space. Then (1) V has a finite basis and (2) any two bases have the samenumber of elements.

Def: Let V be an F-vector space and dim V = || for any basis of V (finite or infinite).Cor: Let V be a finite-dimensional vector space and W a subspace ofV. Then dim W dim V and W = V if and only

if dim W = dim V.Prop:

41 Let F be a finite field. Then |F| = pn for p Z a prime and n 1.Prop: Let : V W be a linear transformation of F-vector spaces. Suppose dim V = n < . Let K = ker and

C = im. Then dimF V = dimF K+ dimF C.Def: The rank of is rank() := dim C and the nullity of is nullity() := dim K.Cor: Let : V W be an F-linear transformation. Suppose dim V = dim W < . Then the following are equivalent:

1. is an isomorphism.

2. is injective.

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6 MATRIX ALGEBRA Algebra Study Guide

3. is surjective.

4. takes a basis of V to a basis of W.

The proof follows from the rank + nullity theorem.

6 Matrix Algebra

6.1 Intro to Matrices

Def: Let R be a commutative ring and m, n N. An m n matrix over R is an array (aij) with m rows, ncolumns, and mnth entry amn.

Let Mmn(R) denote the set of m n matrices over R. We add them in the usual way. Then (Mmn(R), +) is anabelian group.

Define an R-action on Mmn(R) by r (aij) = (raij) for all r R. Thus, Mmn(R) is an R-module.So for 1 i m and 1 j n let Eij denote the m n matrix with 1 = aij and 0 elsewhere.Then {Eij : 1 i m, 1 j n} is an R-basis for Mmn(R). Then Mmn(R) = Rmn.Let A = (aij) be an m n matrix and b = (bj1) be an n 1 matrix. Define

A b = c1...

cm

,

where ci =n

j=1aijbj1.

Prop: Let R be a commutative ring, and M an R-module with B = {e1, , cn} M. The following are equivalent:1. B is a basis for M.2. Given any R-module N and x1, , xn N, there exists a unique R-module homomorphism : M N such

that (ei) = xi for all i.

6.2 Transition Matrices

Def: Let R be a commutative ring, and let M and N be R-modules. Define HomR(M, N) = {f : M N : f R-linear}.Rmrk: HomR(M, N) is an R-module with R-action: for r R, f : M N, define r f : M N by m rf(m).Rmrk: If M = N, then HomR(M, M) is a ring with identity (and multiplication given by composition). HomR(M, M)

is usually denoted EndR(M) (the endomorphism ring).Prop: Let R be a commutative ring, M an R-module, = {e1, , en} M. The following are equivalent:

1. is a basis for M

2. Given any R-module N and x1, , xn N there exists a unique R-homomorphism : M N such that(ei) = xi for all i.

Def: Let R be commutative, M, N free R-modules of ranks m and n, respectively. Let ={

1,

, m}

andE

={E1, , En} be bases for M, N, respectively. Given an R-homomorphism : M N we have for 1 j m, thereexist unique elements aij R such that (j) =

mi=1

aijEi. The n m matrix (aij) is called the matrix of withrespect to and Eand denoted by ME ().

Prop: Let R be commutative, and M and N free R-modules of ranks m, n respectively. Fix bases B, Efor M, N. Thenthe map from HomR(M, N) Mnm(R) taking MEB() is an R-module homomorphism.

Prop: Let : M N, : L M be R-module homomorphisms. Let , B, Ebe bases for L, M, and N respectively.Then ME ( ) = MEB ()MB ().

Cor: If M is free of rank n, then the map EndR(M) Mnn(R) given by MBB () is a ring isomorphism.Cor: Let F be a field and A Mn(F). TFAE:

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6 MATRIX ALGEBRA Algebra Study Guide 6.3 Hom and Duality

1. A is invertible.

2. Ax = 0 if and only ifx = 0 for all x Mn1(F).3. For all y Mn1(F) there exists x Mn1(F) such that Ax = y.

Def:42 Let A, B

M

nn(R). Then A and B are called similar if there exists an invertible n

n matrix P such that

A = P1BP.E.g.: Let F be a field and let V = {f F[x] : deg f 2}. Let B = 1, x , x2 and E= 1, 1 + x, 1 + x + x2. Then

MEB (I) =

1 1 00 1 1

0 0 1

and

MBE(I) =

1 1 10 1 1

0 0 1

6.3 Hom and Duality

Def: Let R be commutative, M an R-module. The dual module (or dual space) of M is

M := HomR(M, R).

In the case that R is a field, the elements of M are called linear functionals.Prop: Let F be a free R-module of rank n. Let {e1, , en} be a basis for F. For each i {1, , n}, define maps

ei := F R

by ei (ej) = ij (the Dirac delta function). Then {e1, , en} is a basis for F called the dual basis for {e1, , en}.In particular, F is free of rank n.

Def: Let M be an R-module, m M. Define a map evm : M R by f f(m). Note: evm is R-linear. Therefore,evm

HomR(M

, R) = M. This gives a natural homomorphism from M to M, i.e., : M

M given by(m) = evm.

Rmrk: For finite-dimensional vector spaces, V = V = V.

6.4 Determinants

Def:43 Let R be a commutative ring and M an R-module with n N. Then an n-multilinear form on M is a

function f : Mn R such that for all i and for all m1, , mi, , mn M (fixed) the function i : M R givenby

x f(m1, , mi1, x , mi+1, , mn)is an R-module homomorphism. We say f is alternating if for all m1, , mn M there exists i such thatmi = mi+1 im plies f(m1, , mn) = 0.

Prop: Let f be an alternating n-multilinear form on M. Let m1, , mn M. Then1. f(m1, , mi1, mi+1, mi, mi+2, , mn) = f(m1, , mn) (i.e., the value of on an n-tuple is negated if

two adjacent components are interchanged).

2. Sn, f(m(1), , m(n)) = sign()f(m1, , mn).3. Ifmi = mj for some i = j, f(m1, , mn) = 0.4. R and i = j, f(m1, , mi + mj , , mn) = f(m1, , mn) (where mi + mj is in the ith position).

Proof. 1. Let : MM R be given by (x, y) = f(m1, , mi1, x , y , mi+2, , mn). then (x+y, x+y) = 0as f is alternating. By the fact that is an R-linear map, (x + y, x + y) = (x, x + y) + (y, x + y) =(x, x) + (x, y) + (y, x) + (y, y) = 0 = (x, y) + (y, x).

