# MATH 1530 ABSTRACT ALGEBRA Selected solutions to problems. a + b = a + b,

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1 MATH 1530 ABSTRACT ALGEBRA Selected 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. So as desired. a + b = a + b, 1

2 Problem Set 3 2. Write out the full multiplication table for the group S 3. It should be a 6 6 table whose entries are elements of S 3, and whose rows and columns are indexed by the elements of S 3. 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 S n has order at most n? Either prove it, or give a counterexample. Solution. No. We saw in the Youtube lecture that the order of an element σ S n is the least common multiple of all of the numbers appearing as lengths of some cycle in σ, when σ is expressed in cycle notation. Thus the order of (12)(345) S 5 is Let (G, ) and (H, ) be groups, and let φ: G H be an isomorphism. Let φ 1 : H G denote 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 map g : B A with f g = id B and g f = id A, where id A : A A and id B : B B denote the respective identity maps. This fact implies that the inverse of a bijective map is again a 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 and φ(g ) = h ; such g, g exist and are unique since φ is bijective. Then note φ(gg ) = hh since φ is a homomorphism. Then φ 1 (hh ) = gg = φ 1 (h)φ 1 (h ), 2

3 as desired. 6. Prove that isomorphism is an equivalence relation on groups. Solution. It s reflexive since id G : G G is an isomorphism from G to itself. The previous problem shows that it is symmetric. For transitivity, it suffices to show that a composition of isomorphisms is again an isomorphism. In other words, we show that if φ: G H and ψ : 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 are bijections. 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

4 Problem Set 4 2. How many homomorphisms Z Z/3Z are there? How many homomorphisms Z/3Z Z are there? Justify your answers briefly. Solution. There are three homomorphisms Z Z/3Z, since we showed in class that for any 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 groups G and H, the map sending everything in G to 1 H is always a homomorphism (check this), so there is at least one such homomorphism. On the other hand, suppose φ: Z/3Z Z is any homomorphism; we ll show that φ(n) = 0 for 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 So φ(1) = 0. 0 = φ(0) = φ( ) = φ(1) + φ(1) + φ(1). 5. Does R have any subgroups isomorphic to Z 2? Prove your answer. Solution. Yes, let H = {a + b 2 : a, b Z}. We claim H R is a subgroup isomorphic to Z 2. I will omit the proof that H is a subgroup. 1 Let us now consider the map φ: Z 2 H given by (a, b) 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 to show (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. We conclude d b = 0, and therefore a c = 0, as desired. Therefore φ is an isomorphism. 1 In fact, it follows from the fact that H = im(φ) where φ is defined below; but we had not isolated and proved the fact that im(φ) is always a subgroup when this problem was assigned.) 4

5 6. Does Q have any subgroups isomorphic to Z 2? Prove your answer. Solution. We claim the answer is no. Suppose on the contrary that φ: Z 2 H was an isomorphism from Z 2 to some subgroup H. Let us write the rational numbers φ(1, 0) and φ(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 0. Then φ(b, 0) = φ(0, a) = ab/n. Since φ is injective, we conclude (b, 0) = (0, a), so a = b = 0. φ(1, 0) 0 and φ(0, 1) 0 by injectivity of φ. But this contradicts that 5

6 Problem Set 5 2. (a) Let G be a group acting on a set A. The stabilizer of an element a A, denoted G a, is defined to be the set G a = {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 on the power set P(R 2 ) of R 2. (b) Which elements of P(R 2 ) have stabilizer equal to G? Justify your answer briefly. (c) (optional, extra) Does there exist any S P(R 2 ) with infinite cyclic stabilizer? Solution. Note 1 G a. Next, if g, h G a, then we have (gh 1 ) a = g(h 1 a) = g a = a, so gh 1 G a. (In general, if h G and h b = c for some b, c A, then applying h 1 on both sides yields that b = h 1 c. Then in particular h a = a implies h 1 a = a.) Thus G a is a subgroup by the subgroup criterion. For part (b), suppose S R 2 has G S = G. If p S then every point obtained by rotating p about the origin must also be in S; in other words, the entire locus of points {x R 2 : 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 consists precisely of the elements of G whose conjugacy classes have size 1. 6

