MATH 1530 ABSTRACT ALGEBRASelected solutions to problems

Problem Set 2

2. Define a relation ∼ on R given by a ∼ b if a− b ∈ Z.

(a) Prove that ∼ is an equivalence relation.

(b) Let R/Z denote the set of equivalence classes of ∼. Prove that the binary operation +on R/Z given by

a+ b = a+ b

is well-defined.

(c) Is (R/Z,+) a group?

Solution. (I will omit the proof that R/Z is a group.) Given a, b, c ∈ R, note:

(a) a− a = 0 ∈ Z;

(b) a− b ∈ Z implies b− a = −(a− b) ∈ Z; and

(c) if a− b ∈ Z and b− c ∈ Z then a− c = (a− b)− (b− c) ∈ Z;

so ∼ is an equivalence relation.

Next, we show that the binary operation + on R/Z given by a + b = a+ b is well defined.Suppose a = a′ and b = b′ for some a, b, a′, b′ ∈ R. We wish to show a+ b = a′ + b′.

The assumptions imply a− a′ ∈ Z and b− b′ ∈ Z, so

(a+ b)− (a′ + b′) = (a− a′) + (b− b′) ∈ Z.

Soa+ b = a′ + b′,

as desired.

1

Problem Set 3

2. Write out the full multiplication table for the group S3.

It should be a 6× 6 table whose entries are elements of S3, and whose rows and columns areindexed by the elements of S3. Please write all elements using cycle notation.

Solution.

· id (12) (13) (23) (123) (132)

id id (12) (13) (23) (123) (132)(12) (12) id (132) (231) (23) (13)(13) (13) (123) id (231) (12) (23)(23) (23) (132) (123) id (13) (12)

(123) (123) (13) (23) (12) (132) id(132) (132) (23) (12) (13) id (123)

3. Is it true that for all n ≥ 1, every element of Sn has order at most n? Either prove it, or givea counterexample.

Solution. No. We saw in the Youtube lecture that the order of an element σ ∈ Sn is theleast common multiple of all of the numbers appearing as lengths of some cycle in σ, when σis expressed in cycle notation.

Thus the order of (12)(345) ∈ S5 is 6.

5. Let (G, ∗) and (H, ◦) be groups, and let φ : G → H be an isomorphism. Let φ−1 : H → Gdenote the inverse of φ. In other words, φ−1(h) = g whenever φ(g) = h.

Prove that φ−1 is also an isomorphism.

Solution. Let us use the fact from class:

Fact. A map f : A → B is bijective if and only if there exists an inverse, i.e. a mapg : B → A with f ◦ g = idB and g ◦ f = idA, where idA : A → A and idB : B → B denotethe respective identity maps. This fact implies that the inverse of a bijective map is againa bijection. So φ−1 is a bijection, and it remains to show φ−1 is a homomorphism of groups.

Indeed, given h, h′ ∈ H, let g, g′ ∈ G be the unique elements of G such that φ(g) = h andφ(g′) = h′; such g, g′ exist and are unique since φ is bijective. Then note φ(gg′) = hh′ sinceφ is a homomorphism.

Thenφ−1(hh′) = gg′ = φ−1(h)φ−1(h′),

2

as desired.

6. Prove that isomorphism is an equivalence relation on groups.

Solution. It’s reflexive since idG : G→ G is an isomorphism from G to itself. The previousproblem shows that it is symmetric. For transitivity, it suffices to show that a compositionof isomorphisms is again an isomorphism. In other words, we show that if φ : G → H andψ : H → K are isomorphisms, then so is ψ ◦ φ : G→ K.

The fact that ψ ◦ φ is a bijection follows from the fact that compositions of bijections arebijections. I omit the proof. To check that ψ ◦ φ is a homomorphism: given g, g′ ∈ G, note

(ψ ◦ φ)(gg′) = ψ(φ(g)φ(g′)) = (ψ ◦ φ)(g)(ψ ◦ φ)(g′),

where we used in turn the facts that ψ and φ are homomorphisms.