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6 MATRIX ALGEBRA Algebra Study Guide 6.4 Determinants

2. Recall: every Sn can be written as a product of transpositions of the form (i i + 1). We do induc-tion on the number of transpositions in such a decomposition. If = (i i + 1), were done by part 1.Now let = (i i + 1), where is a product of n 1 such transpositions. Then f(m(1), , m(n)) =f(m(1), , m(i1), m(i+1), m(i), , m(n)) = f(m(1), , m(n)) (by 1) = sign()f(m1, , mn) =sign()f(m1, , mn).

3. Let be any permutation which fixes i and sendsj to i+1. Then f(m1, , mn) = sign()f(m(1), , m(n)) =0, as mj is now in the i + 1

st component, giving that f equals 0 by the definition of an alternating n-multilinearform.

4. f(m1, , mi + mj , , mn) = f(m1, , mn) + f(m1, , mj , mn) = f(m1, , mn) + 0(by 3, as theith and jth components are equal in the second summand).

Prop: Let f be an alternating n-multilinear form on M. Suppose for j = 1, , n and wj =n

i=1ijvi, where wi M,

vi M, ij R. Then

f(w1,

, wn) = Sn sign()(1)1 (n)n f(v1, , vn).

Def: Let R be a commutative ring and n N. A determinant on R is a function det : Mnn(R) R such that detis an alternating n-multilinear function on the columns of R (M = Rn) and such that det(Inn) = 1.

Notation: If A Mnn(R), let A1, , An denote the columns of A, i.e., det(A1, , An) is alternating multilinear.Thm: There exists a unique determinant function on R. If A = [aij ] then

det(A) =

Snsign()a(1)1 a(n)n.

Prop:44 Let A, B Mnn(R). Then det(AB) = det(A) det(B).

Def: Let A Mmn(R). Say A = [aij ]. The transpose of A, denoted At, is the n m matrix whose ijth entry is aji.Prop: Let A Mnn(R). Then det(A) = det(At).Thm: (Cramers Rule) Let A Mnn(R) with columns A1, , An. Let [x1, , xn] Mn1(R). Let [y1, , yn] =

x1A1 + + xnAn (so AX = Y). Then xi det A = det(A1, , Ai1, Y , Ai+1, , An).Def: Let A = [aij ] Mnn(R). Then the ijth cofactor of A is (1)i+j det(Aij), where Aij is the n 1 n 1

submatrix of A obtained by deleting the ith row and jth column.

Prop: (Cofactor Expansion) Let A = [aij ]. Fix i {1, , n}. Then det A =n

i=1(1)i+j det(Aij).

Def: Let A Mnn(R). Define the adjoint of A, denoted adj A, to be the n n matrix whose ijth entry is(1)i+j det(Aji).

Thm: Let A Mnn(R). Thenadj(A) A = A adj(A) = det(A)I.

Cor: 45 Let A Mnn(R). Then A is invertible IFF det(A) R.Def: Let R be a domain, A Mnn(R). Then the column rank of A is the maximum number of linearly independent

columns of A, (and similarly for row rank).Prop: Let R be a domain, A Mnn(R). TFAE:

1. col rank(A) = n

2. row rank(A) = n

3. det(A) = 0.Def: Let R be commutative, A Mmn(R). Let 1 r min {m, n}. An r r minor of A is the determinant of an

r r submatrix of A obtained by deleting any m r rows and n = r columns.Def: Define the rth fitting ideal of A, denoted Ir(A), to be the ideal of R generated by all the r r minors of A.Rmrk:

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7 MODULES OVER A PID Algebra Study Guide

1. I0(A) = R and Ir(A) = 0 for r min {m, n} by convention.2. Ir(A) = Ir(A

t) for all r 0.3. Ir+1(A) Ir(A) for all r 0.

4. For any A, B, Ir(AB) Ir(A) Ir(B).Thm: Let R be a domain and A an arbitrary m n matrix over R. Then col rank(A) = sup {r : Ir(A) = (0)}.

7 Modules over a PID

7.1 Basic Theory

Def:46 Let M be an R-module. M is called Noetherian ifM satisfies the ascending chain condition on submodules,

i.e., given any ascending chainN0 N1 N2

of submodules of M there exists t such that for all n 0, Nt = Nt+n, Nt = Nt+1 = Nt+2 = .We say R is Notherian as a ring if it is Noetherian as an R-modules (i.e., R satisfies ACC on ideals).Prop: Let M be an R-module. The following are equivalent:

1. M is Noetherian

2. Every submodule ofM is finitely generated.

3. Every nonempty set of submodules ofM has a maximal element.

Cor: R is a Noetherian ring if and only if for all ideals I of R, I is finitely generated.Prop: Let M N be R-modules. Then N is Noetherian if and only if M and N/M are Noetherian.Cor: Let M1, , Mn be R-modules. Then M1 M2 Mn is Noetherian if and only if each Mi is Noetherian.

Proof. By induction on n (M1 M2)/M1 = M2.

Prop: Let R be a Noetherian ring. Then every finitely-generated R-module is Noetherian.Prop: Let R be a domain and M a free R-module of rank n. Then every set ofn+1 elements ofM is linearly dependent.Def: Let R be a domain and M an R-module. A set x1, , xn M is said to be a maximal linearly independent set

if {x1, , xn} is linearly independent and {x1, , xn, z} is linearly dependent for all z M. The rank of M isdefined to be

rank M = sup {n : x1, , xn M forming a maximal LI set}.Prop:

47 Let R be a domain and let L M be R-modules. Then

rank M = rank L + rank M/L.

Cor: rank M1 rank Mn =n

i=1rank Mi.

Proof. Induction and (M1 M2)/M1 = M2.Def: Let R be a domain and M an R-module. Define the torsion submodule of M to be

TorR(M) := {m M : rm = 0 for some r R \ {0}}.

If M = TorR(M), M is called torsion. If TorR(M) = 0, M is said to be torsion-free.Rmrk: Let R be a domain.

