The Stacks project

Lemma 10.43.6. Let $k$ be a field. Let $S$ be a reduced $k$-algebra. Let $K/k$ 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. $\square$

Comments (4)

Comment #5082 by Kazuki Masugi on

Why does the proof need finite-ness of ? Lemma 10.42.2(Tag: 00EW) says that any reduced ring can be embedded in the product of fields.

Comment #5293 by on

The reason is that we are moving a product past a tensor product which only works for finite products. OK?

Comment #9035 by Liam on

In Lemma 030U. Why we can choose x1,…,xr+1∈K as in Lemma 030Q. Does separable + finitely generated property impliy separably generated ?

Comment #9184 by on

Look, this is explained in the proof. Namely, in the first paragraph the extension is separable. By definition this means any finitely generated subextension is separably generated! Hence after replacing as in the proof, we may apply Lemma 10.42.3.

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