The Stacks project

First line of proof: $\mathcal{O}_{X_\eta}$ should be $\mathcal{O}_{X_\eta,x}$. But in fact it is not very clear whether one is proving that $X$ is reduced or that $X_\eta$ is. I propose to change the first 3 lines as follows:

Assume the special fibre is reduced. Let $x \in X$ be any point, and let us show that $\mathcal{O}_{X,x}$
is reduced. (This will prove that $X$ and $X_\eta$ are 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 $\pi \in R$ be a uniformizer. (The rest is unchanged)

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

@#8076 Thanks and fixed here.

@#8088 OK, I added this fun curiosity in a later chapter. See this commit.

I noticed an additional small optimization to this lemma: the flatness hypothesis can be weakened to instead require only the set-theoretic statement, namely, that the irreducible components of $X$ map to the generic point (rather than all associated points). This came up for me in a situation where I knew that $X_s$ and $X_\eta$ were reduced, was not sure whether $X$ was flat, and specifically wanted a technical lemma to imply that $X$ was reduced (en route to proving flatness, in fact).

Here's how the proof goes under the weakened hypothesis: after the same initial setup, we have $x' \in X$ lying over $s$ and we wish to show $A = \mathcal{O}_{X, x'}$ is reduced. Now by the set-theoretic hypothesis, $\pi$ is not in any minimal prime ideal of $A$. The equation $a = \pi^n a'$ thus implies (since $a$ is nilpotent) that $a'$ is in every minimal prime ideal, hence $a'$ is again nilpotent. This contradicts reducedness of $A/\pi A$ as in the current proof.

Typos (?) : in the proof the element $a\in A$ needs to be assumed to be non-zero. Idem for Lemma 38.19.7.

Thank you Jake! I've added the improvement here and I fixed the mistake with zero vs nonzero. It may be that the same result holds in the non-Noetherian case Lemma 38.19.7, but I didn't see immediately how to prove it, so I've left it as is for now.

Jarod Alper observed to me that there is a cleaner way to finish the proof. By hypothesis $X^{\text{red}}$ is flat over $R$. So, with $A$ as in the current proof, tensoring $0 \to \sqrt{0} \to A \to A/\sqrt{0} \to 0$ with $R/\pi \otimes_R -$, the sequence stays exact (because $A/\sqrt{0}$ is $R$-flat) and yields $0 \to R/\pi \otimes_R \sqrt{0} \to A/\pi \to A/(\pi, \sqrt{0}) \to 0$. But $A/\pi$ is reduced, so $R/\pi \otimes_R \sqrt{0} = A/\pi A \otimes_A \sqrt{0} = 0$. By Nakayama, $\sqrt{0} = 0$, so $A$ is reduced.

Going to leave as is.