Lemma 37.26.3. 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 37.24.7 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 37.26.1 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 37.26.2 proves that g(V') is the open we are looking for. \square
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