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Answers and Hints CHAPTER 1 m Exercise () GH) O (HD O GUI. (4) (15534). $7. Exercise. When H,,..., H, are submodules of a module G over a ring R, then the following is a necessary and sufficient condition for G to be the directsum of H,,...,H,. G=H++H, and (Ho + +H, NH, = t0) for i=2,...,r. mo 10. q. 12. MW Z=(a),h=%Z)/g, then the solutionsof x! =1lin Z are d”, ExERCISES $1 , CF. exercise 1.2.4. . Symmetric group of degree 3. . Since (ab)” = b”, (b”) = (b), we have (b) E (ab). Similarly, (a) € (ab) and (a, b) E (ab). , 2h h a af, s2 . Use exercise 1.2.8. Show that if G is solvable and N is a normal sub- group, then G/N is solvable. As for the solvability of N, consider G=-6,6,=[6,,,0,]), N9=N, N,=[N,, Ns then we can show N, € G, by induction on í. Note that this proof can be ap- plied to show that if N is a subgroup of a solvable group, then N is solvable, The center Z(G, x:::xG,) of G, x:::xG, coincides with Z(G,) x «xZ(6,) and (6, xx 6/26, x 0x6) = (G/Z(6) xx (6/26). KH HG/H) =2,then HHG) = 2; hence, ac G,a É H implies G=HuUHa=HUaH. Thus aH = Ha. As for (1), use exercise 1.1.7. As for (2), use (1) and the hint, 227 228 ANSWERS AND HINTS 83 3. (D(D)If 7, J areidealsofaring R,then 1+J(InNI)cIiJcIng. (ii) R=(1,+L) Ci+I and Ii+L=R. (2) Use (1) and an induction argument on a. 84 2. The rational number field. 85 L(DIfa,b,cdeR,a£oO, b£O,and ab =0, then, setting / = ax+c, g=bx+d,wehave degfg< Ii
1 and consider finite fields K, such that H(K) = p”. Then K CKkc--,and K=J,K, isa field such that *(K) = oo. Take a natural number m > 1 which is relatively prime to n, and let q be an element of degree m over 7, then « is not in any K,, consequently o isnotin K. Thus, K is not algebraically closed. CHAPTER III. EXERCISES $2 3. To find a counterexample to (iv) implies (iii), take L = K(VD), M = K(£- 2) with the rational number field K, a primitive nth root € of unity and an odd number n>1. 83 2. One remark: It is not a right answer that, considering a prolongation D' of Dto L, we take the restriction of D' to M, because D'M CM 234 ta ANSWERS AND HINTS may not be true. If M has a separating transcendence base, then the existence of the required prolongation is obvious. So, the characteristic O case is finished, and we assume that the characteristic is p £ O, In general, if we fix a p-base B, then there is a one-to-one correspondence between derivations of L and elements of Hom,.(B, L) (the set of mappingsof B to L). Now, we choose B so that L?(K) = L/(K NB), L(M) = LP(M NB). We define a prolongation D' of D byletting D'b be Db if be KnB and any element of M otherwise. 84 + We can reduce the problem to the case where L,, L, are finitely gener- ated. Then we can use separating transcendence bases. . [K,: K)< (L/K) is easy by considering LG, K,. ML/K) <[L: K(x,,..., X,)l; follows from Theorem 3.4.3 and the fact that Ke): Kº (xs X)J2 IL: K(x,,..., X,)],. For this last fact, use Theo- rem 2.7.3, (ii). 85 - To prove sufficiency, note that L is separable because LB, L'isan inte- gral domain for any purely inseparable extension L/. K is algebraically closed in L, because L&, L' is an integral domain even when L” is separable over K. Cf. Theorem 3.5.2, (ii). . Use Corollary 3.4.5. (and Theorem 3.5.2, (ii). . Use exercise 3.2.2. 86 . (1) For a counterexample, let x, y be algebraically independent elements over a field K,andset R=K[x,y,y/x,3/X,...,)/x",...]