D-MATH Commutative Algebra HS 2018 Prof. Paul Nelson Solutions Sheet 8 Integrality Let R be a commutative ring, k an algebraically closed field. 1. Let R be a subring of a ring S such that S is integral over R. Let b be a maximal ideal of S and set a = b \ R, the corresponding maximal ideal of R. Find a counterexample to show that the localization Sb is not necessarily integral over Ra. (Hint: consider the subring k[X2−1] of k[X], where k is a field, and let b = (X−1). Can the element 1=(X + 1) be integral?) Solution. Let k be a field of characteristic different from 2, so that X +1 6= X −1 in k[X]. Define R = k[X2 − 1] and S = k[X]. Then X 2 S satisfies the polynomial equation Y 2 − (1 − (X2 − 1)) = 0 in R[Y ], hence S is integral over R, being generated by an integral element over R. Now consider the ideal b = (X − 1) in S; then a = b \ R = (X2 − 1). Since X + 1 2= b, we have that 1=(X + 1) is an element of Sb. We claim that it cannot be integral over Ra. For the purpose of contradiction, assume it is: then we have that n i X 1 a s−1 = 0 i i X + 1 i=0 −1 for some ai 2 R, si 2 R r a, with ansn = 1. Now rewrite the former expression as follows: n n−i X ai(X + 1) = 0 : s (X + 1)n i=0 i n Qn Pn Multiply the previous equality by (X + 1) i=0 si, so that you get i=0 aiti(X + n−i Q 2 1) = 0 in S, where ti = j6=i sj 2 R r (X − 1), and an = 1. It follows that 0 X +1 divides each term of the previous sum except possibly for antn(X +1) = tn. 2 This forces tn 2 R \ (X + 1) = (X − 1), contradiction. 2. Let R; S be rings, with R a subring of S and S integral over R. (a) Prove that, if x 2 R is a unit in S, then it is also a unit in R. (b) Show that the Jacobson radical of R is the contraction of the Jacobson radical of S. (Hint: use 5:8 and 5:10 of [1].) 1 Solution. (a) Let x 2 R be invertible in S, so that x does not belong to any maximal ideal n of S. Let now m be a maximal ideal of R. By 5:10 of [1], there exists a prime ideal n of S such that n \ R = m. By 5:8 in [1], such an n is maximal in S. Since x2 = n, a fortiori x2 = m, so that x does not belong to any maximal ideal of R, and hence it is invertible in R. (b) Applying 5:8 and 5:10 of [1], we get that for any maximal ideal m of R there exists a unique maximal ideal n of S such that m = n \ R. Denote by N the set of such maximal ideals of S. Then \ \ \ \ Jac(R) = m = (n \ R) = R \ n ⊃ R \ n m maximal n2N n2N n maximal in S = R \ Jac(S) : On the other hand, denoting by M the set of maximal ideals of R of the form m = n \ R for some maximal ideal n of S, we get \ \ \ R \ Jac(S) = R \ n = R \ n = m ⊃ Jac(R) ; n⊂S maximal n⊂S maximal m2M reaching the conclusion. 3. (a) Let R be a subring of an integral domain S, and let S0 denote the integral closure of R inside S. Let f; g 2 S[X] be monic polynomials with fg 2 S0[X]. Prove that both f and g belong to S0[X]. (Hint: consider a field containing S over which f and g split into linear factors.) (b) Prove the same result of the previous point without the assumption that S is an integral domain. Solution. (a) Let k be a field of fractions of S (which is well-defined since S is an integral domain), and consider Ω to be a splitting field of the product fg over k. We may thus write, in Ω[X], Y Y f(X)g(X) = (X − ξi) (X − ηj) ; i j where the ξi are the roots of f and the ηj are the roots of g. Each of these roots is integral over S0 in Ω. Moreover, each coefficient of f (resp. of g) is a polynomial in the ξi (resp. ηj), so that it belongs to the integral closure S00 of S0 in Ω (because the integral closure of S0 is a subring of Ω). Thus, f; g 2 (S \S00)[X]; finally, it