1. IfM is free, M is torsion-free.

2. Submodules of torsion-free modules are torsion-free.

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7 MODULES OVER A PID Algebra Study Guide 7.1 Basic Theory

3. R = Z, Z/2 is a torsion Z-module. (2) is torsion-free, as it is isomorphic to Z (as groups).

4. R = Q[x], I = (x, y). Then I is torsion-free but not free (as I is not principal).

5. IfM is a finitely-generated torsion module then there exists r R \ {0} such that rM = 0.Def: Let R be a commutative ring and M an R-module. Define the annihilator of M to be

AnnR(M) := {r R : rM = 0}.It is clear that AnnR(M) is an ideal of R.

To restate (5), if M is finitely-generated module and R is a domain, M is torsion if and only if AnnR(M) = (0).Thm: Let R be a PID and M a free R-module of rank m < and L a submodule of M. Then L is free of rank m.

In fact, there exists a basis y1, , yn of M and a1, , a R such that1. {a1y1, , ay} is a basis for L.2. ai|ai+1 for all i 1.

Thm:48 Let R be a PID, M a finitely-generated R-module of rank r. Then there exist nonzero nonunits a1, , at R

such that ai

|ai+1 for all i 1 and

M= Rr R/(a1) R/(at).The ideals (a1), , (at) are called the invariant factors of m.

Cor: Let R be a PID and M a finitely-generated R-module of rank m. Then

1. M is free iff M is torsion-free.

2. M = F TorR(M), where F is free of rank m.Thm: Let R be a PID and M a finitely-generated R-module of rank r. Then there exist prime elements p1, , pr R

not necessarily distinct and non-negative integers 1, , r such thatM= Rr R/(p11 ) R/(prr ).

The ideals (p11 ),

, (prr ) are called the elementary divisors of M.

Thm: 49 The elementary divisors and invariant factors are unique.Cor: Let R be a PID and M1, M2 are finitely-generated R-modules.

1. M1 = M22. rank(M1) = rank(M2) and M1 and M2 have the same set of elementary divisors.

3. rank(M1) = rank(M2) and M1 and M2 have the same invariant factors.

Rmrk: Let M be a finitely-generated torsion R-module, where R is a PID. Then AnnR(M) = (at), where (at) is thelargest invariant factor of M, i.e., M= R/(at).

Def: 50 Let A, B Mn(F). We say A and B are similar if and only if A = P BP1.Prop: Let S, T EndF(V). TFAE:

1. S, T are similar2. For all bases B of V, MB(S) is similar to MB(T)3. For all bases B of V there exists a basis B of V such that MB(S) = MB(T).

Prop: If A, B Mn(F) are similar, that det(A) = det(B).Def: Let V be a finite-dimensional F-vector space. Let T EndF(V). Define

det T := det(MB(T))

for all bases B of V.Rmrk:

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7 MODULES OVER A PID Algebra Study Guide 7.2 Rational Canonical Form

1. det ST = det Sdet T for all S, T EndF(V).2. det S = 0 S is an isomorphism3. IfS, T are similar, then det S = det T.

Def:

Let V be a finite-dimensional F-vector space. A nonzero v V is called a eigenvector of T EndF(V) ifT v = v

for some F.If such a v exists, is called an eigenvalue of T. If is an eigenvalue, E(T) := {v V : T v = v} is called the

eigenspace of T corresponding to .Prop: Let T EndF(V) and F. Then is an eigenvalue for T if and only if det(I T) = 0.Rmrk: The eigenvalues of a linear transformation T are the roots of cT(x).

7.2 Rational Canonical Form

Rmrk: 51 Let F be a field and x a variable.

1. dimF F[x] = , with 1, x , x2, a basis.2. dimF(F[x]/(f(x))) = n, where deg f = n, with

1, x, xn1 a basis.

3. dimF EndF(V) = (dimF V)2 (as EndF(V) = HomF(V, V) = Mn(F)).

Let T EndF(V). For all c0, c1, , cn F, c0 + c1T + + cnTn EndF(V), where c F is identified withcI EndF(V).

For a fixed T EndF(V) there exists a mapT : F[x] EndF(V)

given by T(f(x)) = f(T). Note that T is a ring homomorphism and an F-linear transformation.By comparing dimensions, ker T = (0). Hence, ker T = (f(x)) for some nonzero f F[x]. Therefore, deg f(x) n2.Def: Let T

EndF(V). Define the minimal polynomial of T to be the unique monic polynomial mT(x) such that

ker T = (mT(x)).Let T EndF(V) and T : F[x] EndF(V) the ring homomorphism from before. Thus, V is a left F[x]-module via

T, i.e., for v V,x v := T(v),

i.e., as x T, x v T(v).Notation: Let VT denote the F[x]=module V via T.Rmrk: Let {v1, , vn} be an F-basis for V. In particular, V = F v1 + + F vn. Then VT = F[x]v1 + + F[x]vn.

Therefore, VT is a finitely-generated F[x]-module. Note: V = VT as F-vector spaces.Also note mT(x)VT = 0, so VT is torsion (mT(x) v = mT(T) v = 0(v) = 0 for all v V). By the structure theorem,

VT = F[x]/(a1(x)) F[x]/(at(x)),

where the ai(x)s are monic (and hence unique), with ai|ai+1 for all i. The ais are called the invariant factors of T.Rmrk: Let f(x) F[x]. Then f(T) = 0 f (mT(x)) f(x)VT = 0 f AnnF[x] VT f(x) (at(x)) (mT(x)) = (at(x)) mT(x) = at(x) .

Prop: Let S, T EndF(V). TFAE:1. S and T are similar.

2. VS= VT as F[x]-modules.3. S and T have the same invariant factors.

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7 MODULES OVER A PID Algebra Study Guide 7.2 Rational Canonical Form

7.2.1 Rational Canonical Form and Minimal/Characteristic Polynomials

Def:52 Let x be an indeterminate over F. The polynomial det(xI T) is the characteristic polynomial of T

and will be denoted cT(x). IfA is an n n matrix with coefficients in F, det(xI A) is called the characteristicpolynomial of A and will be denoted cA(x).

Def: The unique monic polynomial which generates the ideal Ann(V) in F[x] is called the minimal polynomial of T

and will be denoted mT(x). The unique monic polynomial of smallest degree which when evaluated at the matrix Ais the zero matrix is called the minimal polynomial of A and will be denoted mA(x).