3

Problem Set 4

2. How many homomorphisms Z → Z/3Z are there? How many homomorphisms Z/3Z → Zare there? Justify your answers briefly.

Solution. There are three homomorphisms Z → Z/3Z, since we showed in class that forany group G and any g ∈ G, there is a unique homomorphism φ : Z→ G such that φ(1) = g.

There is one homomorphism Z/3Z → Z. We mentioned in class that for any pair of groupsG and H, the map sending everything in G to 1H is always a homomorphism (check this), sothere is at least one such homomorphism.

On the other hand, suppose φ : Z/3Z → Z is any homomorphism; we’ll show that φ(n) = 0for all n ∈ Z/3Z, so φ was in fact the one homomorphism we already identified.

It suffices to show that φ(1) = 0, since then φ(n) = nφ(1) = 0. We note

0 = φ(0) = φ(1 + 1 + 1) = φ(1) + φ(1) + φ(1).

So φ(1) = 0.

5. Does R have any subgroups isomorphic to Z2? Prove your answer.

Solution. Yes, letH = {a+ b

√2 : a, b ∈ Z}.

We claim H ≤ R is a subgroup isomorphic to Z2. I will omit the proof that H is a subgroup.1

Let us now consider the mapφ : Z2 → H

given by (a, b) 7→ a+ b√

2. This map is a homomorphism: indeed

φ((a, b) + (c, d)) = φ((a+ c, b+ d)) = (a+ c) + (b+ d)√

2 = φ((a, b)) + φ((c, d)).

It is surjective by construction. As for injectivity, suppose φ((a, b)) = φ((c, d)); we wish toshow (a, b) = (c, d). We have

a+ b√

2 = c+ d√

2 implies a− c = (d− b)√

2.

The left hand side is an integer and the right hand side is irrational unless d − b = 0. Weconclude d− b = 0, and therefore a− c = 0, as desired. Therefore φ is an isomorphism.

1In fact, it follows from the fact that H = im(φ) where φ is defined below; but we had not isolated and provedthe fact that im(φ) is always a subgroup when this problem was assigned.)

4

6. Does Q have any subgroups isomorphic to Z2? Prove your answer.

Solution. We claim the answer is no. Suppose on the contrary that φ : Z2 → H was anisomorphism from Z2 to some subgroup H. Let us write the rational numbers φ(1, 0) andφ(0, 1) with a common denominator, say

φ(1, 0) = a/n, φ(0, 1) = b/n

for some a, b, n ∈ Z with n 6= 0. Then

φ(b, 0) = φ(0, a) = ab/n.

Since φ is injective, we conclude (b, 0) = (0, a), so a = b = 0. But this contradicts thatφ(1, 0) 6= 0 and φ(0, 1) 6= 0 by injectivity of φ.

5

Problem Set 5

2. (a) Let G be a group acting on a set A. The stabilizer of an element a ∈ A, denoted Ga, isdefined to be the set

Ga = {g ∈ G : g · a = a}.

Prove that the stabilizer is a subgroup of G.

In class on Thursday February 23, you considered the action of the group G of rotations onthe power set P(R2) of R2.

(b) Which elements of P(R2) have stabilizer equal to G? Justify your answer briefly.

(c) (optional, extra) Does there exist any S ∈ P(R2) with infinite cyclic stabilizer?

Solution. Note 1 ∈ Ga. Next, if g, h ∈ Ga, then we have

(gh−1) · a = g(h−1 · a) = g · a = a,

so gh−1 ∈ Ga. (In general, if h ∈ G and h · b = c for some b, c ∈ A, then applying h−1 onboth sides yields that b = h−1 · c. Then in particular h · a = a implies h−1 · a = a.) Thus Gais a subgroup by the subgroup criterion.

For part (b), suppose S ⊆ R2 has GS = G. If p ∈ S then every point obtained by rotating pabout the origin must also be in S; in other words, the entire locus of points

{x ∈ R2 : ||x|| = ||p||}

lies in S. Thus S is a union of such loci, i.e. it is a union of circles centered at the origin,possibly together with the origin itself.

