The Stacks project

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. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0H2T. Beware of the difference between the letter 'O' and the digit '0'.