Lemma 10.143.4 (00U3)—The Stacks project

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 extensions 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. $\square$

Comments (7)

Comment #5985 by Manolis Tsakiris on March 19, 2021 at 03:19

Hi, i would just like to point out that Proposition 00KJ does not explicitly say that an Artinian ring is the product of its localizations at its maximal ideals. It would be actually nice to include that statement in the proposition. Eric Wofsey has given a nice scheme-theoretic proof here: https://math.stackexchange.com/questions/2384868/any-artinian-ring-is-direct-product-of-its-localizations-at-the-maximal-ideals

Comment #5991 by Johan on March 19, 2021 at 15:22

@#5985: Do you mean there is something wrong with the proof? Because the proof of this lemma doesn't use the fact that an Artinian local ring is a product of the localizations at its maximal ideals. I'm asking because if you are commenting on Proposition 00KJ, why are you leaving a comment on the page for this lemma?

Comment #5996 by Manolis Tsakiris on March 22, 2021 at 03:49

Dear Johan, my point is the following. We have $S_m \cong k$ for every maximal ideal of $S$ . We also have that $S$ is Artinian. Now, the last sentence of the proof says that "This implies the result because $S$ is the product of the localizations at its maximal ideals by Lemma 00J6 and Proposition 00KJ again." So it seems to me that the proof uses the fact that "an Artinian ring is the product of its localizations at its maximal ideals". On the other hand, this is not mentioned explicitly in Proposition 00KJ. So i suppose my comment is a joint comment for the present tag and Proposition 00KJ.

Comment #5999 by Johan on March 23, 2021 at 00:04

@#5996. Yes, you are totally right. Sorry! OK, I will fix this the next time I go through all the comments.

Comment #6159 by Johan on May 01, 2021 at 20:01

Okay, this is now fixed. I decided to work around the issue you raise and I tried to "simplify" the proof of the lemma. See changes.

There does not seem to be a way of including in the statement of a result every possible way this result will be used in the future as the discussion above shows with regards to Proposition 10.60.7. In fact, I think the tendency to write very long lemmas with many equivalent formulations of the same thing actually often backfires I think. An example is Lemma 10.153.3 characterizing henselian local rings. Another one is Nakayama's lemma 10.20.1.

Comment #8555 by Manolis Tsakiris on July 11, 2023 at 05:28

Two very minor comments now:

Perhaps write $d: k' \cdot \bar{f} \rightarrow k' \cdot dx$ to indicate the class of $f$ mod $(f^2)$ , instead of $d: k' \cdot f \rightarrow k' \cdot dx$ .

Last sentence of second paragraph of the proof: "extensions" instead of "extension".

Comment #9138 by Stacks project on May 13, 2024 at 10:33

OK, I changed 2 but not 1. See here.

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