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. $\square$

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