Finite generation and Henselization

Question

Now I understand the answers.

I am trying to understand Henselian Weierstrass Theorem in Hironaka's Idealistic exponents of singularity, page 76 - 77. A glance at the paper.

At some point there is a ring $R$, which is Noetherian, Henselian, and local, with maximal ideal $m$. A module $A$ such that

1. $\frac{A}{mA}$ is finitely generated over $\frac{R}{m}$.

2. $A$ is finitely generated over the Henselization $S$ of $R[z]$ with respect to $(m,z)$.

And then he concludes, 'thanks to Hensel's lemma', that

(3). $A$ is finitely generated over $R$.

I don't understand how that works. It should be something standard/simple since it is stated just like that, but I don't see it. Could you explain how the argument goes?

PS: If you are looking at the paper, the notation used is:

• $R$ is what in the paper is called $R_{j+1}$,
• $A$ is what in the paper is $G_j$,
• $m$ is $M_{j+1}$,
• $z$ is $z_{j+1}$,
• and $S$ is what in the paper is called $R_{j}$.

Answer 1

(Note: There is no new content in this answer; it is simply an explanation of parts of user's excellent response. I'm writing it as an answer, because that seems less perverse than splitting it across many comments.)

The key result that is being used here is the following consequence of Zariski's Main Theorem + Hensel's lemma: Suppose that $f:A\to B$ is a map of finite type (between noetherian rings), with $A$ local henselian. Suppose that $\mathfrak{n}\subset B$ is a prime above $\mathfrak{m}$ such that the localization $B_{\mathfrak{n}}$ is quasi-finite over $A$ at the closed point (i.e. $B_{\mathfrak{n}}/\mathfrak{m}B_{\mathfrak{n}}$ is finite over the field $A/\mathfrak{m}$). Then $B_{\mathfrak{n}}$ is finite over $A$, and is in particular itself henselian local.

Here, Zariski's Main Theorem is (or can be) used in the following incarnation (see Raynaud, 'Anneaux locaux henseliens', p. 42): Let $f:A\to B$ be as in our hypotheses in the first paragraph (but don't assume that $A$ is henselian). Let $B'$ be the integral closure of $A$ in $B$ (note that $B'$ is automatically finite over $A$); then there exists $f\in B'$, $f\notin \mathfrak{n}$ such that $B'_f=B_f$. In other words, $\text{Spec }B\to\text{Spec }B'$ is an open immersion in a neighborhood of $\mathfrak{n}$. Therefore, there exists a prime $\mathfrak{n}'\subset B'$ such that $B'_{\mathfrak{n}'}=B_{\mathfrak{n}}$. So, replacing $B,\mathfrak{n}$ by $B',\mathfrak{n}'$, we can assume that $B$ is finite over $A$.

If we now assume that $A$ is in addition henselian, then Hensel's lemma implies that $B$ is a product $B=\prod_{i=1}^rB_{\mathfrak{n}_i}$, where $\mathfrak{n}_i$ ranges over all the primes of $B$ that lie over $\mathfrak{m}$; see for example p. 2 of Raynaud's book. This immediately implies that $B_{\mathfrak{n}}$ is finite over $A$.

Let's see why this implies Hironaka's claim. As in user's answer, let $S/I$ be the quotient of $S$ that acts faithfully on the $R$-module $A$. As observed already in the answer, the map $R\to S/I$ is quasi-finite at the closed point; also, $S/I$ is the henselization of the localization of an $R$-algebra $B$ of finite type at a prime $\mathfrak{n}\subset B$ (this is the point where you use the fact that $I$ is finitely generated; $B$ is what user denotes as $S'/I'$). Now, we can apply the result from the first paragraph to conclude that $B_{\mathfrak{n}}$ is finite over $R$ and is therefore itself henselian. This implies that $S/I=B_{\mathfrak{n}}$, which shows that $S/I$ (and hence $A$) is finite over $R$.

Answer 2

You probably meant to assume $R$ is noetherian. And Hironaka seems to have suppressed his appeal to Zariski's Main Theorem, as we'll see below (or maybe someone else sees a more elementary procedure, which is certainly possible).

Consider the schematic support ${\rm{Spec}}(S/I)$ of the $S$-finite $A$ in ${\rm{Spec}}(S)$ (here, $I$ is the annihilator ideal of $A$ in $S$). Its special fiber over ${\rm{Spec}}(R)$ has underlying reduced scheme that coincides with that of the schematic support of $A/mA$ over $S/mS$, and this latter schematic support is $k$-finite (with $k := R/m$) since $A/mA$ is $k$-finite. Thus, $S/I$ has $k$-finite special fiber over $R$ since such finiteness is insensitive to killing nilpotents.

Since $R$ is noetherian, so $I$ is finitely generated, there is a residually trivial etale neighborhood $({\rm{Spec}}(R'),\xi)$ of $(m,z) \in {\rm{Spec}}(R[z])$ such that $I$ has a finite set of generators coming from $R'$ (via the unique map $R'_{\xi}\rightarrow S$ over $R[z]_{(m,z)}$). Letting $I'\subset R'$ be the ideal generated by those elements of $R'$, we see that $R'/I'$ has henselization $S/I$ at $\xi$ with $k$-finite special fiber over $R$. But henselization is compatible with quotients, so the special fiber of $R'_{\xi}/I'$ over $R$ is an essentially finite type local $k$-algebra with $k$-finite henselization, so $R'_{\xi}/I'$ has $k$-finite special fiber over $R$. In other words, ${\rm{Spec}}(R'/I') \rightarrow {\rm{Spec}}(R)$ is quasi-finite at $\xi$.

By openness of the quasi-finite locus of a finite type map between noetherian schemes, we can localize $R'$ around $\xi$ so that $R'/I'$ is a quasi-finite $R$-algebra. Hence, by Zariski's Main Theorem, ${\rm{Spec}}(R'/I')$ is Zariski-open in a finite $R$-algebra. But $R$ is henselian, so by Hensel's Lemma (in the EGA form) applied to lifting idempotents from the special fiber we know that every finite $R$-algebra is a direct product of finite local $R$-algebras. Consequently, by shrinking some more around the point $\xi$ in the special fiber we can arrange that ${\rm{Spec}}(R'/I')$ is Zariski-open in a finite local $R$-scheme with closed point $\xi$, so this Zariski-open locus is the entire space. In other words, we get to the case that $R'/I'$ is $R$-finite and local. Thus, it is equal to its own henselization, which is $S/I$ by design.

We have proved that $S/I$ is $R$-finite, yet $A$ is an $S/I$-module, so $A$ s $R$-finite. QED