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. $\square$

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