Lemma 10.43.8 (0C2X)—The Stacks project

Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.43: Geometrically reduced algebras
Lemma 10.43.8 (cite)

Lemma 10.43.8. Let $k'/k$ be a separable algebraic extension. Then there exists a multiplicative subset $S \subset k' \otimes _ k k'$ such that the multiplication map $k' \otimes _ k k' \to k'$ is identified with $k' \otimes _ k k' \to S^{-1}(k' \otimes _ k k')$.

Proof. First assume $k'/k$ is finite separable. Then $k' = k(\alpha )$, see Fields, Lemma 9.19.1. Let $P \in k[x]$ be the minimal polynomial of $\alpha $ over $k$. Then $P$ is an irreducible, separable, monic polynomial, see Fields, Section 9.12. Then $k'[x]/(P) \to k' \otimes _ k k'$, $\sum \alpha _ i x^ i \mapsto \alpha _ i \otimes \alpha ^ i$ is an isomorphism. We can factor $P = (x - \alpha ) Q$ in $k'[x]$ and since $P$ is separable we see that $Q(\alpha ) \not= 0$. Then it is clear that the multiplicative set $S'$ generated by $Q$ in $k'[x]/(P)$ works, i.e., that $k' = (S')^{-1}(k'[x]/(P))$. By transport of structure the image $S$ of $S'$ in $k' \otimes _ k k'$ works.

In the general case we write $k' = \bigcup k_ i$ as the union of its finite subfield extensions over $k$. For each $i$ there is a multiplicative subset $S_ i \subset k_ i \otimes _ k k_ i$ such that $k_ i = S_ i^{-1}(k_ i \otimes _ k k_ i)$. Then $S = \bigcup S_ i \subset k' \otimes _ k k'$ works. $\square$

Comments (2)

Comment #11259 by david on February 11, 2026 at 16:04

Why is $\cup S_i$ multiplicatively closed? If $Q_i$ corresponds to the extension $k_i$ and $Q_j$ to $k_j$ , it's not obvious that $Q_i Q_j \in \cup S_i$ .

Comment #11264 by thesnakefromthelemma on February 15, 2026 at 01:41

Re Comment #12259, the idea is presumably to take the multiplicative closure of the union.

One way to shore up the slight handwave at the end is to observe that given diagrams $F, G : \mathcal{I} \to R \text{-alg}$ and natural transformation $T : F \to G$ such that for all $i \in \text{Ob} (\mathcal{I})$ there exists a (multiplicative) subset $S _ i \subset F (i)$ such that the component $T _ i : F (i) \to G (i)$ is the (up to unique iso) localization of $F (i)$ by $S _ i$ , then the map $\text{colim} (T) : \text{colim} (F) \to \text{colim} (G)$ is itself the (up to unique iso) localization of $\text{colim} (F)$ by (the multiplicative subset generated by) the union over $i \in \text{Ob} (\mathcal{I})$ of the image of $S _ i$ under the relevant component $K _ i : F (i) \to \text{colim} (F)$ of the universal cocone of $F$ .

This is (assuming I haven't made some mistake) not at all hard to show by straightforward universal property wrangling; at the moment I can't think of any cleverer argument.

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