Lemma of Hironaka in Hartshorne III.9.12

Question

In the proof, the author consider the normalization $\tilde{A}$ of $A$ and show $\tilde{A}/t \tilde{A}$ is a integral domain. He showed that the localizations at points of Spec $A$ are domains, but we know a non-domain ring can have integral localizations. How should I understand the proof? Thanks a lot.

Answer 1

Retain all notation of the lemma as stated in Hartshorne. Pose $S=A-\mathfrak p$. Since normalization commutes with localization, the integral closure of $A_\mathfrak p$ is $S^{-1}\tilde A$. But $A_\mathfrak p$ is a DVR so in fact $S^{-1}\tilde A=A_{\mathfrak p}$. Therefore $\tilde A$ has precisely one prime $\mathfrak p'$ lying over $\mathfrak p$, and $A_\mathfrak p=\tilde A_{\mathfrak p'}$. Note that as $\tilde A$ is normal it satisfies condition (S2) and therefore $\tilde A/t\tilde A$ has no embedded associated primes. To conclude that $\mathfrak p'$ is the only associated prime of $\tilde A/t\tilde A$, it would suffice to show that every minimal associated prime of $\tilde A/t\tilde A$ contracts to $\mathfrak p$, since we already know that only $\mathfrak p'$ contracts to $\mathfrak p$, and the set of associated primes of $\tilde A/t\tilde A$ is nonempty, as $\tilde A$ is finite over $A$, hence noetherian, since $A$ is japanese.

Let $\mathfrak r\in\operatorname{Spec}\tilde A$ be a minimal associated prime of $\tilde A/t\tilde A$. As $A$ is a localization of an affine ring, it is universally catenary. By incomparability (i.e. all fibers of $A\rightarrow\tilde A$ are of dimension zero), a fortiori $\mathfrak r$ is maximal among those primes meeting $A$ in $A\cap\mathfrak r$; since $A$ is japanese, we can apply Theorem 13.8 of Eisenbud (a variant of Nagata's altitude formula) to conclude $$\operatorname{ht} \mathfrak r=\operatorname{ht} \mathfrak r\cap A+\dim K(A)\otimes\tilde A,$$ but of course $K(A)\otimes\tilde A=K(A)$, so we conclude that $\operatorname{ht} \mathfrak r=\operatorname{ht}\mathfrak r\cap A$. But now of course $\operatorname{ht}\mathfrak r=1$ by Krull's hauptidealsatz, and as $t\in\mathfrak r\cap A$ and $\operatorname{ht}\mathfrak r\cap A=1$, we have that $\mathfrak r\cap A$ is minimal over $t$, again by the hauptidealsatz. But by hypothesis, the only minimal associated prime of $tA$ is $\mathfrak p$. Hence $\mathfrak r\cap A=\mathfrak p$.

We now have enough to infer that $t\tilde A$ is a $\mathfrak p'$-primary ideal of $\tilde A$. Let $S(t\tilde A)$ denote the saturation of the ideal $t\tilde A$ with respect to $\tilde A-\mathfrak p'$ (the contraction in $\tilde A$ of $t\tilde A_{\mathfrak p'}$). As $S\cap\mathfrak p'=\emptyset$, it follows from standard facts about the behavior of primary ideals under saturation (Atiyah-Macdonald Prop. 4.8) that $S(t\tilde A)=t\tilde A$. But $t\tilde A_{\mathfrak p'}=tA_{\mathfrak p}$, and $t$ generates the maximal ideal of $A_\mathfrak p$ by hypothesis; hence $t\tilde A=S(t\tilde A)$ is the contraction of a prime and so we finally see that $\tilde A/t\tilde A$ is a domain. $\blacksquare$

(This proof is adapted from the one in EGA IV 2 (5.12.7).)

Answer 2

Using the notation of the proof, i explain why $B:=\tilde{A}/t\tilde{A}$ is an integral domain. Recall that the zero-divisors of a finitely generated module over a Noetherian ring are the union of its associated primes. Since $B$ has positive depth, the maximal ideal $m$ of $A$ is not an associated prime of $B$. On the other hand $B_q = (A/p)_q$ for every prime $q \neq m$. Hence the associated primes of $B$ coincide with those of $A/p$, i.e. $\operatorname{Ass}_A(B) = \{p\}$. Since $B_p = k(p)$ we have that $B$ has no zero-divisors.

PS: I suggest to migrate this question to MSE.