# Why is a variety etale locally like affine space?

I remember from a talk somebody saying that ''a scheme is etale locally like affine space'' and I wonder what this could mean.

Let $Var/K$ be the site of varieties over a field $K$ with the etale topology. My first guess for a meaning of the above saying was that each $X\in Var/K$ has an etale cover $\{V_j\to X\}$ with $V_j\cong\mathbb{A^n}$. But this is wrong since for example every etale morphism into $X=Spec(K)$ has as a domain a finite disjoint union of spectra of finite separable field extensions of $K$.

What does it mean?

-

Formally, this amounts to the following fact: if $f : X \to Y$ is a morphism of schemes smooth at a point $x$ in $X$, then there exist a natural number $d$, affine open neighbourhoods $U \subseteq X$, $x \in U$, $V \subseteq Y$, $f(x) \in V$, and a commutative diagram of the form $$\begin{array}{ccccc} U & = & U & \to & X \\ \downarrow & & \downarrow & & \downarrow \\ \mathbb{A}^d_V & \to & V & \to & Y \end{array}$$ where $U \to \mathbb{A}^d_V$ is étale. This is Lemma 054L in the Stacks project. In particular, we can take $Y = \operatorname{Spec} k$ here to get the claim.
Dear @Zhen Lin, thank you for the answer. Do you know if one can choose $U$ in such a way that the étale $U\to\mathbb{A}^d_V$ becomes surjective? Perhaps this isn't true or what do you think? –  Ronald Bernard Apr 18 '13 at 16:06
I doubt it. Take $Y = \operatorname{Spec} k$ and $X = \mathbb{A}^2_k \setminus \{ (0, 0) \}$; then we must have $V = Y$ and $d = 2$. It's not clear to me where one would get a étale surjection $U \to \mathbb{A}^2_k$ where $U$ is an affine open subscheme of $X$. –  Zhen Lin Apr 18 '13 at 17:46