1. Examples. We did most of the following in class in passing. Now compile all that data.

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1 SOLUTIONS Math A4900 Homework 12 11/22/ Examples. We did most of the following in class in passing. Now compile all that data. (a) Favorite examples: Let R tr, Z, Z{3Z, Z{6Z, M 2 prq, Rrxs, Zrxs, Z{6Zrxsu. For each R P R, briefly answer the following. (Consider making a big table of data for parts (iii) (vii).) (i) What is Rˆ? Answer. We have prqˆ R t0u pzqˆ t1, 1u pz{3zqˆ t 1, 2u pz{6zqˆ t 1, 5u pm 2 prqqˆ GL 2 prq prrxsqˆ R t0u pzrxsqˆ t 1u For pz{6zrxsqˆ, this is a harder computation, whose answer turns out to be pz{6zqˆ. In general, prrxsqˆ tppxq P Rrxs a 0 P Rˆ, a i nilpotent for i ą 0u. (ii) What is the set of zero divisors in R? (Note: in M 2 prq, if AB 0 then 0 detpabq detpaq detpbq.) Answer. We have R : none Z : none Z{3Z : none Z{6Z : t 2, 3u M 2 prq :tx P M 2 prq detpxq 0u t0u Rrxs : none Zrxs : none Again, this is a harder question for Z{6Zrxs. 1

2 2 (iii) Is R commutative? (iv) Is R finite? (v) Is R a division ring? (vi) Is R an integral domain? (vii) Is R a field? R Z Z{3Z Z{6Z M 2 prq Rrxs Zrxs Z{6Zrxs commutative? yes yes yes yes no yes yes yes identity? yes yes yes yes yes yes yes yes finite? no no yes yes no no no no div ring? yes no yes no no no no no ID? yes yes yes no no yes yes no field? yes no yes no no no no no (b) For each of R R, M 2 prq, and Zrxs, give two specific non-trivial proper subrings. For each of R Z{3Z and Z{6Z, list all subrings. [Hint: a subring of R is also a subgroup of pr, `q.] Answer. R : M 2 prq : Zrxs : Z{3Z : Z{6Z : Q, Z, 2Z M 2 pqq, M 2 pzq, M 2 p2zq Z, xz, 2Z 0, Z{3Z 0, 2Z{6Z, 3Z{6Z, Z{6Z (c) Give an example of a commutative ring without identity and a non-commutative ring without identity. Answer. Commutative ring without identity: 2Z. Non-commutative ring without identity: M 2 p2zq. (d) Give an example of non-constant ppxq, qpxq P Z{6Zrxs such that ppxqqpxq 0. Answer. ppxq 2x, qpxq 3x.

3 3 2. Rings. (a) The center of a ring R is ZpRq tz P R zr rz for all r P Ru (i.e. the center of the underlying multiplicative semigroup). (i) Show that ZpRq is a subring of R containing 0 and 1 (if it exists). Proof. Since 0r 0 r0, we have 0 P ZpRq and ZpRq H. Further, if 1 P R, then 1r r r1, so 1 P ZpRq. Now for x, y P ZpRq, r P R, we have so that x y P ZpRq; and px yqr xr yr rx ry rpx yq, pxyqr xry rpxyq, so that xy P ZpRq. So ZpRq is a subring. (ii) Is ZpRq necessarily an ideal of R? Proof. No. If R is not commutative, then there is some r P R ZpRq. So if 1 P R, then 1 P ZpRq, but r1 r R ZpRq. (iii) Show that the center of a division ring is a field. Proof. Since R is a division ring, for all z P ZpRq 0, there is a z 1 P R. Though it s not immediately clear that z 1 is also in ZpRq, we do know that for all r P R, rz zr. Multiplying both sides on the left and right by z 1 then produces z 1 r rz 1, so that z 1 P ZpRq. Therefore ZpRq is also a division ring. Since it is also commutative, it is a field. Note that subrings of division rings are not always themselves division ring. For example, R is a division ring with subring Z; but Z is not a division ring. (b) Show that if R is an integral domain and x P R with x 2 1, then x 1. Proof. If x 2 1, then 0 x 2 1 px ` 1qpx 1q. Since R is an integral domain, we have x ` 1 0 or x 1 0. So x 1. (c) Show that if a P R with a 2 a, then a 0 or 1, or a is a zero divisor. Proof. If a 2 a, then 0 a 2 a apa 1q. So either a 1 or a is a zero divisor.