7 Solution. The first claim follows from the second, since we showed that if a subgroup is a union of conjugacy classes then it is normal. For the second claim, note that an element x 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 D 12. Draw the lattice of subgroups of D 12 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, r 5 }, {r 2, r 4 }, {r 3 }, {s, sr 2, sr 4 }, {sr, sr 3, sr 5 }. (Having listed these conjugacy classes, you may wish to use this list to check which subgroups are normal. Namely, you simply check whether your subgroup is made up of conjugacy classes.) 7

8 Problem Set 6 1. An element g of a group G is called torsion if it has finite order, and G is called torsion-free if 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 a subgroup 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 are torsion, there exist integers m, n 1 with a m = b n = 1. Then note (ab 1 ) mn = a mn b mn = (a m ) n (b n ) m = 1, where the first equality follows from the fact that A is abelian. Thus the subgroup criterion implies that N is a subgroup. Next, we show A/N is torsion-free. Given an A/N, suppose an is torsion; we wish to show an = N is the identity. Let m 1 be such that (an) m = N. Then a m N = N, implying a m N. In other words, a m itself is torsion in A, so there exists an integer n 1 with (a m ) n = 1. Therefore a mn = 1 and so a is torsion itself. So a N, and an = N as desired. 2. (No proofs necessary) Let N = Z(D 12 ) D 12. (a) List the elements of N. You may use Dummit and Foote Problem 4 on p.28, on Homework 3, or combine your answers from Problems 5 and 6 from Homework 5. Now list the elements of D 12 /N; there should be six of them. Solution. N = {1, r 3 }, and D 12 /N = {N, rn, r 2 N, sn, srn, sr 2 N}. (b) Write out the multiplication table for the group D 12 /N. This should be a 6 6 table, all of whose entries are taken from your list of elements of D 12 /N from part (a). Solution. N rn r 2 N sn srn sr 2 N N N rn r 2 N sn srn sr 2 N rn rn r 2 N N sr 2 N sn srn r 2 N r 2 N N rn srn sr 2 N sn sn sn srn sr 2 N N rn r 2 N srn srn sr 2 N sn r 2 N N rn sr 2 N sr 2 N sn srn rn r 2 N N 8

9 3. Number the three diagonals of a regular hexagon as shown. Let D 12 act on the set {1, 2, 3} via g i = j if g D 12 takes the diagonal i to diagonal j. Let φ: D 12 S 3 be the permutation representation of this action. Use φ to prove that D 12 /N = S 3. Feel free to appeal to your geometric intuition. (In light of this fact, you may wish to compare your multiplication table for D 12 /N with your multiplication table for S 3 from Homework 3.) Solution. We know that φ is indeed a homomorphism, as stated in the guest lecture. It suffices to check that φ is surjective, and that ker φ = N, for then the desired isomorphism follows from the First Isomorphism Theorem. Let us show that φ is surjective. Let us write r for rotation 60 degrees clockwise, and s for reflection across the diagonal 1. (Please note: we shall follow the conventions of the textbook that products like sr are interpreted as first do r, then do s. ) Then φ(r) = (132) and φ(s) = (23). Now im(φ), being a subgroup of S 3, necessarily contains (132) = {1, (132), (123)}. But it also contains φ(s) = (23). Therefore im(φ) 4, but im(φ) divides 6 by Lagrange s theorem; therefore im(φ) = S 3. Finally we claim ker(φ) = N. For the inclusion, note that 1 and r 3 do indeed send each diagonal to itself (r 3 does so by flipping each diagonal). On the other hand, since rigid motions send adjacent vertices to adjacent vertices, any rigid motion in ker(φ) that flips one diagonal flips them all, and any rigid motion in ker(φ) that fixes one diagonal fixes them all. This shows the claim. 9