Prop: The minimal polynomial mT(x) is the largest invariant factor of V. All the invariant factors of V divide mT(x).Def: Let a(x) = xk + bk1xk1 + + b1x + b0 be any monic polynomial in F[x]. The companion matrix of a(x) is

the k k matrix with 1s down the first subdiagonal, b0, b1, , bk1 down the last column and zeros elsewhere.The companion matrix of a(x) will be denoted by Ca(x). I.e.,

Ca(x) =

0 0 b01 0 b10 1 b20 0

. . ....

......

. . ....

0 0 1 bk1

Given f F[x]/(a(x)), Ca(x) is the matrix corresponding to T, which is the linear transformation given by multiplicationof f by x.

Thus, if V = F[x]/(a1(x)) F[x]/(am(x)), the matrix for the linear transformation T (given by multiplication byx) has as matrix the direct sum of the companion matrix for the invariant factors, i.e.,

Ca1(x)

Ca2(x). . .

Cam(x)

Def:A matrix is said to be in rational canonical form if it is the direct sum of companion matrices for monicpolynomials a1(x), , am(x) of degree at least one with a1(x)|a2(x)| |am(x). The polynomials ai(x) are called

the invariant factors of the matrix. Such a matrix is also said to be a block diagonal matrix with blocks thecompanion matrices for the ai(x).

Thm: Let A and B be n n matrices over the field F. Then A and B are similar if and only ifA and B have the samerational canonical form.

Cor: Let A and B be two n n matrices over a field F and suppose F is a subfield of the field K.1. The rational canonical form of A is the same whether it is computed over K or over F. The minimal and

characteristic polynomials and the invariant factors ofA are the same whether A is considered as a matrix overF or as a matrix over K.

2. The matrices A and B are similar over K if and only if they are similar over F.

Lma: Let a(x) F[x] be any monic polynomial.1. The characteristic polynomial of the companion matrix of a(x) is a(x).

2. IfM is the block diagonal matrix

M =

A1 0 00 A2 0...

.... . .

...0 0 Ak

given by the direct sum of the matrices A1, , Ak, then the characteristic polynomial of M is the product ofthe characteristic polynomials of A1, , Ak.

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7 MODULES OVER A PID Algebra Study Guide 7.3 Jordan Canonical Form

Prop: Let A be an n n matrix over the field F.

1. The characteristic polynomial ofA is the product of all the invariant factors of A.

2. The minimal polynomial of A divides the characteristic polynomial of A.

3. The characteristic polynomial of A divides some power of the minimal polynomial of A. In particular thesepolynomials have the same roots, not counting multiplicities.

7.3 Jordan Canonical Form

Rmrk:53 Let T EndF(V) and dim V < . We assume F contains all eigenvalues of T. Equivalently, cT(x) factors

completely into linear factors in F[x]. Then cT(x) = (x 1) (x n) F[x], where the is are eigenvalues, notnecessarily distinct. Then the elementary divisor form for T is

VT = F[x]/(p1(x)m1) F[x]/(ps(x)ms).

Note that the pi(x)s are irreducible and monic, and hence unique.Rmrk: Note that cT(x) = a1(x) at(x) = p1(x)m1 ps(x)ms . Note that pi(x) = x i, as the roots of the pis are in

F and are the eigenvalues of T. Hence,

VT = F[x]/((x 1)m1) F[x]/((x s)ms),

where the is are not necessarily distinct.Rmrk: Consider the special case VT = F[x]/((x )m). Then B =

1, x , , (x )m1 is a basis for VT as an

F-v.s. Note dimF VT = m. In general, T((x )i) = x(x )i = ( + x )(x )i = (x )i + 1 (x )i+1.Also, T((x )m1) = (x )m1. Thus, the Jordan block corresponding to (x )m is

J(x)m = [T]B =

0 0 01 0 00 1 0...

. . . 0

0 0 0 1

In general, ifVT = F[x]/((x 1)m1) F[x]/((x s)ms),

with Bi =

1, x i, , (x i)mi1

and B = n1 Bi, then

JCF(T) =

J(x1)m1 0 00 J(x2)m2 0

...

0 0. . . 0

0 0 J(xs)ms

E.g.: Let A = [1]

M3(Q). Then cA(x) = x

2(x

3) and mA(x) = x(x

3). The EDs are x,x,x

3, so

JCF =

0 0 00 0 0

0 0 3

Prop: Suppose A Mn(F) and cA splits over F. Then A is similar to a diagonal matrix if and only if mA has norepeated roots.

Prop: If F contains all eigenvalues for A Mn(F), then A is similar to a lower (upper) triangular matrix.

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8 FIELD THEORY Algebra Study Guide

8 Field Theory

8.1 Introduction

Def:54 Let R be a commutative ring with 1. There is a natural ring homomorphism : Z R given by

(m) = m

1R. Then ker = (n)Z for a unique n

0. The characteristic of R is n, where ker = (n), and we

write char R := n.Rmrk: If R is a domain, then char R = 0 or p for some prime p > 0.Def: Note that, with definitions as above, im R. Further, im is contained in every subring of R containing 1R.

We call im the prime subring of R. If char R = n > 0, then im =Z/(n), and if char R = 0, im =Z.Def: Let F be a field. The prime subfield of F is defined to be the quotient field of the prime subring of F. The

prime subfield is contained in any subfield of F.Rmrk: If CharF = p > 0, Zp is the prime subfield. If char F = 0, the prime subfield of F is (isomorphic to) Q.Def: Let E be a field and F a subfield. We call E/F a field extension. The degree of E/F, denoted [E : F], is

defined to be dimF E. E/F is finite if [E : F] < .Prop: If E/F is finite, then E/F is algebraic.Def: Let E/F be a f.e. and E. Define F[] := {f() : f(x) F[x]}. F[] is the smallest subring of E containing

both F and . Define F() := Q(F[]).Prop:

Let E/F be a field extension and E. TFAE:1. is algebraic over F.

2. F() = F[].

3. [F() : F] < .Prop: Let E/F be a f.e. and 1, , n E. TFAE:

1. 1, , n are all algebraic.2. F(1, , n) = F[1, , n].3. [F(1, , n) : F] < .

Cor: Let E/F be a f.e. and , E. If , are algebraic over F, then , , / are all algebraic over F(assuming

= 0).