4. Let p be any prime number. Prove that every group of order p is isomorphic to Z/pZ.

Solution. Let G be a group of order p. Since p ≥ 2, G has some nonidentity element, say x.Then 〈x〉 is a subgroup of G whose cardinality is greater than 1 and divides p (by Lagrange).Therefore 〈x〉 = G, so G is cyclic, and every cyclic group of order p is isomorphic to Z/pZ,as shown in class.

5. We showed in an earlier lecture that

Z(G) = {x ∈ G : gx = xg for all g ∈ G}

is a subgroup of G, called the center of G. Prove that Z(G) is normal, and that it consistsprecisely of the elements of G whose conjugacy classes have size 1.

6

Solution. The first claim follows from the second, since we showed that if a subgroup isa union of conjugacy classes then it is normal. For the second claim, note that an elementx ∈ G has conjugacy class {x} if and only if

gxg−1 = x for all g ∈ G,

equivalently, if gx = xg for all g ∈ G, equivalently if x ∈ Z(G), as desired.

6. List the conjugacy classes of the dihedral group D12. Draw the lattice of subgroups of D12

and indicate in your lattice which subgroups are normal. (No proofs necessary.)

Solution. I will leave the full drawing of the lattice to you. The conjugacy classes are

{1}, {r, r5}, {r2, r4}, {r3}, {s, sr2, sr4}, {sr, sr3, sr5}.

(Having listed these conjugacy classes, you may wish to use this list to check which subgroupsare normal. Namely, you simply check whether your subgroup is made up of conjugacyclasses.)

7

Problem Set 6

1. An element g of a group G is called torsion if it has finite order, and G is called torsion-freeif its only torsion element is the identity.

Let A be an abelian group and let N be the set of its torsion elements. Prove that N is asubgroup and that A/N is torsion-free.

Solution. 1 has order 1, so 1 ∈ N . Now given a, b ∈ N , let us show ab−1 ∈ N . Since a, b aretorsion, there exist integers m,n ≥ 1 with am = bn = 1. Then note

(ab−1)mn = amnb−mn = (am)n(bn)−m = 1,

where the first equality follows from the fact that A is abelian. Thus the subgroup criterionimplies that N is a subgroup.

Next, we show A/N is torsion-free. Given aN ∈ A/N , suppose aN is torsion; we wish to showaN = N is the identity. Let m ≥ 1 be such that (aN)m = N . Then amN = N , implyingam ∈ N .

In other words, am itself is torsion in A, so there exists an integer n ≥ 1 with (am)n = 1.Therefore amn = 1 and so a is torsion itself. So a ∈ N , and aN = N as desired.

2. (No proofs necessary) Let N = Z(D12) E D12.

(a) List the elements of N . You may use Dummit and Foote Problem 4 on p.28, on Home-work 3, or combine your answers from Problems 5 and 6 from Homework 5.

Now list the elements of D12/N ; there should be six of them.

Solution. N = {1, r3}, and

D12/N = {N, rN, r2N, sN, srN, sr2N}.

(b) Write out the multiplication table for the group D12/N . This should be a 6 × 6 table,all of whose entries are taken from your list of elements of D12/N from part (a).

Solution.

· N rN r2N sN srN sr2N

N N rN r2N sN srN sr2NrN rN r2N N sr2N sN srNr2N r2N N rN srN sr2N sNsN sN srN sr2N N rN r2NsrN srN sr2N sN r2N N rNsr2N sr2N sN srN rN r2N N

8

3. Number the three diagonals of a regular hexagon as shown. Let D12 act on the set {1, 2, 3}via g · i = j if g ∈ D12 takes the diagonal i to diagonal j.

Let φ : D12 → S3 be the permutation representation of this action. Use φ to prove that

D12/N ∼= S3.