. Then, yoy/x,... EXR=I,and RI=XK. (ii) For (+) in the case 1” = (0), we shall show that R/Iº is Noetherian by induction on s. Itis soif s=1,and we assume that s > 1, Let J bean ideal of R/I and let & be the natural homomorphism of R/T to R/I7!. Then by our induction hypothesis, J' = &J is finitely generated, and there exist by, be J, =D SARL). Then J = bAR/I9) + (Jn (1/1). Since 17 !/F is finitely generated as an R/I-module, its submodule Jn (17!/T) is finitely generated. Thus, J is finitely generated, . 1 is not a primary ideal, because xy El, x é 1,)" & 1 foral n. . M$I is the unique prime ideal of R/1. Hence, x É M (x e R) implies x mod 7 is invertiblein R/Z. Let f=0,N---nNQ, be a shortest expression of 1 as an intersection of primary ideals. For the only if part, P = /Q, implies there is c € 236 ANSWERS AND HINTS Xx= ax, (X, 1 =4x,) (for some x, ER). xRC x,RC--- and for some m, X,R=X,,R. Then x, =ax,,| —ax,2z (forsome ze R) and x,(l-az)=0. Since aR£R, I-az 0, and therefore, x, =0. Thus, x = O. For the latter half, if a — p,---P, with prime elements P;, then there existsan i for which pc P and P=pKIX,,..., Xl. . It is advised to start with the latter half. If P is a maximal ideal, then consider the field K(X,,..., X,]/P. Let a, be the residue class of X, modulo P and set K, = K(a,,...,4, |). Let f(X) be the mini- mal polynomial for a, over K, and let g(X,,..., X,) be the monic polynomial in X, obtained from /; by replacing coefficients by their representatives in K[X,,...,X, ,J and X by X,. Then P is gen- erated by g,,..., &,- For the first half, we can assume that + > 0. Let Y,,..., Y, be the elements obtained by applying Theorem 3.8.2 to K[X] and P. Then K[X] = K[X,,...,X,,Y.,,::.,Y] and PK(L o FX... X] is a maximal ideal. We apply the latter half to this maximal ideal. . If 7 isa prime ideal, then trans. degp R'/1 = trans.deg R/(1 NR), which implies ht7/ = ht(//NR). In the general case, we can adapt our proof of Corollary 3.7.3. 59 . RIhJC a!'R and a !R isa finitely generated R-module. 810 . For the first half, use the zero-point theorem of Hilbert and the fact that VU(A) + A,))) = A,NA,. For the last part, consider the polynomial ring P=R[x,,...,x,] (n > 2) over the real number field R. Then, with 4,=V(x), 4,=V(-x)+ Dix) +1), we have MA)+ HA) = x PAHO ox) + 1)P=VMA)FIAS), but ANA, is empty. . Let W be a component of VNH, and let h be an element which defines H. We apply Theorem 3.8.7 to O[X,,..., X,]/I(V), and we have htI(W)/I(V) < 1, because I(W)/I(V) is a minimal prime divisor of the principal ideal generated by hA modulo 7(V). Hence, htH(W) < htI(V)+ 1 by Corollary 3.8.5. . (i) It has sufficiently many pointsin R if r>0. If r=0, then there is no R-rational point. (ii) It has sufficiently many points in R. g11 . Consider x) + +22. .Set K=R(t,,...,t) and L=C(t,,...,t;) with algebraically indepen- dent elements £,,..., t, over the real number field R. Here C denotes the complex number field. K isnota C,-field for any n and L isa C,field but nota C, field. ANSWERS AND HINTS 237 g12 1. Use exercise 3.12.1. 2. For the only if part, if (p, q) isa K-rational point, then K(x, y) = K(t) with £ = (x—p)/(y-4) . (Indeed, x = p+i(y-9) ; hence, a(p+ (pa) + by? =c= ap? + bg”. From this relation, we have ye K(t).) If (p, q) with p £ O is a nontrivial solution of ax + by? = 0, then (gp! ,0) is a nontrivial solution of a +bY? = cz? (with x=27", py=xY), we sec the existence of similar t and K(z, Y) = K(x, y). For the if part, assume that x = f(1)/g(t), Y = h(i)/k(t) with polynomials /, g, A, k. If for some pe K, g(D) 0, k(p) £ 0, then (A(p)/g(p), h(p)/k(p)) isa K-rational point. In the other case, factor g, k as g(1) = Ego) » k(1) =U“k(1) with g,, k, such that g,(0), k,(0) are different from 0, We can assume that s>u>0 (>0 follows from the nonexistence of p). Then f(0) £ O and FIGO EAOR + b(tT"h(t ()/k, (97 = ct. By setting t=0, we have a nontrivial solution for ax? + by? =0. REMARK. The proof above shows that if *H(K) is infinite, then ac +by? = c hasa K-rational point ifffor some te L. L=K(!). 