Definition 57.26.1. A morphism of schemes is *étale* if it is smooth of relative dimension 0.

## 57.26 Étale morphisms

For more details, see Morphisms, Section 29.34 for the formal definition and Étale Morphisms, Sections 41.11, 41.12, 41.13, 41.14, 41.16, and 41.19 for a survey of interesting properties of étale morphisms.

Recall that an algebra $A$ over an algebraically closed field $k$ is *smooth* if it is of finite type and the module of differentials $\Omega _{A/k}$ is finite locally free of rank equal to the dimension. A scheme $X$ over $k$ is *smooth* over $k$ if it is locally of finite type and each affine open is the spectrum of a smooth $k$-algebra. If $k$ is not algebraically closed then a $k$-algebra $A$ is a smooth $k$-algebra if $A \otimes _ k \overline{k}$ is a smooth $\overline{k}$-algebra. A ring map $A \to B$ is smooth if it is flat, finitely presented, and for all primes $\mathfrak p \subset A$ the fibre ring $\kappa (\mathfrak p) \otimes _ A B$ is smooth over the residue field $\kappa (\mathfrak p)$. More generally, a morphism of schemes is *smooth* if it is flat, locally of finite presentation, and the geometric fibers are smooth.

For these facts please see Morphisms, Section 29.32. Using this we may define an étale morphism as follows.

In particular, a morphism of schemes $X \to S$ is étale if it is smooth and $\Omega _{X/S} = 0$.

Proposition 57.26.2. Facts on étale morphisms.

Let $k$ be a field. A morphism of schemes $U \to \mathop{\mathrm{Spec}}(k)$ is étale if and only if $U \cong \coprod _{i \in I} \mathop{\mathrm{Spec}}(k_ i)$ such that for each $i \in I$ the ring $k_ i$ is a field which is a finite separable extension of $k$.

Let $\varphi : U \to S$ be a morphism of schemes. The following conditions are equivalent:

$\varphi $ is étale,

$\varphi $ is locally finitely presented, flat, and all its fibres are étale,

$\varphi $ is flat, unramified and locally of finite presentation.

A ring map $A \to B$ is étale if and only if $B \cong A[x_1, \ldots , x_ n]/(f_1, \ldots , f_ n)$ such that $\Delta = \det \left( \frac{\partial f_ i}{\partial x_ j} \right)$ is invertible in $B$.

The base change of an étale morphism is étale.

Compositions of étale morphisms are étale.

Fibre products and products of étale morphisms are étale.

An étale morphism has relative dimension 0.

Let $Y \to X$ be an étale morphism. If $X$ is reduced (respectively regular) then so is $Y$.

Étale morphisms are open.

If $X \to S$ and $Y \to S$ are étale, then any $S$-morphism $X \to Y$ is also étale.

**Proof.**
We have proved these facts (and more) in the preceding chapters. Here is a list of references: (1) Morphisms, Lemma 29.34.7. (2) Morphisms, Lemmas 29.34.8 and 29.34.16. (3) Algebra, Lemma 10.142.2. (4) Morphisms, Lemma 29.34.4. (5) Morphisms, Lemma 29.34.3. (6) Follows formally from (4) and (5). (7) Morphisms, Lemmas 29.34.6 and 29.28.5. (8) See Algebra, Lemmas 10.158.7 and 10.158.5, see also more results of this kind in Étale Morphisms, Section 41.19. (9) See Morphisms, Lemma 29.24.10 and 29.34.12. (10) See Morphisms, Lemma 29.34.18.
$\square$

Definition 57.26.3. A ring map $A \to B$ is called *standard étale* if $B \cong \left(A[t]/(f)\right)_ g$ with $f, g \in A[t]$, with $f$ monic, and $\text{d}f/\text{d}t$ invertible in $B$.

It is true that a standard étale ring map is étale. Namely, suppose that $B = \left(A[t]/(f)\right)_ g$ with $f, g \in A[t]$, with $f$ monic, and $\text{d}f/\text{d}t$ invertible in $B$. Then $A[t]/(f)$ is a finite free $A$-module of rank equal to the degree of the monic polynomial $f$. Hence $B$, as a localization of this free algebra is finitely presented and flat over $A$. To finish the proof that $B$ is étale it suffices to show that the fibre rings

are finite products of finite separable field extensions. Here $\overline{f}, \overline{g} \in \kappa (\mathfrak p)[t]$ are the images of $f$ and $g$. Let

be the factorization of $\overline{f}$ into powers of pairwise distinct irreducible monic factors $\overline{f}_ i$ with $e_1, \ldots , e_ b > 0$. By assumption $\text{d}\overline{f}/\text{d}t$ is invertible in $\kappa (\mathfrak p)[t, 1/\overline{g}]$. Hence we see that at least all the $\overline{f}_ i$, $i > a$ are invertible. We conclude that

where $I \subset \{ 1, \ldots , a\} $ is the subset of indices $i$ such that $\overline{f}_ i$ does not divide $\overline{g}$. Moreover, the image of $\text{d}\overline{f}/\text{d}t$ in the factor $\kappa (\mathfrak p)[t]/(\overline{f}_ i)$ is clearly equal to a unit times $\text{d}\overline{f}_ i/\text{d}t$. Hence we conclude that $\kappa _ i = \kappa (\mathfrak p)[t]/(\overline{f}_ i)$ is a finite field extension of $\kappa (\mathfrak p)$ generated by one element whose minimal polynomial is separable, i.e., the field extension $\kappa (\mathfrak p) \subset \kappa _ i$ is finite separable as desired.

It turns out that any étale ring map is locally standard étale. To formulate this we introduce the following notation. A ring map $A \to B$ is *étale at a prime $\mathfrak q$* of $B$ if there exists $h \in B$, $h \not\in \mathfrak q$ such that $A \to B_ h$ is étale. Here is the result.

Theorem 57.26.4. A ring map $A \to B$ is étale at a prime $\mathfrak q$ if and only if there exists $g \in B$, $g \not\in \mathfrak q$ such that $B_ g$ is standard étale over $A$.

**Proof.**
See Algebra, Proposition 10.142.16.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (4)

Comment #237 by Keenan Kidwell on

Comment #243 by Johan on

Comment #4677 by Peng DU on

Comment #4804 by Johan on