The Stacks project

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 $k_ i = S_ i^{-1}(k_ i \otimes _ k k_ i)$. Let $S$ be the multiplicative closure of $\bigcup S_ i \subset k' \otimes _ k k'$. Clearly, the multiplication maps sends every element of $S$ to an invertible element of $k'$. Using that $k' \otimes _ k k'$ is the union of the rings $k_ i \otimes _ k k_ i$ we see that every element in the kernel of the multiplication map is mapped to zero in $S^{-1}(k' \otimes _ k k')$. Using exactness of localization the result follows. $\square$