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Outline Solutions Outline Solutions ["finitely generated" is sometimes abbreviated as f.g.] 1.1 1. By R.6, (a + 1)(b + 1) = a(b + 1) + b + 1 = ab + a + b + 1; expanding the second bracket first, we have (a + l)b + a + 1 = ab + b + a + 1, and a + b = b + a follows by cancellation. 2. The unit element is (0,1). 3. For a ring homomorphism we require ax + ay = a(x + y), ax.ay = axy, a.1 = 1; hence we require a = 1. 4. Products with 0,1 are clear. Z/(3) = {O, 1, 2}, where 22 = 1, Z/(4) = {O, 1,2, 3}, where 22 = 0,2.3 = 3.2 = 2,32 = 1. The first is a field , but not the second. 5. Routine verification (try 2 factors first) . 6. If the cyclic group has order m, this is just Z/(m). 7. Last part: AB S;; A, AB S;; B, hence AB S;; An B . Now 1 = a + b, where a E A, b E B, so if c E A n B, then c = aca + bca + acb + bcb E AB. 8. If X S;; Ai' then X S;; n~, hence (X) S;; nAi , where (X) is the ideal generated by X . Since X S;; (X), (X) is one of the Ai and we have equality. 9. (n). resp. (n)/(m), where nlm. 10. If x E X, Y E ~R(X), then xy = yx, so x E ~R(~R(X)); if X is commutative, then xy = yx for all x, y E X . 203 204 Introduction to Ring Theory 1.2 1. If x.y = x + y + xy, then (x.y).z = x + y + xy + z + xz + yz + xyz = x .(y.z), x .o = O.x = x and when x =1= -1, then x + y + xy = 0 has the solution -(1 + X)-lx. 2. Elements: 0, 1, a, a2, where a3 = 1. Addition: a + a = 0 (characteristic 2), 1 + a = a2 (it cannot be 0, 1 or a) . 3. If a E R is regular, then na = n1.a = 0 holds iff nl = 0, so a has the same order as 1, and this can be infinite or any positive integer. 4. aba = a in an integral domain implies ba = 1. For a counter-example in general take the monoid acting on infinite sequences (Ul, u2, . ) by the rules (Ul , U2,· .. )a = (0, Ul , U2,·· .) , (Ul, U2,· · .)b = (U2, U3' ·. ·) and take the ring generated by a and b. 5. If aba = 1, then ab is a left inverse and ba a right inverse of a, hence ab = ba is the inverse. 6. Put Me = {x E Mlx.l = ex}, where e = 0,1. Then Mo, Ml satisfy V.I-V.3 and x = (x - x .l) + x .l, so M = Mo + M l; clearly Mo n Ml = O. 7. Routine. 8. i, iii, iv, vi, vii. 9. Dimension n + 1 (basis:l, x, x2, ... , xn). 10. 1, x , x2 , . • . are linearly independent. 11. If v = Au, then v' = Qv = QAu = QAP-lU' . By reduction to Smith normal form A takes the form I ED 0 = diag(I, ... ,1,0, .. ,0). Since the rank remains unchanged, it is equal to the number of 1 's. 12. A has rank m iff its rows are linearly independent, i.e. XA = 0 =} X = O. When this is so, the m rows of A span pm, i.e. AA I = I for some A I ,and the converse is clear. Moreover, the rank is then m, so n ;::: m. 1.3 1. If Di(e) is the matrix with (i, i)-entry e and the rest zero, then for any diagonal matrix A = diag(al, ... , an), ADi(e) = Di(aie), Di(e)A = Di(ea;). 2. Routine. 3. If Au = Bu for all u , then Aei = Bei' i.e. the i-th columns of A and B agree. If Ul, ... , Un is a basis, then every vector U has the form U = L liUi, hence AUi = BUi for all i implies Au = Bu. 4. Routine. 5. The determinant of a triangular matrix is the product of the diagonal elements. 6. C = (C;j), where eij = 0 unless i < j , but there are only n - 1 steps from 1 to n, so any product of n or more e's is zero. Outline Solutions 205 7. Take an invertible matrix P with first column u such that Cu = 0 (e.g. u could be a non-zero column of ck, with the largest k such that C k =I- 0). Then p-1CP has the first column zero; now use induction on n. 8. If A = (aij), X = (Xij), then tr AX = Ers asrXrs = tr XA. Taking Xij = 1 and all other entries of X zero, we find tr AX = aji. 9. Put e12 = u, e21 = v, ell = UV, e22 = vu and check (1.14). 10. Routine. 11. bi-labl-jbk-Iabl-I = 8kjbi-Iabl-l. Rest clear. 