Feel free to appeal to your geometric intuition. (In light of this fact, you may wish to compareyour multiplication table for D12/N with your multiplication table for S3 from Homework 3.)

1

3

2

Solution. We know that φ is indeed a homomorphism, as stated in the guest lecture. Itsuffices to check that φ is surjective, and that kerφ = N , for then the desired isomorphismfollows from the First Isomorphism Theorem.

Let us show that φ is surjective. Let us write r for rotation 60 degrees clockwise, and s forreflection across the diagonal 1. (Please note: we shall follow the conventions of the textbookthat products like sr are interpreted as “first do r, then do s.”) Then φ(r) = (132) andφ(s) = (23).

Now im(φ), being a subgroup of S3, necessarily contains 〈(132)〉 = {1, (132), (123)}. Butit also contains φ(s) = (23). Therefore | im(φ)| ≥ 4, but | im(φ)| divides 6 by Lagrange’stheorem; therefore im(φ) = S3.

Finally we claim ker(φ) = N . For the inclusion ⊇, note that 1 and r3 do indeed send eachdiagonal to itself (r3 does so by flipping each diagonal). On the other hand, since rigidmotions send adjacent vertices to adjacent vertices, any rigid motion in ker(φ) that flips onediagonal flips them all, and any rigid motion in ker(φ) that fixes one diagonal fixes them all.This shows the claim.

9

Problem Set 8

1. In class today (March 21), you formed groups of five students each with four students remain-ing, and then groups of seven students each with three students remaining. In your opinion,how many students came to class today?

Solution. Since the natural map Z/35Z→ Z/5Z×Z/7Z discussed in class sends 24 to (4, 3),we conclude from the Chinese Remainder Theorem that the number of students present musthave been 24 (mod 35). I think there were 24 + 35 = 59 students present.

2. Let m,n ≥ 1 be integers. Prove that the set of numbers appearing as the order of someelement of Z/mZ× Z/nZ is

{d ∈ Z>0 : d divides lcm(m,n)}.

Solution. Let us first establish a lemma: for any group G, if g ∈ G has order n, and n = cdfor integers c, d ≥ 1, then gd has order c. Indeed, (gd)c = 1 but for any positive numberc′ < c, note dc′ < dc = |g|, so (gd)c

′= gdc

′ 6= 1.

Let’s write ` = lcm(m,n) for short. We claim that |(1, 1)| = `. Indeed, note that (`, `) = (0, 0)since m|` and n|`. On the other hand, if (d, d) = (0, 0) for d ∈ Z, then m|d and n|d, hence`|d (as established in class). Hence |(1, 1)| = `, and by the lemma, every positive divisor of `occurs as the order of some element of the form (d, d).

Conversely, if (a, b) has order d, then d necessarily divides ` since (a`, b`) = (0, 0).

3. Let G be a group and let∆ = {(g, g) : g ∈ G} ⊆ G×G.

Prove that ∆ ≤ G×G and that ∆ is normal if and only if G is abelian.

Solution. ∆ is often called the diagonal. I leave it to you to check ∆ is a subgroup. I alsoleave it to you to check that G × G is abelian if and only if G is abelian. Thus the “if”direction follows since every subgroup of an abelian group is normal. Conversely, suppose ∆is normal, and given g, h ∈ G, let us prove gh = hg, or equivalently g = hgh−1. We have

∆ 3 (h, 1)(g, g)(h, 1)−1 = (hgh−1, g),

implying that hgh−1 = g as desired.

10

4. Let H,K be groups and let φ : K → Aut(H) be a homomorphism. Recall that G = H oφ Kcontains subgroups

H̃ = {(h, 1) : h ∈ H} E G K̃ = {(1, k) : k ∈ K} ≤ G

isomorphic to H and K respectively. Prove that K̃ E G if and only if φ is trivial.

(The map φ being trivial means that φ(k) = idH for all k ∈ K.)