Prop: Let E/F be a f.e. Let L = { E : algebraic over F}. Then L is a subfield of E with F L E (L is thealgebraic closure of F in E).

Prop:55 If F E L are fields, then [L : F] = [L : E][E : F].

E.g.: [C : R] = 2.Def: Let E/F be a f.e. and let E be algebraic over F. Consider the natural ring homomorphism : F[x] F[]

given by f(x) = f(). Then ker = (p(x)) for some non-constant monic p F[x]. Then p is uniquely-determinedby . Denote p(x) = Min(, F).

Prop: Let E/F be a f.e. and E algebraic over F. Let p(x) F[x] be monic and p() = 0. TFAE:1. p(x) = Min(, F)

2. p(x) Irr(F[x])3. degp(x) = F[() : F]

4. Iff(x) F[x] \ {0} and f() = 0 then deg f(x) degp(x).Rmrk: If f() = 0 and f(x) F[x], then Min(, F) = p(x)|f(x).Rmrk: If [F() : F] = n, then F() = F[] = F[x]/(p(x)) with degp = n,

1, x, , xn1 is an F-basis ofF[x]/(p(x)),

which implies

1, , , n1 is an F-basis of F[x]/(p(x)).E.g.: Find [Q(i, 3

2) : Q]. Then [Q(i) : Q] = 2 and [Q( 3

2) : Q] = 3; further [Q(i, 3

2) : Q(i)] 3 and [Q(i, 32) :

Q( 3

2)] 2. Since i / Q( 32) R, [Q(i, 32) : Q( 32)] > 1, so [Q(i, 32) : Q] = 6.Rmrk:

56 IfE/F is algebraic, E/F is not necessarily finite. Example: Q = { C : algebraic over Q} and n2 Q.Rmrk: Let E/F be an alg. extension. Then E/F is finite if and only ifE is f.g. as a field over F (IFF E = F(1, , n)

for some 1, , n F.Prop: Let F E L be fields. Then L/F is algebraic if and only if L/E and E/F are algebraic.

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8 FIELD THEORY Algebra Study Guide 8.2 Composite Fields

8.2 Composite Fields

Def: Let E and F be subfields of a field L. The composite field EF of E and F is

EFTLT a fieldT

Suppose:

E

EF??

K

??F

Then we have the following remarks.Rmrk: EF = E(F) = F(E).Rmrk: IfE = K(1, , n), then EF = F(1, , n). If additionally F = K(1, , l), then EF = K(1, , n, 1, Rmrk: If E/K is algebraic then EF/F is algebraic.Rmrk: If E/F is algebraic then EF = {e1f1 + + enfn : ei E, fi F, n N}.Rmrk: [EF : F] [E : K].

8.3 Splitting Fields

Prop:57 Let F be a field and f(x) F[x] irreducible. Then there exists a field E containing F such that f(x) has

a root in E. If E is a root off(x) then

F() = F[x]/(f(x)).

Cor: Let F be a field and f(x) F[x] with deg f 1. There exists a field E containing a root of F.Def: Let f(x) F[x] and E an extension field of F. We say f(x) splits in E if f(x) is the product of linear factors in

E[x].Prop: Let f(x) F[x] \ F. Then there exists a field E containing F such that f(x) splits over E.Def: Let f(x)

F[x] with deg f > 0. A splitting field for f(x) (over F) is a field E containing F such that f(x)

splits in E but f(x) does not split in any proper subfield of E containing F.Prop: Let f(x) F[x] \ F. Then a splitting field for f(x) over F exists. In particular, let L be a field containing F in

which f(x) splits. Say f(x) = c(x 1) (x n) for some c F, i L. Then E = F(1, , n) is a splittingfield for f(x) over F.

Lma: Let be a primitive pth root of unity where p is prime. Then [Q() : Q] = p 1.

8.4 Algebraic Closure

Def:58 A field F is algebraically closed if every nonconstant polynomial in F[x] splits into linear factors in F[x].

Rmrk: Let F be a field. TFAE:

(1) F is algebraically closed

(2) Every nonconstant polynomial in F[x] has a root in F

(3) a field properly containing F which is algebraic over FDef: Let F be a field. An algebraic closure of F is an algebraically closed field containing F which is algebraic over

F.Prop: Let F L be fields where L is algebraically closed. Let F = { L : is algebraic over F}. Then F is an

algebraic closure of F.Thm: Let F be a field. Then there exists an algebraic closure of F.Def:

59 Let E/F and L/K be field extensions and : E L and : F K be field homomorphisms. We say extends (or is defined over ) if|F = .

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8 FIELD THEORY Algebra Study Guide 8.5 Splitting Fields

Rmrk: : Let : F K be a field isomorphism. For f(x) = cnxn + + c1x + c0 F[x], let f(x) = (cn)xn +(cn1)xn1 + + (c0) K[x]. Note: (f g) = fg and (f + g) = f + g. Thus, : F[x] K[x] is anisomorphism of polynomial rings sending f to f.

Lma: Let : F K be an isomorphism of fields. Let F and K be algebraic closures of F and K respectively.Let f(x) F[x] be an irreducible polynomial and K be a root of f(x). Then there exists an isomorphism : F() K() which extends such that () = , i.e.,

F()

// K()

F // K

Thm: Let E/F and L/K be field extensions. Let : F K be a nonzero field map. Suppose E/F is algebraic and Lis algebraically closed. Then there exists : E L which extends .

Cor: 60 Let F be a field and say E1, E2 are two algebraic closures ofF. Then there exists an isomorphism : E1 E2which fixes F (i.e., (x) = x for all x F).

8.5 Splitting Fields

Def: Let F be a field and S F[x] \ F. A splitting field for S (over F) is a field E containing F such that everyf S splits in E[x] and E is minimal with respect to this property.

Rmrk: Let S F[x]\F as above and let F be any algebraic closure ofF. Let T = F : is a root of some f S.Then E = F(T) is a splitting field for S.

Exercise: Suppose E is a splitting field for a subset S F[x] \ F. Then E is a splitting field for f F[x] if and only ifE/F is finite.

Prop: Let F be a field and S F[x] \ F. Let E1 and E2 be splitting fields for S over F. Then there exists anisomorphism : E1 E2 fixing F.