3. Use the preceeding exercise and the act that K(z) isa C,-field. (Con- sider the homogeneous form f(z Dx? +lz mr f(o)U? in X,Y,U) SA 1. Take a valuation ring V of K(x,,..., x,) containing K and such that V, has prime ideals P DP D--DP,D (0) with the property that P,= 49», for each i. Then adapt Lemma 3.4.2, considering VN K(X,, 05X). 2. Here is a proof of the theorem stated in the hint. If n =, then there is nothing to prove. Assume that n > r. We use the normalization theorem for polynomial rings using exercise 3.8.1, and we sec that there are linear combinations y,,...,)Y,., Of Xj,..., X, with coefficients in K such that L[x,,..., x,] is integral over L[y,,...,Y, ,]. Choose 4,,-.-,9,€ L suchthat L=K(a,,...,a,). Then, x,,..., x, arein- tegralover K[a,,...,a,,),,.--s Yp.y]. Thereis ce K such that, when we write 4,,...,4, in fractional forms of x,,..., x,, no denominator is divisible by y, —c. Consider the ring V=K[x,,... Xalo ot, a Since a e V, V contains Kla,,...,a,,X,...,X,). Set P = Q-oVn kia... ax... ,x,] and Q = O -o9JVn Kla,, A Pp cs Yao]. Since the field of fractions of R = Ka... X,.coX)/P is VMy, — c)V, we see that trans.deg R=n-—1. R isintegralover S = Kla,,...,0,, Pp ee» Y,-,] and trans.deg;S = n— 1. Nowfor T = Kla,,...,a]/ (QnKla,,...,a]), trans.deg T > (n-1)-(n—r— 1), because v;,-cEQ. But, trans.deg Kla,,...,a] = trans.deg, L = r, and ANSWERS AND HINTS 239 neighborhood of f(P) contains the identity element. 3. Consider the two-dimensional vector space V = ((a, b)la,b Ee R) over the real number field R. This is a topological group under the usual topology (as a Euclidean space). H = [(0, b)|b E RJ is a closed normal subgroup. Then the mapping f/ defined by f(a,b) = a is the one as stated. F=((a,bJO1) imply a”-5º Ent! R, and therefore, n af ben tR. (ii) If we take a,, db, for a, be R/7R, as in (i), then ab, isa representative of (ab)? " (iii) is casy. (iv) is difficult, and the reader is advised to see some book, for instance, N. Jacobson, Lectures on abstract algebra, II. REMARK. In general, Witt vectors and Witt rings are defined over an ar- bitrary commutative ring K with f,, 8, in (iv), which define the addition and multiplication. It is known that the Witt ring of length infinity is a valuation ring if and only if K isa perfect field. 4. Show that m, = Di au; (0, € K) (1=1,2,...) form a Cauchy sequence which converges to O 1f a; (i=1,2,...) form a Cauchy sequence which converges to O, for each j = 1,...,n. The if part follows from [|555,u;| < 5 lb;u;ll-= 3 v(bllu;ll (note that the in- equality follows from the fact that M is a metric space). For the only if part, use an induction argument on n. Assume that 4,,,..., Gp» + is not a Cauchy sequence converging to O. Then there is a positive num- ber & such that v(a,) > é for infinitely many i. Choosing a suitable subsequence of (m,), we may assume that v(a,,) > é for alli. Then (an'mj) is a Cauchy sequence converging to O, where the coefficient of u, in each term is 1. Thus, we may assume that q, =, =:=1. Let «(i) be