Solution. If φ is trivial, then k · h = h for all h ∈ H, k ∈ K (here · denotes the action of kon h via the map φ. Then for all h ∈ H, k, k′ ∈ K, we have

(h, k)(1, k′)(h, k)−1 = (h, kk′)(h−1, k−1) = (hh−1, kk′k−1) = (1, kk′k−1) ∈ K̃,

so K̃ E G.

Conversely, assume K̃ E G. Given h ∈ H, k ∈ K, we wish to show k · h = h. Note

(h−1, 1)(1, k)(h−1, 1)−1 = (h−1, k)(h, 1) = (h−1(k · h), k).

This last element is in K̃ by assumption, so h−1(k · h) = 1, so k · h = h as desired.

5. Consider the group of rigid motions of R2, i.e. rotations, reflections, translations, and compo-sitions of these. (Optional, ungraded: express this group as a nontrivial semidirect product.)

For each of the following subsets S of R2, let G be the group of rigid motions of R2 thatpreserves S. Exhibit G as a nontrivial semidirect product in each case. Brief explanationsmay be helpful, but full proofs are not necessary.

(a) S = {(x, y) ∈ R2 : x2 + y2 = 1}, the unit circle

Solution.

R/2πZ o Z/2Z where the nontrivial element of Z/2Z acts on the group of rotationsR/2πZ by negation. (Same as the cylinder example in class.)

(b) S = Z2, the integer lattice points

Solution.

One possible solution: Z2 oD8, where D8 is identified with the group of rotations andreflections preserving the square {(x, y) ∈ R2 : −1 ≤ x, y ≤ 1} (these are precisely therigid motions of the plane that fix the origin and preserve Z2), and acts on Z2 accordingto the natural action of this group on the plane R2.

11

(c) S = {(x, y) ∈ R2 : −1 ≤ y ≤ 1}, an infinite horizontal strip.

Solution. Z/2Zo (R×Z/2Z), where the second factor records translations and reflec-tion across the x-axis, and the first factor records reflection across the y-axis. Here, theaction Z/2Z→ Aut(R× Z/2Z) sends the nonidentity element of Z/2Z to the automor-phism of R× Z/2Z acting by negation on R and trivially on Z/2Z.

(There are other equally good ways to describe the same group, such as (Z/2Z×Z/2Z)oφ

R, with φ defined appropriately.)

6. Let p and q be distinct primes. Prove that any group of order pq is a semidirect product oftwo of its proper subgroups.

Solution. By Sylow’s theorem such a group G has subgroups K and H of order p and qrespectively. Assume without loss of generality that p < q; then nq ≡ 1 mod q and nq|pimplies nq = 1, hence H is normal.

Moreover H ∩K = 1, since (|H|, |K|) = 1, and the fact that HK/H ∼= K/(K ∩H) (SecondIsomorphism Theorem) implies |HK| = |H||K| so HK = G.

As shown in class, these hypotheses imply that G ∼= H oK.

12

Problem Set 9

4. Let I be the ideal of Z[x] consisting of polynomials whose constant term, coefficient of x, andcoefficient of x2 are zero. Identify, with proof, the nilradical of Z[x]/I.

Solution. Write f = f + I for short. We claim the nilradical is (x), that is, the nilradical iscomprised of the classes of polynomials with zero constant term. If f has zero constant term,

then f3 ∈ I, so f3

= 0 in Z[x]/I. Conversely, if f has constant term a0 6= 0, then fn hasconstant term an0 6= 0, so f

n 6= 0 ∈ Z[x]/I for any n ≥ 0.

5. Let I be an ideal in a ring R, and let π : R → R/I be the natural projection. Prove thefollowing universal property of quotients: If ϕ : R → S is any ring homomorphism such thatI ⊆ ker(φ), then there exists a unique homomorphism ϕ : R/I → S such that

ϕ = ϕ ◦ π.

Solution. If such a ϕ exists, then given any a ∈ R, the equation above implies

ϕ(π(a)) = ϕ(a+ I) = ϕ(a).

Thus ϕ is uniquely determined, if it exists.