Prop: Let E/F be an algebraic extension and : E E be a field homomorphism fixing F. Then is an isomorphism.Thm: Let E/F be an algebraic extension. The following are equivalent:

1. E be a splitting field for some set S

F[x]

\F.

2. Every irreducible f(x) F[x] which has a root in E splits completely over E.3. Let E be an algebraic closure of E and : E E be any field homomorphism fixing F. Then (E) = E, i.e.,

is an automorphism of E.

Def: An algebraic extension E/F satisfying 1-3 above is called a normal extension.Def:

61 Let F be a field, f(x) F[x], and F a root off(x). We say is a repeated root of f(x) if (x )2|f(x)in F[x].

Def: Let F be a field and x an indeterminate over F. Define an F-linear transformation Dx : F[x] F[x] byDx(x

n) = nxn1 for n 1 and Dx(1) = 0.Exercise: Let E/F be a field extension and f, g F[x]. Then gcdF[x](f, g) = gcdE[x](f, g).Prop: Let F be a field and f(x) F[x] \ F. Then f(x) has a repeated root if and only if gcd(f, f) = 1.Def: Let F be a field. A polynomial f(x)

F[x]

\F is separable if f(x) has no repeated roots. Let

F. We say

is separable over F if Irr(, F) is separable. An algebraic extension E/F is called separable if every element ofE is separable over F.

Def: F is called a perfect field if every algebraic extension of F is separable.Prop: Any field of characteristic 0 is perfect.Prop: Any finite field is perfect.Prop: Let F be a finite field. Then (F, ) is a cyclic group.Rmrk: 62 If F is a finite field then |F| = pn for p = char F.Thm: Let p be a prime and n > 0. Let E be a splitting field for xp

n x over Fp. Then |E| = pn. Moreover, any otherfield of order pn is also a splitting field for xp

n x over Fp. Therefore, any two fields of order pn are isomorphic.Prop: Let F be a finite field. Then every algebraic extension of F is normal.

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9 GALOIS THEORY Algebra Study Guide 8.6 Cyclotomic Stuff

8.6 Cyclotomic Stuff

Def: Let n > 0 be an integer, w = e2i/n C a primitive nth root of unity. Define the nth cyclotomic polynomial

n(x) :=

1ingcd(i,n)=1(x wi) =

C annth root of unity(x ) C[x].

Rmrk: 1(x) = x 1, 2(x) = x + 1Rmrk: xn 1 =

dNd|n

d(x)

Rmrk: x3 1 = 1(x) = 3(x) so 3(x) = x3 1

x 1 = x2 + x + 1, 4(x) =

x4 112

= x2 + 1, 5(x) =x5 1x 1 =

x4 + x3 + x2 + x + 1, etc.Rmrk: degn = (n), where is Eulers -function.Rmrk:

63 Let F be a field and R a subring of F. Let f, g R[x] with f monic. Suppose f|g in F[x]. Then f|g in R[x].Lma: n(x) Z[x] for all n 1.Thm: n(x) is irreducible over Q for all n.Cor: If w

C is a primitive nth root of unity, [Q(w) : Q] = (n) = deg n(x) and n = Irr(w,Q).

9 Galois Theory

9.1 Basic Definitions

Def: Let E/F be a field extension. Define Aut(E) := { : E E : is a field isomorphism} to be the automor-phism group of E (which is a group under composition). Define Aut(E/F) := { Aut(E) : |F = 1F}.

Rmrk: If L is the prime subfield of E then Aut(E) = Aut(E/L).Rmrk: Let Aut(E/F) and f(x) F[x]. If E is a root off(x) then () is a root of f(x).E.g.:

64 Aut(Q( 3

3)) = 1, as the other roots of x3 2 are non-real.E.g.: Let E be the splitting field of x6 + 3 over Q. Then Aut(E) = S3 or Aut(E) = C6.Lma: Any algebraic extension of a perfect field is perfect.Thm:

Let F be a perfect field and : F F a field isomorphism (so F is also perfect). Let f(x) F[x] and E be asplitting field for f(x) over F. Let E by a splitting field for f(x) over F.Let E = { : E E : is an isomorphism such that |F = }. Then |E | = [E : F].

Cor: Let F be perfect and f(x) F[x] and E a splitting field for f over F. Then | Aut(E/F)| = [E : F].Def: Let E/F be algebraic. Then E/F is called Galois if E/F is normal and separable. If E/F is Galois, then the

Galois group of E/F is Aut(E/F), usually denoted Gal(E/F).Rmrk: Weve proved if F is perfect and E/F is a finite normal extension then |Gal(E/F)| = [E : F].Rmrk: Suppose F L E and E/F is Galois. Then E/L is also Galois but L/F may not be.E.g.:

65 Let E = Q(

2,

3). Note that E is a splitting field for (x2 2)(x2 3) which implies E/Q is normal. Further,|Gal(E/Q)| = [E : Q] = 4. If Gal(E/Q) then is determined by (2) and (3). There are only 4 possibilities,and hence they are all elements of Gal(E/Q).

Prop: Let C be a primitive nth root of unity. Then Q()/Q is Galois and Gal(Q()/Q) =Zn . In particular, if nis prime, then Gal(Q()/Q) is cyclic of order n 1.

9.2 The Fundamental Theorem of Galois Theory

Thm: (Fundamental Theorem of Galois Theory) Let K/F be a Galois extension and set G = Gal(K/F). Then thereis a bijection between the subfields E of K containing F and the subgroups H of G given by the correspondencesE {the elements of G fixing E} and H {the fixed field of H} which are inverse to each other.

Prop: Let E/F be a finite Galois extension and let G = Gal(E/F). Then EG = F (where EG := {u E : (u) = u G}Cor: Let E/F be a finite Galois extension and L an intermediate field. Then EGal(E/L) = L.Cor: Let E/F be a finite Galois extension. The map (L) = Gal(E/L) for L {intermediate fields of E/F} is 1-1 and

hence there are only finitely many intermediate fields.