4 4 3. Homomorphisms. (a) Show that 2Z and 3Z are isomorphic as groups but not as rings. [Hint: Note that 2Z is generated by 2. So try to build a bijective homomorphism and consider the image of 2.] Proof. Recall that, up to isomorphism, there is only one infinite cyclic group. So since 2Z and 3Z are both infinite and cyclic, they must be isomorphic as groups. There are exactly two group isomorphisms, corresponding to the image of 2: since 2 is a generator of 2Z, it must map to one of the generators of 3Z, namely 3 or 3. As rings, if there is an isomorphism between 2Z and 3Z, that isomorphism must also be an isomorphism of their underlying group structure. So if ϕ : 2Z Ñ 3Z is an isomorphism, then either ϕp2q 3 or ϕp2q 3. If ϕp2q 3, then 6 3 ` 3 ϕp2q ` ϕp2q ϕp2 ` 2q ϕp4q ϕp2 2q ϕp2qϕp2q 3 3 9, which is a contradiction. Similarly, if ϕp2q 3, then 6 3 ` p 3q ϕp2q ` ϕp2q ϕp2 ` 2q ϕp4q ϕp2 2q ϕp2qϕp2q p 3q p 3q 9, which is also a contradiction. Therefore, no such isomorphism exists. (b) Let R be the set of (weakly) upper-triangular matrices in M 2 pzq. Prove that ˆa b ϕ : R Ñ Z ˆ Z defined by ÞÑ pa, dq 0 d is a surjective homomorphism and calculate its kernel. Proof. Note that since a and d are free in the codomain, this map is surjective. To check that is it a homomorphism, we compute ˆˆa b ˆˆa1 b ϕ ` ϕ 0 d 0 d 1 pa, dq ` pa 1, d 1 q pa ` a 1, d ` d 1 q ˆˆa ` a 1 b ` b ϕ ˆˆa b ˆa1 b 0 d ` d 1 ϕ ` 0 d 0 d 1 and ˆˆa b ˆˆa1 b ϕ ϕ 0 d 0 d 1 pa, dqpa 1, d 1 q paa 1, dd 1 q ˆˆaa 1 bd ϕ 1 ` ab ˆˆa b ˆa1 b 0 dd 1 ϕ 0 d 0 d 1. Finally, * "ˆa b kerpϕq P R 0 d ˇ pa, dq p0, 0q "ˆ0 b 0 0 ˇˇˇˇ b P Z*.

5 5 (c) Show that if ϕ : R Ñ S is a surjective homomorphism of rings, then the image of ZpRq is contained in ZpSq. Explain why it is necessary to assume that ϕ is surjective. Proof. For s P S, if ϕ is surjective, then there is some r P R such that ϕprq s. Then for any z P ZpRq, ϕpzqs ϕpzqϕprq ϕpzrq ϕprzq ϕprqϕpzq sϕpzq. So ϕpzq P ZpSq. Therefore ZpRq Ď ZpSq. If ϕ is not surjective, then the same does not necessarily hold. For example, let R be the set of diagonal matrices in M 2 prq, let S M 2 prq, and let ϕ be the inclusion map. Then ZpRq R, which is not contained in ZpSq tλi λ P Ru (where I is the identity matrix). (d) Let R and S be rings with identities 1 R and 1 S, respectively. Let ϕ : R Ñ S be a non-zero ring homomorphism. Prove that if ϕp1 R q 1 S, then ϕp1 R q is a zero divisor in S. 4. Ideals. Proof. Note that ϕp1 R q ϕp1 R ˆ 1 R q ϕp1 R qϕp1 R q. So by 2(c), either ϕp1 R q 1, ϕp1 R q 0, or ϕp1 R q and p1 S ϕp1 R qq are both zero divisors. However, for any r P R t1 R u, ϕprq ϕp1 R ˆ rq ϕp1 R qϕprq. So p1 S ϕp1 R qqϕprq 0 for all r. Therefore, since ϕ is not the zero-map, either 1 S ϕp1 R q or p1 S ϕp1 R qq is a zero divisor. But if 1 S ϕp1 R q is a zero-divisor, then ϕp1 R q 0 (since 1 S 1 S 0 is not a zero-divisor). So ϕp1 R q is either 1 S or is a zero divisor. (a) Decide which of the following are ideals in Z ˆ Z: tpa, aq a P Zu, tpa, aq a P Zu, tp2a, 0q a P Zu. Proof. We have tpa, aq a P Zu is a subgroup, but is not an ideal since for any a P Z 0, we have p1, 2qpa, aq pa, 2aq R tpa, aq a P Zu. Similarly for tpa, aq a P Zu. However, p0, 0q P tp2a, 0q a P Zu, so tp2a, 0q a P Zu 0; and for any a, b, c P Z, and p2a, 0q p2b, 0q p2a 2b, 0 0q p2pa bq, 0q P tp2a, 0q a P Zu pb, cqp2a, 0q p2a, 0qpb, cq p2ab, 0q P tp2a, 0q a P Zu. So tp2a, 0q a P Zu is an ideal. (b) Which of the following are subrings of Zrxs? Which of the following are ideals of Zrxs?

