Lemma 10.142.4. 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.**
If $k \to k'$ 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.

Conversely, suppose that $k \to S$ is étale. Let $\overline{k}$ be an algebraic closure of $k$. Then $S \otimes _ k \overline{k}$ is étale over $\overline{k}$. Suppose we have the result over $\overline{k}$. Then $S \otimes _ k \overline{k}$ is reduced and hence $S$ is reduced. Also, $S \otimes _ k \overline{k}$ is finite over $\overline{k}$ and hence $S$ is finite over $k$. Hence $S$ is a finite product $S = \prod k_ i$ of fields, see Lemma 10.52.2 and Proposition 10.59.6. The result over $\overline{k}$ means $S \otimes _ k \overline{k}$ is isomorphic to a finite product of copies of $\overline{k}$, which implies that each $k \subset k_ i$ is finite separable, see for example Lemmas 10.43.1 and 10.43.3. Thus we have reduced to the case $k = \overline{k}$. In this case Lemma 10.139.2 (combined with $\Omega _{S/k} = 0$) we see that $S_{\mathfrak m} \cong k$ for all maximal ideals $\mathfrak m \subset S$. This implies the result because $S$ is the product of the localizations at its maximal ideals by Lemma 10.52.2 and Proposition 10.59.6 again. $\square$

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