is clear that S \S00 = S0, because S0 is integrally closed in S. 2 (b) We claim that there exists an overring S1 ⊃ S such that both f and g split into linear factors x − ξi, x − ηj inside S1[X]. If such a ring S1 exists, then we are done: indeed, both the ξi and the ηj would be roots of fg, and as such they are integral over S0. The coefficients of f and g, being polynomials 0 in the ξi and the ηj respectively, have also to be integral over S . However, this coefficients are in S by assumption, so that they are in S0 because S0 is integrally closed in S. It now remains to construct such a ring S1. Define S1;1 to be the ring S[X]=(f(X)), so that f has a root ξ1 in S1;1. Then f(Y ) factors inside S1;1[Y ] as a product f(Y ) = (Y − ξ1)f1(Y ). Apply the same procedure, so as to get the ring S1;2 = S1;1[Y ]=(f1(Y )). After a finite number of steps, you ~ reach a ring S1 ⊃ S over which f is a product of linear factors. Do the same ~ for g over S1[X], and get the desired ring S1. 4. Let R be a subring of a ring S, and denote by S0 the integral closure of R inside S. Prove that the subring S0[X] is the integral closure of R[X] inside S[X]. (Hint: apply Exercise 3 of the current sheet. You may also look at the hint in [1], Exercise 9 Chapter 5.) Solution. Define C to be the integral closure of R[X] inside S[X]. Then X 2 R[X] ⊂ C, and S0 is integral over R, hence over R[X], so that S0[X] ⊂ C. It now remains to show that S0[X] is integrally closed in S[X]. Thus, let f 2 S[X] be a 0 n Pn−1 i root of a monic polynomial with coefficients in S [X], say f + i=1 gif + g0 = 0 0 0, with gi 2 S [X] for all 0 6 i 6 n − 1. Since g0 2 S [X], we have that n−1 Pn−1 i−1 0 f(f + i=1 gif ) 2 S [X] is a product of monic polynomials in S[X]. The previous exercise shows that f 2 S0[X]. 5. Let R be an integral domain containing a subring S isomorphic to k[T ] for some field k. Show that, if R is finitely generated as an S-module, then R is free as an S-module. Solution. Since S isomorphic to k[T ] for a field k, S is a principal ideal domain. Thus R is a finitely generated module over a principal ideal domain. The classifi- cation theorem for such modules gives us that n M R ' S=(di) i=1 where (d1) ⊃ (d2) ⊃ · · · ⊃ (dn) are proper ideals of S, and the above isomorphism is intended in the category of S-modules. Since R is an integral domain, we must have di = 0 for all 1 6 i 6 n. It follows that R is free of rank n over S. 6. The setting is the same as in the previous exercise. 3 (a) Assume that R = k[X; Y ]=(X2 − Y 3), and let T = XmY n for some integers m; n > 1. Show by exhibiting a basis that the rank of the free S-module R is 3m + 2n. (b) Let R; S be again as in the assumptions of Exercise 5. Denote by R¯ the integral closure of R. Apply theorem 4:14 of [2] to show that R¯ is again finitely generated over S (hence a free S-module), and prove that it has the same rank as R. Solution. (a) Let f¯ be an element of R, and write it as ¯ X ¯ i ¯ j f = αi;jX Y ; i;j2N ¯ ¯ where αi;j 2 k and X; Y denote the projections of X; Y under the quotient map k[X; Y ] ! k[X; Y ]=(X2 − Y 3). Since X¯ 2 = Y¯ 3, we may also write ¯ X ¯ n ¯ X ¯ n f = βnY + X γnY : n2N n2N Therefore, we may assume without loss of generality that T = XY n. An arbitrary elementg ¯ of k[T ] ⊂ R takes the form X ¯ ¯ n j X ¯ (2n+3)j ¯ X ¯ n+(2n+3)j g¯ = δj(XY ) = δ2kY + X δ2k+1Y : j2N k2N k2N Comparing the expressions for f¯ andg ¯ it is immediate to deduce that R has rank 2n + 3 over k[T ] with free basis given by the elements Y¯ l + X¯Y¯ l ; l = 0;:::; 2n + 2 : (b) Let K(R) be the fraction field of R.