Now we check that ϕ(a + I) = ϕ(a) is a well-defined homomorphism. Given a, a′ ∈ R, ifa + I = a′ + I then a− a′ ∈ I ⊆ ker(ϕ), hence ϕ(a) = ϕ(a′). This shows well-definedness. Ileave it to you to check that ϕ is a homomorphism of rings.

13

Problem Set 10

1. Let R be an integral domain, and let f, g ∈ R. Prove that (f) = (g) if and only if f = ga forsome unit a.

Solution. Let us prove the ⇒ direction; I leave the other direction to you. If (f) = (g) thenf = ga and g = fb for some a, b ∈ R. Thus f = abf and f(1 − ab) = 0. Now if f = 0 thennecessarily g = 0 and we are done; if f 6= 0 then 1− ab = 0 since R has no zero divisors. Soa and b are units.

2. Let K be a field, and consider the ring K[[x]] of formal power series.

(a) Prove that K[[x]] is an integral domain.

Solution. Suppose f, g ∈ K[[x]] are nonzero; we wish to show fg is nonzero. Definethe order of a nonzero power series f to be the smallest exponent appearing in f .

Let d and e denote the orders of f and g respectively, and let ad and be be the coefficientof xd in f and the coefficient of xe in g, respectively. Note ad, be 6= 0.

Then the coefficient of xd+e in fg is adbe, which is nonzero since K is a field and hencehas no zerodivisors.

Thus fg 6= 0, since we have exhibited a nonzero coefficient in it.

(b) Prove that the ideals of K[[x]] are 0 or are of the form (xn) for some integer n ≥ 0.

Solution. Let I ⊆ K[[x]] be an ideal. If I = 0 we are done. Otherwise, let f ∈ I be anonzero element of smallest order d.

We claim I = (xd). The inclusion I ⊆ (xd) follows from the fact that every element ofI has order at least d and hence is divisible by xd.

The inclusion I ⊇ (xd) follows from the fact that we may factor f = xd · g for g ∈ K[[x]]a power series with nonzero constant term. Such power series are units, by a previoushomework problem. Therefore xd ∈ I, as desired.

(c) Which of the ideals of K[[x]] are principal? maximal? prime? Prove your answers.

Solution. Each ideal is generated by 1 element (including the 0 ideal which is generatedby 0), so they are are all principal.

Note (x) ⊃ (x2) ⊃ · · · ⊃ 0 and (x) is thus the unique maximal ideal.

14

Note that (x) is therefore prime, being maximal; and 0 is prime since K[[x]] is an integraldomain. On the other hand, we claim that for d ≥ 2 the ideal I = (xd) is not prime,since x · xd−1 ∈ I but x 6∈ I and xd−1 6∈ I, which I leave to you to verify.

5. This is a problem in beginning algebraic geometry. Given an ideal I ⊆ R[x, y], we let thevanishing locus or variety of I be the subset of R2

V (I) = {(a, b) ∈ R2 : f(a, b) = 0 for all f ∈ I}.

(a) Prove that if I = (f1, . . . , fn) then V (I) = {(a, b) ∈ R2 : fi(a, b) = 0 for all i = 1, . . . , n}.(b) Draw pictures of V (I) for I = (y(y − x2)) and for I = (x− y, y − x3).(c) Prove that if I1 ⊆ I2 are ideals of R[x, y] then V (I1) ⊇ V (I2).

(d) Using part (c) to help, prove the identities

V (I + J) = V (I) ∩ V (J) and V (IJ) = V (I ∩ J) = V (I) ∪ V (J).

Check your results in part (b) accordingly.

Solution. I will prove the second statement of (d) only and leave the rest to you. This isthe statement that

V (IJ) = V (I ∩ J) = V (I) ∪ V (J).

To prove it, let us begin by noting

IJ ⊆ I ∩ J ⊆ I and IJ ⊆ I ∩ J ⊆ J.

So by part (c), we haveV (IJ) ⊇ V (I ∩ J) ⊇ V (I) ∪ V (J).

