Lemma 37.25.3. Let $f : X \to Y$ be a morphism of schemes. Assume $Y$ irreducible with generic point $\eta$ and $f$ of finite type. If $X_\eta$ is not geometrically reduced, then there exists a nonempty open $V \subset Y$ such that for all $y \in V$ the fibre $X_ y$ is not geometrically reduced.

Proof. Apply Lemma 37.23.7 to get

$\xymatrix{ X' \ar[d]_{f'} \ar[r]_{g'} & X_ V \ar[r] \ar[d] & X \ar[d]^ f \\ Y' \ar[r]^ g & V \ar[r] & Y }$

with all the properties mentioned in that lemma. Let $\eta '$ be the generic point of $Y'$. Consider the morphism $X' \to X_{Y'}$ (which is the reduction morphism) and the resulting morphism of generic fibres $X'_{\eta '} \to X_{\eta '}$. Since $X'_{\eta '}$ is geometrically reduced, and $X_\eta$ is not this cannot be an isomorphism, see Varieties, Lemma 33.6.6. Hence $X_{\eta '}$ is nonreduced. Hence by Lemma 37.25.1 the fibres of $X_{Y'} \to Y'$ are nonreduced at all points $y' \in V'$ of a nonempty open $V' \subset Y'$. Since $g : Y' \to V$ is a homeomorphism Lemma 37.25.2 proves that $g(V')$ is the open we are looking for. $\square$

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