Lemma 10.43.6. Let $k$ be a field. Let $S$ be a reduced $k$-algebra. Let $k \subset K$ be either a separable field extension, or a separably generated field extension. Then $K \otimes _ k S$ is reduced.

**Proof.**
Assume $k \subset K$ is separable. By Lemma 10.43.4 we may assume that $S$ is of finite type over $k$ and $K$ is finitely generated over $k$. Then $S$ embeds into a finite product of fields, namely its total ring of fractions (see Lemmas 10.25.1 and 10.25.4). Hence we may actually assume that $S$ is a domain. We choose $x_1, \ldots , x_{r + 1} \in K$ as in Lemma 10.42.3. Let $P \in k(x_1, \ldots , x_ r)[T]$ be the minimal polynomial of $x_{r + 1}$. It is a separable polynomial. It is easy to see that $k[x_1, \ldots , x_ r] \otimes _ k S = S[x_1, \ldots , x_ r]$ is a domain. This implies $k(x_1, \ldots , x_ r) \otimes _ k S$ is a domain as it is a localization of $S[x_1, \ldots , x_ r]$. The ring extension $k(x_1, \ldots , x_ r) \otimes _ k S \subset K \otimes _ k S$ is generated by a single element $x_{r + 1}$ with a single equation, namely $P$. Hence $K \otimes _ k S$ embeds into $F[T]/(P)$ where $F$ is the fraction field of $k(x_1, \ldots , x_ r) \otimes _ k S$. Since $P$ is separable this is a finite product of fields and we win.

At this point we do not yet know that a separably generated field extension is separable, so we have to prove the lemma in this case also. To do this suppose that $\{ x_ i\} _{i \in I}$ is a separating transcendence basis for $K$ over $k$. For any finite set of elements $\lambda _ j \in K$ there exists a finite subset $T \subset I$ such that $k(\{ x_ i\} _{i\in T}) \subset k(\{ x_ i\} _{i \in T} \cup \{ \lambda _ j\} )$ is finite separable. Hence we see that $K$ is a directed colimit of finitely generated and separably generated extensions of $k$. Thus the argument of the preceding paragraph applies to this case as well. $\square$

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