10 Problem Set 8 1. In class today (March 21), you formed groups of five students each with four students remaining, 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 must have been 24 (mod 35). I think there were = 59 students present. 2. Let m, n 1 be integers. Prove that the set of numbers appearing as the order of some element 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 = cd for integers c, d 1, then g d has order c. Indeed, (g d ) c = 1 but for any positive number c < c, note dc < dc = g, so (g d ) c = g dc 1. Let s write l = lcm(m, n) for short. We claim that (1, 1) = l. Indeed, note that (l, l) = (0, 0) since m l and n l. On the other hand, if (d, d) = (0, 0) for d Z, then m d and n d, hence l d (as established in class). Hence (1, 1) = l, and by the lemma, every positive divisor of l occurs as the order of some element of the form (d, d). Conversely, if (a, b) has order d, then d necessarily divides l since (al, bl) = (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 also leave 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 implying that hgh 1 = g as desired. (h, 1)(g, g)(h, 1) 1 = (hgh 1, g), 10

11 4. Let H, K be groups and let φ: K Aut(H) be a homomorphism. Recall that G = H φ K contains subgroups H = {(h, 1) : h H} G K = {(1, k) : k K} G isomorphic to H and K respectively. Prove that K G if and only if φ is trivial. (The map φ being trivial means that φ(k) = id H for all k K.) Solution. If φ is trivial, then k h = h for all h H, k K (here denotes the action of k on h via the map φ. Then for all h H, k, k K, we have so K G. (h, k)(1, k )(h, k) 1 = (h, kk )(h 1, k 1 ) = (hh 1, kk k 1 ) = (1, kk k 1 ) K, Conversely, assume K 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 K by assumption, so h 1 (k h) = 1, so k h = h as desired. 5. Consider the group of rigid motions of R 2, i.e. rotations, reflections, translations, and compositions of these. (Optional, ungraded: express this group as a nontrivial semidirect product.) For each of the following subsets S of R 2, let G be the group of rigid motions of R 2 that preserves S. Exhibit G as a nontrivial semidirect product in each case. Brief explanations may be helpful, but full proofs are not necessary. (a) S = {(x, y) R 2 : x 2 + y 2 = 1}, the unit circle Solution. R/2πZ Z/2Z where the nontrivial element of Z/2Z acts on the group of rotations R/2πZ by negation. (Same as the cylinder example in class.) (b) S = Z 2, the integer lattice points Solution. One possible solution: Z 2 D 8, where D 8 is identified with the group of rotations and reflections preserving the square {(x, y) R 2 : 1 x, y 1} (these are precisely the rigid motions of the plane that fix the origin and preserve Z 2 ), and acts on Z 2 according to the natural action of this group on the plane R 2. 11

12 (c) S = {(x, y) R 2 : 1 y 1}, an infinite horizontal strip. Solution. Z/2Z (R Z/2Z), where the second factor records translations and reflection across the x-axis, and the first factor records reflection across the y-axis. Here, the action Z/2Z Aut(R Z/2Z) sends the nonidentity element of Z/2Z to the automorphism 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) φ R, with φ defined appropriately.) 6. Let p and q be distinct primes. Prove that any group of order pq is a semidirect product of two of its proper subgroups. Solution. By Sylow s theorem such a group G has subgroups K and H of order p and q respectively. Assume without loss of generality that p < q; then n q 1 mod q and n q p implies n q = 1, hence H is normal. Moreover H K = 1, since ( H, K ) = 1, and the fact that HK/H = K/(K H) (Second Isomorphism Theorem) implies HK = H K so HK = G. As shown in class, these hypotheses imply that G = H K. 12

