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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