natural numbers such that (1) < (2) < ---. Then d, = Man M€ Eija Ku, , and (d,) is a Cauchy sequence converging to O. Therefore, by our induction hypothesis, tc; = jo alt =1,2,..) is a Cauchy sequence converging to O for each j >2. It follows now that tai=1,2, ...j is à Cauchy sequence. Set a; = lim, 4: Then m+au,++aiu, =lim,.,m,= 0, which contradicts the linear independence of u,,...,4,- 56 1. Let B be a transcendence base of C over Q. Since *(B) is infinite, 240 ANSWERS AND HINTS Q(B) = Q(B, 1). Hence, there is an injection of C(t) to C. 2. Either its cardinality is greater than the cardinality of continuum or the characteristic is different from 0. 87 1. Let R be such a valuation ring, and let P be a nonmaximal, nonzero prime ideal, Let 0% a €e P,andlet b be an element of the maximal ideal that is notin P. Then b "ae Rand aRcblaRCc-.c b"aRCbTIaRC-.. 2. The intersection of integrally closed integral domains is an integrally closed integral domain. .a-bePifa>b. 4. For the first half, use Theorem 4.7.2, (vi) and the fact that RC SNK € K. For the latter half, show that if P is a prime ideal of R, then VPSnR=P. é Consider P= (alva >h forall he H) for an isolated subgroup H. 7. (i) Use exercise 4.7.1 for the only if part. (ii) For the only if part; take a, E P, such that a, Rp = = PRp-. For each o element b of the field of fractions of R$letm,,...,m, pé the integers such that bRp = ar 6 Rp (i=1,.., mM. Then the mapping wb > (m,,...,m,) gives the required isomorphism. 8. (i) Use an induction argument on n. Let ww, be the valuation defined by R, - Let B beasubsetof P, such that (w, “lb € B) is maximal among linearly independent subsets of w,(P,) over Q. Consider the valuation v defined by R/P,. Our induction hypothesis shows that there is an order isomorphism &, from the value group of v into the (n — 1)-ple direct sum R$---&R. Then we define & as follows. Let x be a nonzero element of K. Since w,x is linearly dependent on (w,b]b e BJ, there is a natural number m such w,(x”) = w,(bj'---b:) with b, E B. Now elx) = (w(by bom, di(u(x” [bj --- bs modulo Pj))/m). 9. It suffices to show that x,,..., X, (€ R) are algebraically independent if () wx, == wx,=0 and x modP,...,x, mod P are alge- braically independent over k and if (ii) wx,,,...., Wx, are linearly independent over the rational number field. Assume for a moment that EG xm=0 (c., €k). Denoteby 5, Coe -«xim the Pra m partial sum on the terms such that w(c, ci, Es . “) is the least. Then we have a contradiction from w(D,,c, Ra : cede) >w (one term in this partial sum) (this ineguality follows + from the fact that wa < wb implies w(a + b) = wa). 88 LIf if, then RIR] =K. (In this case, we say thai R,,..., R, (or, 242 ANSWERS AND HINTS being the rational number field). Let K be an ordered field with an order such that 2 <0. 84 2. Adapt the proof of Theorem 5.4.2. CHAPTER VI. ExERCISES 83 2. Let x, y be algebraically independent elements over a field k of charac- teristic 43, andset K=k(x,)), R=klx, y]. Let z be an algebraic element over K defined by P+xz+y =0,andset L=K(z)=k(x, 2), Ri=klx,y, 2]=klx, 2). (1) P=(x-DR,, P'=(x- DR +ZR,. (ii) P=(z-ax)R, witha root o of Xº+X+41, and we assume that ack, P=xRj+2ZRç. 84 1. The condition (i) implies that the order of the Galois group G is a mul- tiple of 3 and the condition (ii) implies that *(G) is an even number. Since G isa subgroup of S,, wchave G=5S,. 85 2. An exampleis f(x) =x +64! 12x) 415x)-17x+14 from xº+x—1 (mod 3), x(0+x+1) (mod 2) and (x + 3X A De Dl mod 7). 
