Lemma 38.19.7. Let X \to \mathop{\mathrm{Spec}}(R) be a proper flat morphism where R is a valuation ring. If the special fibre is reduced, then X and every fibre of X \to \mathop{\mathrm{Spec}}(R) is reduced.
Proof. Assume the special fibre X_ s is reduced. Let x \in X be any point, and let us show that \mathcal{O}_{X, x} is reduced; this will prove that X is reduced. Let x \leadsto x' be a specialization with x' in the special fibre; such a specialization exists as a proper morphism is closed. Consider the local ring A = \mathcal{O}_{X, x'}. Then \mathcal{O}_{X, x} is a localization of A, so it suffices to show that A is reduced. Let a \in A and let I = (\pi ) \subset R be its content ideal, see Lemma 38.19.6. Then a = \pi a' and a' maps to a nonzero element of A/\mathfrak mA where \mathfrak m \subset R is the maximal ideal. If a is nilpotent, so is a', because \pi is a nonzerodivisor by flatness of A over R. But a' maps to a nonzero element of the reduced ring A/\mathfrak m A = \mathcal{O}_{X_ s, x'}. This is a contradiction unless A is reduced, which is what we wanted to show.
Of course, if X is reduced, so is the generic fibre of X over R. If \mathfrak p \subset R is a prime ideal, then R/\mathfrak p is a valuation ring by Algebra, Lemma 10.50.9. Hence redoing the argument with the base change of X to R/\mathfrak p proves the fibre over \mathfrak p is reduced. \square
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