12. Routine (try the 2 x 2 case first). 1.4 1. First part is routine. A can be defined as Zj(m)-module, provided that A has exponent dividing m (Le. mA = 0). 2. The condition for linear dependence of (ai' a2), (bl , b2) is: bi = Aai for some A E QX . Example: (4,6),(10,15). 3. Given x EMs and a E R, define x.a = x(af). 4. Routine. 5. Routine. 6. The bottom row reads: M jker 1 ~ im I; now use Theorem 1.21. 7. If M is generated by X = X' U X", where X' is the finite subset of X involved in the defining relations, then M = F' EB F", where F' is generated by X' with the defining relations for M and F" is free on X". 8. To get a left inverse for /3, take a free generating set of M" and map them to their inverse images under /3; any such mapping can be extended to a homomorphism. 9. Routine. 10. The direct sum of two cyclic groups of coprime orders rand s is cyclic of order rs. 1.5 1. Routine. 2. If f has an inverse g, then Ig = 1, gl = 1; if g' is another inverse, then g' = g'lg = g. 3. For any ring R, the mapping n f-t n.l R is a homomorphism; it is determined by the image of 1 which must be l R , so it is unique. Similarly, a f-t 0 is the only homomorphism to O. 4. The least, resp., greatest element. 5. Any skeleton of Ens has members of arbitrary cardinal and so cannot be small. 6. For each set S of n elements choose a fixed bijection from S to n and use it to translate mappings between sets. 206 Introduction to Ring Theory 7. If S, T, U are functors from I to A, and f: S ~ T, g: T ~ U are natural transformations, then fg : S ~ U is a natural transformation from S to U and the identity has the properties of a neutral element. 8. f: A ~ B can be factorized as f = gh, where 9 : A ~ Z, h : Z ~ B, with zero object Z; then g, h are uniquely determined by A, B resp., by the defin­ ition of Z as zero object. Of the four categories only Gp has a zero object. 9. The composition of morphisms that include Z among its source or target is uniquely defined, by definition of Z. 10. The initial object is easily seen to be Ip : P ~ P. 2.1 1. An endomorphism is A-linear and k is (isomorphic to) a subfield of A. 2. Routine. 3. If a E ~(A) and a f:. 0, then aA = A, so ab = 1 for some b, which must be the inverse of a. For any x E A, x = abx = xab = axb, hence bx = xb. 4. Routine. Substitute from (2.2) in (2.12). 5. Routine. If A has a unit element e, then in AI, el = Ie = e. 6. By Theorem 2.1, Al is embedded in kn +b hence so is A. 7. Routine. 8. (xy)ab = xab.y + x.yab + xa.yb + xb.ya; when we form [a, b], the last two terms cancel. 9. (exp d)-I = exp-d. 10. Associativity is clear. The rest follows from (x 1\ y) 1\ Z = (x 1 z)(S2 + S) 1 y - (y 1 z)(S2 + S) 1 x. 11. Routine. 12. Routine. 2.2 1. GdCi+I ~ kn, as right kn-module. 2. A + (B n A) = (A + B) n A by Theorem 2.6; A contains the left- and is contained in the right-hand side. 3. If f is injective but not surjective on M, then im f is a proper submodule isomorphic to M, giving rise to a strictly descending sequence im fn. For a surjective but not injective g, ker gn is a strictly ascending chain. 4. n must be a prime power. 5. If fi: Mi ~ Mi +I , then i(Mi) - i(Mi+1) = i(ker fi) - i(ker fi+I); now form the alternating sum. 6. Given N c M, take composition series of N and M / N. 7. The index (Z: Gi ) grows without bound, so any n fails to lie in some q. 8. Use Theorem 2.6. Outline Solutions 207 9. If (A+B)nC=AnC+BnC, then (A+C)n(B+C) = An(B+ C) + Cn(B+ C) = AnB+AnC+CnB+C ~ A nB + C. The opposite inclusion is clear. 10. If f: P -> M, then N n Pf = Nf-1 f for any submodule N of M, hence (A+ B)f-lf = (A + B) n Pf = AnPf + BnPf = Af-lf + Bf-1 f, hence (A + B)f-1 = (A + B) f-1 ff-1 ~ Af-1 + Bf-1 • For the converse apply the inclusion mapping of C in M.
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