Lemma 10.143.4. Let $k$ be a field. A ring map $k \to S$ is étale if and only if $S$ is isomorphic as a $k$-algebra to a finite product of finite separable extensions of $k$.

**Proof.**
We are going to use without further mention: if $S = S_1 \times \ldots \times S_ n$ is a finite product of $k$-algebras, then $S$ is étale over $k$ if and only if each $S_ i$ is étale over $k$. See Lemma 10.143.3 part (11).

If $k'/k$ is a finite separable field extension then we can write $k' = k(\alpha ) \cong k[x]/(f)$. Here $f$ is the minimal polynomial of the element $\alpha $. Since $k'$ is separable over $k$ we have $\gcd (f, f') = 1$. This implies that $\text{d} : k'\cdot f \to k' \cdot \text{d}x$ is an isomorphism. Hence $k \to k'$ is étale. Thus if $S$ is a finite product of finite separable extension of $k$, then $S$ is étale over $k$.

Conversely, suppose that $k \to S$ is étale. Then $S$ is smooth over $k$ and $\Omega _{S/k} = 0$. By Lemma 10.140.3 we see that $\dim _\mathfrak m \mathop{\mathrm{Spec}}(S) = 0$ for every maximal ideal $\mathfrak m$ of $S$. Thus $\dim (S) = 0$. By Proposition 10.60.7 we find that $S$ is a finite product of Artinian local rings. By the already used Lemma 10.140.3 these local rings are fields. Hence we may assume $S = k'$ is a field. By the Hilbert Nullstellensatz (Theorem 10.34.1) we see that the extension $k'/k$ is finite. The smoothness of $k \to k'$ implies by Lemma 10.140.9 that $k'/k$ is a separable extension and the proof is complete. $\square$

## 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Your email address will not be published.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 (5)

Comment #5985 by Manolis Tsakiris on

Comment #5991 by Johan on

Comment #5996 by Manolis Tsakiris on

Comment #5999 by Johan on

Comment #6159 by Johan on