So I've spent quite a lot of time trying to work this proof out myself and I'm getting nowhere. The proof is Proposition 6.6 in Chapter II of Hartshorne. Let $X$ be a noetherian, integral, separated scheme which is regular in codimension $1$. Let $\mathbb{A}^{1}$ denote $\text{Spec } \mathbb{Z}[t]$. The proof is to show that $X \times \mathbb{A}^{1}$ (with the product taken over $\text{Spec }\mathbb{Z}$) is noetherian, integral, separated, and regular in codimension $1$. The noetherian, integral, and separated parts are easy if you know some basic change-of-base results. My question is about the regular in codimension $1$ part of the proof.

We can immediately reduce to the affine case. Let $X = \text{Spec }B$. Then the projection morphism takes the form $$ \pi: \text{Spec } B[t] \longrightarrow \text{Spec }B $$ corresponding to the inclusion of rings $$ B \hookrightarrow B[t] $$ If $p \in \text{Spec }B[t]$ is a point of codimension $1$, say it corresponds to a prime ideal $\mathfrak{p} \subset B[t]$ of height $1$. Hartshorne deduces that there are two cases:

1) $\pi(p)$ is a codimension $1$ point in $\text{Spec } B$;

2) $\pi(p)$ is the generic point of $\text{Spec } B$.

So far I understand. But this is as far as I understand. He claims that in the first case, $p$ is the generic point of $\pi^{-1}(p)$. But why is $\pi^{-1}(y)$ even a closed set, let alone irreducible? Moreover, why is it then immediate that $\mathcal{O}_{p} \simeq \mathcal{O}_{\pi(p)}[t]_{\mathfrak{m}_{\pi(p)}}$ as he claims?

He claims in the second case that the local ring $\mathcal{O}_{p}$ is a localization of $K[t]$ at some maximal ideal, where $K$ is the function field of $X$. This claim in itself was not clear to me, but I got an answer in another question here.


Firstly, $\pi$ is a continuous map. This implies that $\pi^{-1}(\pi(p))$ is a closed set. Non-irreducible schemes still have generic points, so you do not need to worry about irreducibility here.

The claim that $\mathcal{O}_{\operatorname{Spec} B[t],p} \cong \mathcal{O}_{\operatorname{Spec} B,\pi(p)}[t]_{\mathfrak{m}_{\pi(p)}}$ follows from the product description of $\operatorname{Spec} B[t]$. You can think of this as saying that localizing at $p$ is the same thing as localizing in the $B$ direction and then localizing in the $t$ direction. Writing out the right hand side of the isomorphism as a composition of tensor products (remember $B[t] = B\otimes_{\Bbb Z} \Bbb Z[t]$ and that localizations can be thought of as tensor products) can show you the result a bit more explicitly if you like.

  • $\begingroup$ Thank you for this reply! Although one thing that wasn't immediately clear to me: Why is $\pi(p)$ a closed point? Indeed if it was then $\pi^{-1}(\pi(p))$ would be a closed set, but $\pi(p)$ is a point of codimension $1$. So unless the Krull dimension of $B$ is $1$, I don't see why it would be closed. $\endgroup$ – Luke May 13 '18 at 7:42

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