6 6 (i) The set of integer polynomials with a 1 0. Answer. This is a subring, since it is closed under subtraction and multiplication. It is not an ideal, however, since it contains 1, but not x ˆ 1 x. (ii) The set of integer polynomials with a 0 P 3Z. Answer. This is an ideal. It is closed under subtraction, and if ppxq has constant term a 0 P 3Z and qpxq has constant term b 0 P Z, then ppxqqpxq has constant term 3a 0 b 0 P 3Z. (iii) Zrx 2 s (the set of integer polynomials with a i 0 for all i odd). Answer. This is a subring, since it is the image of the evaluation map (evaluate x to x 2 ). However, it is not an ideal, since 1 P Zrx 2 s, but x ˆ 1 R Zrx 2 s. (c) Let I be an ideal of R and let S be a subring of R. Show that I X S is an ideal of S (it need not be an ideal of R). Proof. Again, since I and S are subgroups, so is I X S. Further, for any x P I X S and s P S, since and since x P I, we have sx P I because I is an ideal, x P S, we have sx P S because S is closed under multiplication. So sx P I X S. So I X S is an ideal of S. [Note that I X S is not necessarily an ideal of R. For example, let R be a ring with a subring S that is not an idea (see above for examples). Then I R is an ideal of R, but I X S S is not an ideal in R. ]

7 7 5. Nilpotent elements. We call an element x P R nilpotent if x n 0 for some n P Z ą0. (a) Show that if x is nilpotent, then x does not have finite multiplicative order. Give an example. Answer. Assuming 1 0, if x is nilpotent, then either x 0 (in which case x n 0 for all n, so there is no power of x that is equal to 1), or x 0 and x n 0 for some n. In the latter case, let n be the smallest such power for which x n 0. Then x is a zero divisor (since x ˆ x n 1 0 but x, x n 1 0), so x cannot be a unit. But if x m 1, then x is a unit with x 1 x m 1. And a zero-divisor cannot be a unit. For example, in Z{4Z, 2 is nilpotent, since And therefore 2 m m m 2 0 for any m ą 2. So 2 m 1. [The key to keeping track of multiplicative and additive orders versus nilpotency is to keep track of which identity goes with which binary operation. A nilpotent element x has x n 0; an element x of additive order n means xlooooomooooon ` ` x 0; an element x of multiplicative order n means x n 1. ] n (b) Explain why the only nilpotent element of any integral domain is 0. Answer. As shown in the previous problem, non-zero nilpotent elements are all zero-divisors. (c) Prove that if R is commutative, then the nilradical, defined by NpRq tx P R x is nilpotent u, is an ideal of R. [You should use the Binomial Theorem given in Exercise to show closure under addition or subtraction.] Answer. Since 0 is nilpotent, NpRq H. For x, y P NpRq, let a, b P Z ą0 such that x a y b 0. Then a`b ÿ ˆa ` b px yq a`b x a`b i y i. i i 0 But since for each i 0,..., a`b, either a`b i ě a or i ě b. So either x a`b i 0 or y i 0. So px yq a`b 0, and so x y P NpRq. Similarly, for any r P R, prxq a pxrq a x a r a 0, so rx, xr P NpRq.

8 8 (d) Prove that if R is commutative, then the only nilpotent element of R{NpRq is 0. Conclude that NpR{NpRqq 0. Answer. Let x P R{NpRq be nilpotent, with x n 0. Then NpRq 0 x n x n ` NpRq. So x n P NpRq. But if x n P NpRq, then there is some m for which 0 px n q m x nm. So x is nilpotent, and so x P NpRq. Thus x 0. (e) Show that, in M 2 prq, x ˆ and y ˆ are both nilpotent, but x ` y is not. Conclude that the nilradical NpRq is not necessarily an ideal if R is not commutative. Answer. We have x 2 y 2 0, so x and y are nilpotent. However, x ` y 1, which is a unit (it is its own inverse) and therefore not nilpotent. So NpM 2 pzqq is not a subgroup, let alone an ideal. (f) An ideal N is called nilpotent if N n t0u for some n P Z ą0. Prove that the ideal where p is prime. I pz{p 2 Z is a nilpotent ideal of Z{p 2 Z, Answer. For any x, ȳ P pz{p 2 Z, x kp ` p 2 Z, ȳ lp ` p 2 Z, for some k, l P Z. So xȳ pkp ` p 2 Zqplp ` p 2 Zq pkpqplpq ` p 2 Z klp 2 ` p 2 Z p 2 Z 0. So for all z P I 2, for some x i, y i P I, we have Therefore I 2 t 0u. z x 1 ȳ 1 ` ` x m y m 0 ` ` 0 0.