13 Problem Set 9 4. Let I be the ideal of Z[x] consisting of polynomials whose constant term, coefficient of x, and coefficient of x 2 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 is comprised of the classes of polynomials with zero constant term. If f has zero constant term, then f 3 I, so f 3 = 0 in Z[x]/I. Conversely, if f has constant term a 0 0, then f n has constant term a n 0 0, so f n 0 Z[x]/I for any n Let I be an ideal in a ring R, and let π : R R/I be the natural projection. Prove the following universal property of quotients: If ϕ: R S is any ring homomorphism such that I 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 Thus ϕ is uniquely determined, if it exists. ϕ(π(a)) = ϕ(a + I) = ϕ(a). Now we check that ϕ(a + I) = ϕ(a) is a well-defined homomorphism. Given a, a R, if a + I = a + I then a a I ker(ϕ), hence ϕ(a) = ϕ(a ). This shows well-definedness. I leave it to you to check that ϕ is a homomorphism of rings. 13

14 Problem Set Let R be an integral domain, and let f, g R. Prove that (f) = (g) if and only if f = ga for some unit a. Solution. Let us prove the direction; I leave the other direction to you. If (f) = (g) then f = ga and g = fb for some a, b R. Thus f = abf and f(1 ab) = 0. Now if f = 0 then necessarily g = 0 and we are done; if f 0 then 1 ab = 0 since R has no zero divisors. So a 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. Define the 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 a d and b e be the coefficient of x d in f and the coefficient of x e in g, respectively. Note a d, b e 0. Then the coefficient of x d+e in fg is a d b e, which is nonzero since K is a field and hence has no zerodivisors. Thus fg 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 (x n ) for some integer n 0. Solution. Let I K[[x]] be an ideal. If I = 0 we are done. Otherwise, let f I be a nonzero element of smallest order d. We claim I = (x d ). The inclusion I (x d ) follows from the fact that every element of I has order at least d and hence is divisible by x d. The inclusion I (x d ) follows from the fact that we may factor f = x d g for g K[[x]] a power series with nonzero constant term. Such power series are units, by a previous homework problem. Therefore x d 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 generated by 0), so they are are all principal. Note (x) (x 2 ) 0 and (x) is thus the unique maximal ideal. 14

15 Note that (x) is therefore prime, being maximal; and 0 is prime since K[[x]] is an integral domain. On the other hand, we claim that for d 2 the ideal I = (x d ) is not prime, since x x d 1 I but x I and x d 1 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 the vanishing locus or variety of I be the subset of R 2 V (I) = {(a, b) R 2 : f(a, b) = 0 for all f I}. (a) Prove that if I = (f 1,..., f n ) then V (I) = {(a, b) R 2 : f i (a, b) = 0 for all i = 1,..., n}. (b) Draw pictures of V (I) for I = (y(y x 2 )) and for I = (x y, y x 3 ). (c) Prove that if I 1 I 2 are ideals of R[x, y] then V (I 1 ) V (I 2 ). (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 is the 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 have It remains only to prove V (IJ) V (I J) V (I) V (J). V (IJ) V (I) V (J). Suppose there is a point (a, b) R 2 in neither V (I) nor V (J). This means there exists f I and g J with f(a, b) 0 and g(a, b) 0. Therefore (fg)(a, b) 0. Since fg IJ, we conclude (a, b) V (IJ), as desired. 15

16 Problem Set 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 exponent appearing in p. Solution. First, we claim the following, using what you know about K[[x]] from last week s homework: 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 0. If f = 0 then If f 0 and ord(f) < ord(g), then Finally, if f 0 and ord(f) ord(g), then f = 0 g + 0. f = 0 g + f. 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 valuation rings are Euclidean domains). 4. Compute a gcd of 4 + 2i and 5i in Z[i]. Identifying Z[i] with the integer lattice points in the complex 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, generated additively by 2 + i and 1 + 2i. 5. Let R be an integral domain. We defined the field of fractions K, whose elements are equivalence classes of {(a, b): a, b R, b 0} where (a, b) (c, d) if ad = bc. We write a/b for the class 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 and 1 = 1/1 (ungraded). 16

