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
S \otimes _ k K \to (\prod S_\mathfrak p) \otimes _ k K \to \prod S_\mathfrak p \otimes _ k K
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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