Lemma 10.43.7. Let $k$ be a field and let $S$ be a $k$-algebra. Assume that $S$ is reduced and that $S_{\mathfrak p}$ is geometrically reduced for every minimal prime $\mathfrak p$ of $S$. Then $S$ is geometrically reduced.

**Proof.**
Since $S$ is reduced the map $S \to \prod _{\mathfrak p\text{ minimal}} S_{\mathfrak p}$ is injective, see Lemma 10.25.2. If $K/k$ is a field extension, then the maps

are injective: the first as $k \to K$ is flat and the second by inspection because $K$ is a free $k$-module. As $S_\mathfrak p$ is geometrically reduced the ring on the right is reduced. Thus we see that $S \otimes _ k K$ is reduced as a subring of a reduced ring. $\square$

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