I have some problems to prove the exercise 4.13 (on page 39) from J.S. Milne's Etale Cohomology (the "book", not the online accessible script!):

Let $A$ be a Henselian local ring and $k=A/m$ it's residue field. We consider a smooth $A$-scheme $X$ with canonical map $f: X \to A$.

Problem The exercise is to show that the map

$$\operatorname{Hom}(\operatorname{Spec}(A), X)=X(A) \to X(k) = \operatorname{Hom}(\operatorname{Spec}(k), X)$$

between $A$-and $k$-valued points induced by $A \to k$ is surjective. or in other words that each map $\varphi: \operatorname{Spec}(k) \to X$ obtains a lift in $X(A)$.

This is what I tried:

let $x = \varphi(\{*\})$ the unique image of $\varphi$. since $f$ is smooth, there exist an open subscheme $U \subset X$ containing $x$ and an open $V \subset \operatorname{Spec}(A)$ with $f(U) \subset V$ such that the restriction of $f$ to $U$ factorizes as

$$U \xrightarrow{\text{g}} \mathbb{A}^n_V \xrightarrow{\text{h}} V$$

with $g$ étale and $h$ canonical. (this was application of Prop. 3.24(b) following the hint). Assume after restricting if neccessary that $U:=\operatorname{Spec}(B)$ and $V:=\operatorname{Spec}(R)$ are affine. since $V$ is open in $\operatorname{Spec}(A)$, we can assume that $R= A_s$ (i.e. a localization of $A$ at a $a \in A$).

The problem translates to comm. algebra:

abusing notation we have an etale ring map $g:A_s[X_1,\ldots, X_n] \to B$ and $\varphi: B \to k$, which we want to lift to $\bar{\varphi}: B \to A$.


(1) In genral localizations of Henselian ring are not Henselian, thus $A_s$ is not Henselian in general, thus we cannot at this point apply Henselian lifting theorem to lift $A_s[X_1,\ldots, X_n] \to k$ to $A$. We need an argument that we can choose $s \in A$ such that $A_s=A$.

(2) assume we solved problem (1) and have a lift $A_s[X_1,\ldots, X_n] \to A$. can we show that it factorizes through $B$? which characterization of étaleness of $g$ could at this point do it's job?

Could anybody help me how to solve this problem?

Update #1: I think I have solved (1): $f \circ \varphi$ maps $\{*\}$ to the unique closed point $x_{\mathfrak{m}}$ of $\operatorname{Spec}(A)$ and every open $V \subset \operatorname{Spec}(A)$, which contains $x_{\mathfrak{m}}$ is already $\operatorname{Spec}(A)$, since $A$ local. therefore we can assume $V=\operatorname{Spec}(A)$.

What do we know about etale $g:A[X_1,\ldots, X_n] \to B$ and Hensel lifts? Is there a criterion which allows to lift zeros of more then one polynomial simulaneously?

  • $\begingroup$ Does (d) of Theorem 4.1 of Chapter I of Milne's book help? $\endgroup$
    – anon
    Nov 2 '19 at 19:49
  • $\begingroup$ I see, you mean probably 4.2? and étaleness must garantee then that $B$ has the form $A[T_1,..., T_n]/(f_1,...,f_n)$, such that $det(\partial f_i/\partial T_j)_{ij})$ is unit,right? the only cruical point where we have to be careful, is that if we evaluate $det(\partial f_i/\partial T_j)_{ij})$ in some point, that we have to be sure that it can't become zero. If we have it, then with observation (1) 4.2(d) seems to be applyable $\endgroup$
    – user705174
    Nov 2 '19 at 20:08

Locally $f$ factors through $\mathbb{A}^n_A$ (by smoothness), and so we may assume this globally. The element of $X(k)$ gives an element of $\mathbb{A}^n(k)$, which (obviously) lifts to an $A$-morphism $Spec(A)\to \mathbb{A}^n_A$. Form the fibered product of this morphism with the morphism $X\to \mathbb{A}^n_A$ to get a scheme etale over $A$, and apply I 4.2(d) of Milne's book. [I think you were only missing the last step.]

  • $\begingroup$ this solves indeed the problem up to a few subtle points. firstly: you used 4.2(d) after changing base to $A$. My first idea (after reading your comment) was to use 4.2(d'). The only obstruction was that I wasn't pretty sure if for all etale ring maps $A[X_1,...,X_n] \to B$ we know that $B$ must be isomorphic to quotient $A[T_1,..., T_n]/(f_1,...,f_n)$ with $det(\partial f_i/\partial T_j)_{ij}) \in B^*$? is that always true? $\endgroup$
    – user705174
    Nov 4 '19 at 20:09
  • $\begingroup$ secondly, I'm not sure about the reason why we can argue "locally" , i.e. assuming that we have such factorization $X \xrightarrow{\text{g}} \mathbb{A}^n_A \xrightarrow{\text{h}} Spec(A)$, and may assume this globally. is indeed the main ingredient here that $A$ was assumed to be local? the criucal point is that what if the composition $l:Spec(k) \to X \to Spec(A)$ would map $\{*\} \in Spec(k)$ not to unique closed point of $Spec(A)$, but to another point of $Spec(A)$. Then we cannot expect that every open $V \subset Spec(A)$ containing image $l(\{*\})$ already coinsides $\endgroup$
    – user705174
    Nov 4 '19 at 20:09
  • $\begingroup$ with $Spec(A)$ and thus in order to obtain to factorization over $ \mathbb{A}^n_V$ we have to localize $A$ at a certain element $s \in A$ in order to obtain $V=Spec(A_s)$. but $A_s$ is in general not more Hensel, so we cannot argue more with 4.2. Thus I'm not sure why we can here argue "locally". $\endgroup$
    – user705174
    Nov 4 '19 at 20:10
  • $\begingroup$ In the first step, I'm using 3.24(b): as $A$ is local, the only open subset of $Spec(A)$ containing the closed point is the whole space. $\endgroup$
    – anon
    Nov 4 '19 at 20:54

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