It remains only to proveV (IJ) ⊆ V (I) ∪ V (J).

Suppose there is a point (a, b) ∈ R2 in neither V (I) nor V (J). This means there exists f ∈ Iand g ∈ J with f(a, b) 6= 0 and g(a, b) 6= 0. Therefore (fg)(a, b) 6= 0. Since fg ∈ IJ , weconclude (a, b) 6∈ V (IJ), as desired.

15

Problem Set 11

2. Let K be a field. Prove that K[[x]] is a Euclidean domain with respect to the following norm:N(0) = 0, and for all nonzero p ∈ K[[x]], N(p) is the order of p, i.e. the smallest exponentappearing in p.

Solution. First, we claim the following, using what you know about K[[x]] from last week’shomework: if f, g ∈ K[[x]] are nonzero power series, then g|f if and only if ord(f) ≥ ord(g).

I leave the proof of the above claim to you.

Now, suppose f, g ∈ K[[x]] and g 6= 0. If f = 0 then

f = 0 · g + 0.

If f 6= 0 and ord(f) < ord(g), thenf = 0 · g + f.

Finally, if f 6= 0 and ord(f) ≥ ord(g), then

f = h · g + 0

for some h, by the claim above. In each case, the Euclidean property is satisfied.

(By the way, this problem is an instance of the more general statement that discrete valuationrings are Euclidean domains).

4. Compute a gcd of 4 + 2i and 5i in Z[i]. Identifying Z[i] with the integer lattice points in thecomplex plane, draw a picture of the elements of the ideal (4 + 2i, 5i).

Solution. We have

5i = i(4 + 2i) + (2 + i)

4 + 2i = 2(2 + i) + 0

So 2 + i is a gcd, and the ideal (4 + 2i, 5i) = (2 + i) looks like a square lattice, generatedadditively by 2 + i and −1 + 2i.

5. Let R be an integral domain. We defined the field of fractions K, whose elements are equiv-alence classes of {(a, b) : a, b ∈ R, b 6= 0} where (a, b) ∼ (c, d) if ad = bc. We write a/b for theclass of (a, b). We defined

a/b+ c/d = (ad+ bc)/bd, a/b · c/d = (ac)/(bd).

Convince yourself that + and · are well-defined and make K into a field with 0 = 0/1 and1 = 1/1 (ungraded).

16

(a) Prove that the map i : R → K given by i(r) = r/1 is a ring homomorphism sending allnonzero elements to units.

Solution. Given r, r′ ∈ R, we check

i(r + r′) = (r + r′)/1 = r/1 + r′/1 = i(r) + i(r′)

andi(rr′) = (rr′)/1 = i(r)i(r′).

Moreover if r 6= 0 then r/1 is a unit since r/1 · 1/r = 1.

(b) Prove the following universal property of localization: Let S be a commutative ring with1. If f : R→ S is any ring homomorphism sending all nonzero elements of R to units ofS, then there is a unique ring homomorphism f̃ : K → S such that

f = f̃ ◦ i.

Solution. First we establish that f(1) = 1. We have

f(1) = f(1 · 1) = f(1) · f(1).

But f(1) is a unit in S by assumption, so we conclude 1 = f(1).

Next, if f̃ : K → S is a ring homomorphism satisfying the conditions above, thenf̃(r/1) = f(r) for all r 6= 0. This implies

f̃(r/1) · f̃(1/r) = f̃(1) = f(1) = 1,

and since f̃(r/1) = f(r) is assumed to be a unit, the equation above implies f̃(1/r) =f(r)−1. Therefore, for all r, s ∈ R with r 6= 0,

f̃(s/r) = f̃(s/1 · 1/r) = f(s)f(r)−1.

This shows uniqueness of f̃ , if it exists.

