**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](https://docs.google.com/file/d/0B-vOsANdAoEJUUk3TnZhZENUNVU/edit?usp=sharing). 


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 $R\_{j}$.



Finite generation and Henselization
(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 $R$.