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9 GALOIS THEORY Algebra Study Guide 9.3 Radical Extensions

Thm:66 (Primitive Element Theorem) Let E/F be a finite extension with only finitely many intermediate fields. Then

there exists E such that E = F(). Then is called a primitive element for E/F.Cor: Let E/F be a finite Galois extension and L an intermediate field. Then L = F() for some L.Cor: Let F be a perfect field, and E/F a finite extension. Then E = F() for some E.Lma: Suppose E/F is a Galois extension (not necessarily finite). Suppose there exists n 1 such that [F() : F] n

for all E. Then E/F is finite and [E : F] n.Thm: (Artin) Let E be a field and G a finite subgroup of Aut(E). Let F be the fixed field of G, i.e., F = EG.1. E/F is finite Galois, and

2. Gal(E/F) = G.

Rmrk: On FTGT: Gal(E/EH) = H for some H Gal(E/F) (and EH an intermediate field).Rmrk: On FTGT: L1 L2 if and only if Gal(E/L1) Gal(E/L2).Rmrk: On FTGT: H1 H2 if and only if EH1 EH2 .Rmrk: On FTGT: [E : EH] = |H|Rmrk: On FTGT: [EH : F] = [G : H].Thm:

67 Let E/F be a finite Galois extension and G = Gal(E/F). Let L be an intermediate field and H = Gal(E/L).Then

1. L/F is normal if and only if H G.

2. IfL/F is normal, then Gal(L/F) = G/H.Prop: 68 Let F be a perfect field, f(x) F[x] of degree n. Assume f(x) has distinct roots (e.g., if f is irreducible).

Let E be a splitting field for f(x) over F. Then E/F is finite Galois, and Gal(E/F) = H Sn.Cor: [E : F] n!.

9.3 Radical Extensions

Def:69 A finite extension of fields E/F is called radical if there exist elements u1, , un E such that E =

F(u1, , un) and for each i, there exists mi such that umii F(u1, , ui1), i.e., there exists a root towerF = F0

F1

Fn = E, where Fi = Fi1(ui) and u

mii

Fi1. A polynomial f(x)

F[x] is called solvable

by radicals if f(x) splits in some radical extension of F.Prop: Suppose E = F(u), where char F = 0 and un F, where F contains a primitive nth root of unity. Then E/F is

Galois and Gal(E/F) is cyclic.Lma: 70 Let E/F be a finite radical extension and char F = 0. Then E E such that E/F is finite, Galois, and

radical.Thm: 71 Let E/F be Galois, radical with char F = 0. Then the Galois group is solvable.

Proof. Let F = F0 F1 Fn = E, Fi = Fi1(i) and mii Fi1. Let m = m1 mn, and a primitivemth root of 1. Then Gal(E/F) = Gal(E()/F)/Gal(E()/E). Hence if Gal(E()/F) is solvable, so is Gal(E/F),as quotients of solvable groups are solvable. Now Gal(E()/F) is solvable if and only if Gal(E()/F()) is solvableand Gal(F()/F) is solvable. By the homework exercise, Gal(F()/F) is abelian and hence solvable. It is enoughto show that Gal(E()/F) is solvable. (Hence we may assume F (in our original setup) contains a primitive mithroot of unity.) By a previous proposition, Fi/Fi1 is Galois and has cyclic Galois group. let Hi = Gal(E/Fi). Then

we have {1} = Hn Hn1 H0 = Gal(E/F). The quotients Hi1/Hi = Gal(Fi/Fi1), which is cyclic. Thisgives a solvable series for G.

Thm: (Galois) Let F be a field of characteristic 0, f(x) F[x]. Let E be a splitting field for f over F. Then if f(x) issolvable by radicals over F, then Gal(E/F) is solvable.

Proof. Iff(x) is solvable by radicals, there exists a finite radical extension L/F such that E L. By a lemma, thereexists a Galois radical extension L/F with L L. So by the theorem, Gal(L/F) is solvable, and Gal(E/F) =Gal(L/F)/Gal(L/E).

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9 GALOIS THEORY Algebra Study Guide 9.3 Radical Extensions

Prop: Let char F = 0, and let f(x) F[x] be irreducible of degree p, where p is a prime. Let E be the splitting fieldfor f(x) over F. Suppose there exists an automorphism : E E fixing F and all but two roots of f. ThenGal(E/F) = Sp.

E.g.: Let f(x) = x52x38x2 Q[x]. Then f is not solvable by radicals, as complex conjugation is an automorphismofC fixing 3 real roots and transposing 2 others. Then Gal(E/F) = S5, which is not solvable by group theory.

Def:

Let F be a field, x, t1, , tn indeterminates over F. Then the polynomial fn(x) = xn

+ t1x

n

1

+ tn1x + tn F(t1, , tn)[x] is called the general equation of degree n over F. We say fn(x) is solvable by radicals if theroots of fn(x) lie in a radical extension of F(t1, , tn).

Thm: Let E be a splitting field for fn(x) over F(t1, , tn). Then Gal(E/F(t1, , tn)) = Sn.Thm: 72 (Fundamental Theorem of Algebra) C is algebraically closed.

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10 OLD EXAMS Algebra Study Guide

10 Old Exams

10.1 Exam 1, Fall 2007

1. Let G be a group and Z the center of G. Provethat if G/Z is cyclic then G is abelian. Also, givean example (no proof necessary) where this resultfails if the hypothesis is changed to assume G/Z isabelian.

2. Let H be a subgroup of a finite group G and con-sider the set

:=

xHx1 : x G.(a) Prove that || = [G : NG(H)] where NG(H) =

g G : gH g1 = H.(b) Prove that

xG xHx

1 = G if and only ifH = G. (Hint: Use part (a) to show that if

H = G then the number of elements int heunion of the conjugates of H is less than |G|.)

3. Let G be a finite group of whose order is square-free.(That is, no square integer greater than 1 divides|G|.) Let p be the smallest prime dividing |G| andsuppose G contains a subgroup H of index p. Provethat H is the only subgroup of G of index p.

4. Let G be a group with exactly three conjugacyclasses. Prove that G = C3 or G = S3.

10.2 Exam 2, Fall 2007

1. Let G be a finite group and suppose G has a uniquesubgroup of order d for each positive divisor d of|G|. Prove that G is cyclic. (October 24, 2007)

2. Prove that if|G| = 3 57 then G has a cyclic normalsubgroup of order 35.

3. Let G be a finite group and suppose pn divides |G|for some prime p and positive integer n. Prove thatG has a subgroup of order pn.

4. Let G be a finite group and suppose G contains asimple subgroup H = 1 such that [G : H] = 2. Sup-pose further that

|Z(G)

|is odd. Prove that H is the

only nontrivial normal subgroup of G.

