# Lift $k$-valued points $X(k)$ to $X(A)$ of a smooth scheme over Henselian local ring

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 $$\bar{\varphi}: B \to A$$.

Problems:

(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?

• Does (d) of Theorem 4.1 of Chapter I of Milne's book help?
– anon
Nov 2 '19 at 19:49
• 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
– 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.]

• 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?
– user705174
Nov 4 '19 at 20:09
• 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
– user705174
Nov 4 '19 at 20:09
• 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".
– user705174
Nov 4 '19 at 20:10
• 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.
– anon
Nov 4 '19 at 20:54