Finally, we check existence, i.e. that f̃(s/r) := f(s)f(r)−1 really is a well-defined ringhomomorphism with f = f̃ ◦ i. Suppose s′/r′ = s/r, so s′r = sr′. Then f(s′)f(r) =f(s)f(r′), and

f(s′)f(r′)−1 = f(s)f(r)−1,

which shows that f̃ is well-defined. I leave it to you to check that f̃ is a ring homomor-phism with f = f̃ ◦ i.

17

Problem Set 12

1. Let R = {(a1, a2, a3, . . .) : ai ∈ Z}, i.e., R is the ring of infinite tuples of Z, indexed by thepositive integers, with coordinatewise addition and multiplication. For each j = 1, 2, . . . let

Ij = {(a1, a2, a3 . . .) ∈ R : ai = 0 for all i ≥ j.}

(a) Show that the Ij are principal ideals forming an ascending chain I1 ( I2 ( · · · thatdoesn’t stabilize. Conclude that I =

⋃j≥1 Ij an ideal that is not finitely generated.

Solution. Write ej for the element which is j ones followed by all zeroes. Note thatIj = {(a1, a2, . . .) · ej : ai ∈ Z} = (ej), so in particular Ij is a principal ideal.

For each j, we have Ij ( Ij+1 directly from the definitions; in particular ej+1 ∈ Ij+1 \ Ijshows that Ij is properly contained in Ij+1.

Therefore I is not finitely generated, since if instead I = (f1, . . . , fn), then there areindices j1, . . . , jn such that fi ∈ Iji . Let j = max(j1, . . . , jn). Then f1, . . . , fn ∈ Ij . Butthis would imply that Ij = I, contradicting that the ideals Ij+1, Ij+2, . . . are strictlybigger than Ij .

(b) Is I prime?

Solution. No. Note that I consists of those sequences of integers which are eventuallyzero. Then writing a = (1, 0, 1, 0, . . .) and b = (0, 1, 0, 1, . . .), we have ab = 0 ∈ I buta, b 6∈ I.

3. Convince yourself that the polynomial x3 + x + 1 is irreducible in F2[x]. Write out themultiplication table for the 8-element field K = F2[x]/(x3 + x + 1), and check that themultiplicative group of nonzero elements in K is isomorphic to Z/7Z.

Solution. If x3 + x+ 1 were reducible, then one of its factors would have to have degree 1,since it has degree 3, i.e. it would have a solution over F2. But it doesn’t (as seen by pluggingin 0 and 1.)

The elements of K are (the classes of)

0, 1, x, x+ 1, x2, x2 + 1, x2 + x, x2 + x+ 1.

I leave the full multiplication table to you, but here is one example. Let us compute x2 ·(x2+1)in K. We use the division algorithm, dividing x4 + x2 by x3 + x+ 1 to get

x4 + x2 = x(x3 + x+ 1) + x

18

So in K, we have x2 · (x2 + 1) = x.

4. For K a field of characteristic p > 0, we define the Frobenius map e : K → K by e(a) = ap.Show that e is a homomorphism. (To show that e(a+b) = e(a)+e(b) you may wish to appealto the Binomial Theorem.)

Also, compute what e does to each element of the field K from Problem 3.

Solution. We have e(ab) = (ab)p = apbp = e(a)e(b) by commutativity of multiplication.Next, we have

(a+ b)p = ap + (stuff) + bp,

where the stuff in the middle consists of terms of the form(pi

)· aibp−i where(

p

i

)=

p!

i!(p− i)!.

(To be clear, by(pi

)we officially mean 1 + · · · + 1 ∈ K where there are

(pi

)summands. The

integer(pi

)is a multiple of p, since the numerator has a factor of p but not the denominator.

Since k has characteristic p, we conclude that (a+ b)p = ap + bp.

As for the specific field K from problem 3, I computed that e sends the elements

0, 1, x, x+ 1, x2, x2 + 1, x2 + x, x2 + x+ 1

to0, 1, x2, x2 + 1, x2 + x, x2 + x+ 1, x, x+ 1

respectively.

(In particular, you may check that e is actually an automorphism of K whenever K is a finitefield.)

19