5. Let G be a group and K a finite cyclic normal sub-group of G. Prove that G CG(K), where Gis the commutator subgroup of G and CG(K) ={g G : gk = kg for all k K}. (Hint: Let g Gand consider the inner automorphism g of G.Since K is normal, g restricted to K is an au-tomorphism of K. This gives a natural homomor-phism : G Aut(K). Examine the kernel andthe image of this map.)

6. Let G be a group of order 3 7 11. Suppose Gcontains an element of order 21. Prove that G iscyclic.

10.3 Final Exam, Fall 2007

1. Let R be a ring and I1, , In ideals ofR which arepairwise co-prime (i.e., Ii + Ij = R for all i = j).Prove that

I1I2 In = I1 I2 In.2. Let I be an ideal of R and define the radical

I of

I by I := {r R : rn I for some n}.

Prove that

I is an ideal of R and that R/

I hasno nonzero nilpotent elements.

3. Prove that every Euclidean domain is a PID.

4. Prove that Z[5] is a UFD. (Show all details.)5. Prove that the following polynomials are irreducible

over the given field:

(a) x5 2x + 2 over Q.(b) x3 + x2 + 2 over F3.

(c) x4 + x2 + 6 over Q.

6. Let R be a ring and r R such that r is not nilpo-tent. Let be the set of all ideals I of R such thatrn / I for all n 1. Prove that is a maximal ele-ment and that any maximal element of is a prime

ideal.

10.4 Exam 1, Spring 2008

1. Let M be an R-module and : M M a surjectiveR-module homomorphism.

(a) Suppose M is Noetherian. Prove that is anisomorphism.

(b) Give an example of such an M and such that is not an isomorphism.

2. Let R be a domain and M an R-module. Recallthat a subset S of M is called a maximal linearly

independent set of M if S is linearly independentand any subset of M properly containing S is lin-early dependent.

(a) Let T be a linearly independent subset of M.Prove that T is contained in some maximallinearly independent subset of M.

(b) Let T be a linearly independent subset of Mand N the R-submodule of M generated byT. Prove that T is a maximal linearly inde-pendent subset if and only if M/N is torsion.

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10 OLD EXAMS Algebra Study Guide 10.5 Exam 2, Spring 2008

3. Let I be an ideal of R.

(a) Prove that I is a free R-module if and only ifI = (0) or I = (a) for some a R which is nota zero-divisor.

(b) Give an example of a submodule of a free R-module which is not free.

4. Let I be an ideal of R and M a finitely-generatedR-module.

(a) Suppose IM = M. Prove there exists an ele-ment a I such that (1 a)M 0.

(b) Suppose I is finitely generated and I2 = I.Prove that I = (e) for some e R such thate2 = e.

5. Let F be a field and V a finite dimensional F-vector

space. Let W be a subspace of V. Prove thatdim V = dim W + dim V /W.

6. Let A = [aij ] be an n n matrix over a field F.Define the trace of A, denoted tr(A), by tr(A) =n

i=1 aii.

(a) Prove that for n n matrices A, tr(AB) =tr(BA).

(b) If A and B are similar, prove that tr(A) =tr(B).

(c) Let A be a 2 2 matrix which is not a scalarmatrix (i.e., not a scalar multiple of the iden-tity matrix.) Prove that A is similar to aunique matrix of the form:

0 det(A)1 tr(A)

.

7. Let A be a matrix with rational entries whose char-acteristic polynomial is c(x) = x2(x 1)2(x + 1).

(a) Find all possible sets of invariant factors for A.

(b) Find a specific matrix A as above whose min-imal polynomial has degree 3.

8. Find the RCF of the following matrix in M3(Q):1 1 11 1 1

1 1 1

10.5 Exam 2, Spring 2008

1. Let E be a splitting field of x6 + 5 over Q. Find[E : Q]. Justify all details.

2. Let f(x) be an irreducible polynomial of degree nover a field F. Let g(x) be any nonconstant poly-nomial in F[x] and set h(x) = f(g(x)). Prove thatif q(x) F[x] is any irreducible factor of h(x) thendeg q is divisible by n. (Hint: If is a root of h(x)

then g() is a root of f(x).)

3. Let E be a splitting field for x4 4x2 1 over Q.Find the Galois group of E/Q. That is, explicitlydescribe all automorphisms of E and identify thegroup structure (e.g., by showing it is isomorphicto some well-known group).

4. Let E/F be a finite Galois field extension and Kand L intermediate fields. Suppose E = KL andF = K L.

(a) Give an example of fields E, F, K, and L asabove such that [E : F] = [K : F][L : F].(b) If either K or L is normal over F, prove that

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

5. Let be a primitive 7th root of unity and letE = Q(). Find all subfields of E and primitiveelements (over Q) for each.

6. Let E/F be a finite Galois extension and K anintermediate field. Let G = Gal(E/F) and H =Gal(E/K). Let NG(H) denote the normalizer ofH

in G. Prove that NG(H) = {g G : g(K) = K}and NG(H)/H= Aut(K/F).

7. Let Q be the algebraic closure ofQ in C and F amaximal subfield of Q not containing

5 (such a

field exists by Zorns Lemma). Let E be a finitenormal extension of F. Prove that Gal(E/Q) iscyclic.

10.6 Final Exam, Spring 2008

1. Let R be a ring and M an R-module and N a sub-module of M. Prove that M is Noetherian iff Nand M/N are Noetherian.

2. Let R be a domain and M a finitely-generated tor-sion R-module. Prove there exists a nonzero r Rsuch that rM = 0, i.e., r annihilates M. Givean example of a torsion Z-module M such that nononzero element ofZ annihilates M.

3. Consider the matrix A below with rational coeffi-

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10 OLD EXAMS Algebra Study Guide 10.6 Final Exam, Spring 2008

cients.

A =

1 0 0 0 0 0 00 0 0 0 0 0 00 1 1 0 0 0 00 0 0 0 0 0 0

0 0 0 1 0 0 00 0 0 0 1 0 10 0 0 0 0 1 2

.

(a) Find the rank ofA.

(b) Find the invariant factors and the characteris-tic polynomial of A.

(c) Find the eigenvalues of A.

(d) F

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