Lemma 37.26.1. 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 nonreduced, then there exists a nonempty open V \subset Y such that for all y \in V the fibre X_ y is nonreduced.
Proof. Let Y' \subset Y be the reduction of Y. Let X' \to Y' be the base change of f. Note that Y' \to Y induces a bijection on points and that X' \to X identifies fibres. Hence we may assume that Y' is reduced, i.e., integral, see Properties, Lemma 28.3.4. We may also replace Y by an affine open. Hence we may assume that Y = \mathop{\mathrm{Spec}}(A) with A a domain. Denote K the fraction field of A. Pick an affine open \mathop{\mathrm{Spec}}(B) = U \subset X and a section h_\eta \in \Gamma (U_\eta , \mathcal{O}_{U_\eta }) = B_ K which is nonzero and nilpotent. After shrinking Y we may assume that h comes from h \in \Gamma (U, \mathcal{O}_ U) = B. After shrinking Y a bit more we may assume that h is nilpotent. Let I = \{ b \in B \mid hb = 0\} be the annihilator of h. Then C = B/I is a finite type A-algebra whose generic fiber (B/I)_ K is nonzero (as h_\eta \not= 0). We apply generic flatness to A \to C and A \to B/hB, see Algebra, Lemma 10.118.3, and we obtain a g \in A, g \not= 0 such that C_ g is free as an A_ g-module and (B/hB)_ g is flat as an A_ g-module. Replace Y by D(g) \subset Y. Now we have the short exact sequence
with B/hB flat over A and with C nonzero free as an A-module. It follows that for any homomorphism A \to \kappa to a field the ring C \otimes _ A \kappa is nonzero and the sequence
is exact, see Algebra, Lemma 10.39.12. Note that B/hB \otimes _ A \kappa = (B \otimes _ A \kappa ) / h(B \otimes _ A \kappa ) by right exactness of tensor product. Thus we conclude that multiplication by h is not zero on B \otimes _ A \kappa . This clearly means that for any point y \in Y the element h restricts to a nonzero element of U_ y, whence X_ y is nonreduced. \square
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