The Stacks project

Lemma 37.26.6. Let $X \to \mathop{\mathrm{Spec}}(R)$ be a proper flat morphism where $R$ is a discrete valuation ring. If the special fibre is reduced, then both $X$ and the generic fibre $X_\eta $ are reduced.

Proof. Let $x \in X$ be a point in the generic fibre $X_\eta $ such that $\mathcal{O}_{X_\eta }$ is nonreduced. Then $\mathcal{O}_{X, x}$ is nonreduced. 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 $A = \mathcal{O}_{X, x'}$. Let $\pi \in R$ be a uniformizer. If $a \in A$ then there exists an $n \geq 0$ and an element $a' \in A$ such that $a = \pi ^ n a'$ and $a' \not\in \pi A$. This follows from Krull intersection theorem (Algebra, Lemma 10.51.4). 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 $A/\pi A = \mathcal{O}_{X_ s, x'}$. This is a contradiction unless $A$ is reduced, which is what we wanted to show. $\square$


Comments (2)

Comment #8076 by Laurent Moret-Bailly on

First line of proof: should be . But in fact it is not very clear whether one is proving that is reduced or that is. I propose to change the first 3 lines as follows:

Assume the special fibre is reduced. Let be any point, and let us show that
is reduced. (This will prove that and are reduced). Let be a specialization with in the special fibre; such a specialization exists as a proper morphism is closed. Consider the local ring . Then is a localization of , so it suffices to show that is reduced. Let be a uniformizer. (The rest is unchanged)

Comment #8088 by Laurent Moret-Bailly on

Not sure this is useful, but the lemma works for any valuation ring : instead of Krull's intersection theorem, use the fact that has a content ideal (Lemma 0ASX) which is principal by Comment 8087. (I do realize that Lemma 0ASX comes a bit later).


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