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 $R_{j}$.

2 Answers 2


(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 $R$.

  • $\begingroup$ Maybe no new content, but less names that are defined in terms of names, defined in terms of other names, some of which I couldn't find their translation (definition). I think I will be able to understand now. I recognize the language here. Thank you. $\endgroup$
    – O.R.
    Jul 17, 2013 at 23:07
  • $\begingroup$ Unfortunately I got stuck in the part before your explanation. When they say that the support of $A/mA$ over $S/mS$ is $k$-finite because $A/mA$ is finitely generated over $R/m$. Why is that? This support is $\text{Spec}(\frac{S}{mS+I})$. I am assuming being $k$-finite means $\frac{S}{mS+I}$ is finitely generated over $R/m$. So, why $A/mA$ finitely generated over $R/m$ implies $\frac{S}{mS+I}$ finitely generated over $R/m$? $\endgroup$
    – O.R.
    Jul 18, 2013 at 19:48
  • 1
    $\begingroup$ This follows from: Let $A\to B$ be a map of commutative Noetherian rings and let $M$ be a faithful, finitely generated $B$-module. Then $B$ is a finitely generated as an $A$-module if and only it $M$ is. For the non-trivial direction, note that $B$ embeds in the $A$-module $End_A(M)$, and use Noetherianness. $\endgroup$ Jul 18, 2013 at 22:07

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

  • $\begingroup$ Yes, $R$ is Noetherian. I will add it now. Thank you for the explanation. $\endgroup$
    – O.R.
    Jun 18, 2013 at 17:52
  • $\begingroup$ Let me momentarily unselect this as the answer since it is not true that I understand everything in it to be able to say by myself if it is correct or not. Now I need to unravel all the definitions of the concepts in this explanation. $\endgroup$
    – O.R.
    Jun 18, 2013 at 19:56
  • 1
    $\begingroup$ All user36938 is saying is this: Let $M$ be a finitely generated $R$-module, and let $Supp(M)\subset\text{Spec }R$ be the sub-set of primes such that $M_P$ (localization at $P$) is non-zero. By Nakayama's lemma, this is the same as saying that $M_P/PM_P\neq 0$. Claim: This set is compatible with base-change in the obvious sense. That is, if $f:R\to S$ is any map of Noetherian rings, then $Supp(S\otimes_RM)\subset\text{Spec }S$ consists precisely of those primes that lie above primes in $Supp(M)$. $\endgroup$ Jul 17, 2013 at 1:15
  • 1
    $\begingroup$ Indeed, given a prime $Q\subset S$, we have $(S\otimes_RM)_Q/Q(S\otimes_RM)_Q=k(Q)\otimes_RM=k(Q)\otimes_{k(P)}M_P/PM_P$, where $P=f^{-1}(Q)$. Therefore, $Q\in Supp(S\otimes_RM)$, if and only if $P\in Supp(M)$. The geometric interpretation is that you have a coherent sheaf over $\text{Spec }R$ attached to $M$, and $Supp(M)$ is precisely the locus where this sheaf has non-zero fibers. If you think about the behavior of fibers under pull-back for a moment, you'll find that this assertion is not at all surprising, if not obvious. $\endgroup$ Jul 17, 2013 at 1:21
  • 1
    $\begingroup$ @Franklin: I am glad that Keerthi came up with a formulation which is agreeable to you. But since you tell us basically nothing about your background (except that you have some difficulty translating between algebraic and geometric descriptions of certain things), it is to be expected that some may offer an answer which inadvertently does not fit with your background. I recommend learning the "geometric" way of thinking about these things (smoothness, henselization, Zariski's Main Theorem, etc.); it is incredibly illuminating even for purely algebraic results. Look at "Neron Models", Ch. 2. $\endgroup$
    – user36938
    Jul 21, 2013 at 15:44

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