17 (a) Prove that the map i: R K given by i(r) = r/1 is a ring homomorphism sending all nonzero elements to units. Solution. Given r, r R, we check i(r + r ) = (r + r )/1 = r/1 + r /1 = i(r) + i(r ) and i(rr ) = (rr )/1 = i(r)i(r ). Moreover if r 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 with 1. If f : R S is any ring homomorphism sending all nonzero elements of R to units of S, then there is a unique ring homomorphism f : K S such that f = f 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 f : K S is a ring homomorphism satisfying the conditions above, then f(r/1) = f(r) for all r 0. This implies f(r/1) f(1/r) = f(1) = f(1) = 1, and since f(r/1) = f(r) is assumed to be a unit, the equation above implies f(1/r) = f(r) 1. Therefore, for all r, s R with r 0, f(s/r) = f(s/1 1/r) = f(s)f(r) 1. This shows uniqueness of f, if it exists. Finally, we check existence, i.e. that f(s/r) := f(s)f(r) 1 really is a well-defined ring homomorphism with f = f 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 f is well-defined. I leave it to you to check that f is a ring homomorphism with f = f i. 17

18 Problem Set Let R = {(a 1, a 2, a 3,...) : a i Z}, i.e., R is the ring of infinite tuples of Z, indexed by the positive integers, with coordinatewise addition and multiplication. For each j = 1, 2,... let I j = {(a 1, a 2, a 3...) R : a i = 0 for all i j.} (a) Show that the I j are principal ideals forming an ascending chain I 1 I 2 that doesn t stabilize. Conclude that I = j 1 I j an ideal that is not finitely generated. Solution. Write e j for the element which is j ones followed by all zeroes. Note that I j = {(a 1, a 2,...) e j : a i Z} = (e j ), so in particular I j is a principal ideal. For each j, we have I j I j+1 directly from the definitions; in particular e j+1 I j+1 \ I j shows that I j is properly contained in I j+1. Therefore I is not finitely generated, since if instead I = (f 1,..., f n ), then there are indices j 1,..., j n such that f i I ji. Let j = max(j 1,..., j n ). Then f 1,..., f n I j. But this would imply that I j = I, contradicting that the ideals I j+1, I j+2,... are strictly bigger than I j. (b) Is I prime? Solution. No. Note that I consists of those sequences of integers which are eventually zero. Then writing a = (1, 0, 1, 0,...) and b = (0, 1, 0, 1,...), we have ab = 0 I but a, b I. 3. Convince yourself that the polynomial x 3 + x + 1 is irreducible in F 2 [x]. Write out the multiplication table for the 8-element field K = F 2 [x]/(x 3 + x + 1), and check that the multiplicative group of nonzero elements in K is isomorphic to Z/7Z. Solution. If x 3 + 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 F 2. But it doesn t (as seen by plugging in 0 and 1.) The elements of K are (the classes of) 0, 1, x, x + 1, x 2, x 2 + 1, x 2 + x, x 2 + x + 1. I leave the full multiplication table to you, but here is one example. Let us compute x 2 (x 2 +1) in K. We use the division algorithm, dividing x 4 + x 2 by x 3 + x + 1 to get x 4 + x 2 = x(x 3 + x + 1) + x 18

19 So in K, we have x 2 (x 2 + 1) = x. 4. For K a field of characteristic p > 0, we define the Frobenius map e: K K by e(a) = a p. Show that e is a homomorphism. (To show that e(a+b) = e(a)+e(b) you may wish to appeal to 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 = a p b p = e(a)e(b) by commutativity of multiplication. Next, we have (a + b) p = a p + (stuff) + b p, where the stuff in the middle consists of terms of the form ( p i) a i b p i where ( ) p p! = i i!(p i)!. (To be clear, by ( ( p i) we officially mean K where there are p i) summands. The integer ( p i) 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 = a p + b p. As for the specific field K from problem 3, I computed that e sends the elements to respectively. 0, 1, x, x + 1, x 2, x 2 + 1, x 2 + x, x 2 + x + 1 0, 1, x 2, x 2 + 1, x 2 + x, x 2 + x + 1, x, x + 1 (In particular, you may check that e is actually an automorphism of K whenever K is a finite field.) 19

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