Lemma 33.6.8. Let $k$ be a field. Let $X$ be a scheme over $k$.

1. If $x' \leadsto x$ is a specialization and $X$ is geometrically reduced at $x$, then $X$ is geometrically reduced at $x'$.

2. If $x \in X$ such that (a) $\mathcal{O}_{X, x}$ is reduced, and (b) for each specialization $x' \leadsto x$ where $x'$ is a generic point of an irreducible component of $X$ the scheme $X$ is geometrically reduced at $x'$, then $X$ is geometrically reduced at $x$.

3. If $X$ is reduced and geometrically reduced at all generic points of irreducible components of $X$, then $X$ is geometrically reduced.

Proof. Part (1) follows from Lemma 33.6.2 and the fact that if $A$ is a geometrically reduced $k$-algebra, then $S^{-1}A$ is a geometrically reduced $k$-algebra for any multiplicative subset $S$ of $A$, see Algebra, Lemma 10.43.3.

Let $A = \mathcal{O}_{X, x}$. The assumptions (a) and (b) of (2) imply that $A$ is reduced, and that $A_{\mathfrak q}$ is geometrically reduced over $k$ for every minimal prime $\mathfrak q$ of $A$. Hence $A$ is geometrically reduced over $k$, see Algebra, Lemma 10.43.7. Thus $X$ is geometrically reduced at $x$, see Lemma 33.6.2.

Part (3) follows trivially from part (2). $\square$

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