Exercise 22, § 7

\* Let A be a noetherian integral ring, which is local and complete (resp. a finitely generated algebra over a field k), $a \in A$ a noninvertible and nonzero element. We make the following hypotheses: α) Supp(A/aA) possesses a single minimal element $p$. β) One has $aA_p = pA_p$. γ) $A/p$ is an integrally closed ring. We propose in this exercise to prove that, for every maximal ideal m of A containing a, the ring $A_m$ is then integrally closed and that one has $p = aA$ (“Hironaka’s lemma”). In this exercise we shall use the property of catenarity. This property is satisfied by integral, noetherian, local, complete rings, as will be seen in Chapter X. It is also satisfied by integral algebras of finite type over a field (§ 2, No. 4, Theorem 3). a) Prove that $A_p$ is a discrete valuation ring (§ 3, No. 1, Proposition 1 and VI, § 3, No. 6, Proposition 9). b) We make the hypotheses α) and β) and suppose A integrally closed. Prove that one has $p = aA$ (VII, § 1, No. 4, Proposition 8). c) Let A’ be the integral closure of A. Prove that A’ is an integral, noetherian, local and complete ring (resp. that it is a finitely generated algebra over k). (One will remark that A is Japanese (IX, § 4, No. 2, Theorem 2), hence that A’ is integral, noetherian, semi-local and complete and, consequently, local (III, § 2, No. 13, corollary to Proposition 19). In the case where A is a finitely generated algebra over k, one will use V, § 3, No. 2, Theorem 2.) d) Prove that one has $A' \otimes_A A_p = A_p$ (use a) and V, § 1, No. 5, Proposition 16). Deduce that there exists a single prime ideal $p'$ of A’ above p and that one has $A_p = A_{p'}', aA_{p'}' = p'A_{p'}'$. e) Let q be a minimal element of the support of $A'/aA'$. Prove that q is a prime ideal of A’ above p. (Notice that q is of height 1 (§ 3, No. 1, Proposition 1) and that consequently by virtue of catenarity, one has dim(A'/q) = dim(A) - 1. Deduce that one has dim(A/(q ∩ A)) = dim(A) - 1 and consequently that q ∩ A = p.) f) Deduce from e) that aA' is a prime ideal of A’ and that one has aA' = p' = pA' (one may use b)). g) Prove that one has A'_m = A_m for every maximal ideal m of A containing a. (From the inclusions A ⊂ A' ⊂ A_p deduce the inclusions A/p ⊂ A'/p' ⊂ A_p/pA_p. Deduce that A/p and A'/p' have the same field of fractions and consequently, according to γ), that A/p = A'/p' = A' ⊗_A (A/p), the last equality resulting from f). Thus (A'/A) ⊗_A (A/p) = 0. Conclude.) Complete then the proof of Hironaka’s lemma. \*