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I'm having a really hard time understanding the proof of this proposition. $X$ is a noetherian integral separated scheme that is regular in codimension 1. We consider $X\times \mathbb{A}^1$ and the projection $\pi$ onto $X$.

First, Hartshorne says there are two types of codimension 1 points of $X\times \mathbb{A}^1$, type 1 being those which map to a codimension 1 point of $X$ through $\pi$ and type 2 being those which map to the generic point. I'm not clear on why these are the only types. I believe that when $Y$ is a subscheme of $X$ that $\pi^{-1}(Y)=Y\times \mathbb{A}^1$ and is a subscheme of $X\times \mathbb{A}^1$. Is it true that the codimension of $\pi^{-1}(Y)$ is less than or equal to the codimension of $Y$? Is this what Hartshorne is implicitly using? I think if this is true we would also get a 1-1 correspondence between type 1 prime divisors of $X\times \mathbb{A}^1$ and prime divisors of $X$.

Next, if we have that the function field of $X$ is $K$, then we have the function field of $X\times \mathbb{A}^1$ is $K(t)$. I guess Hartshorne is saying that if $f\in K$ then $(f)_X=(f)_{X\times \mathbb{A}^1}$ where on the r.h.s we view $f\in K(t)$. I am confused about why this is true. I guess I need to understand how to compare the image of $f$ in the local rings at generic points of $X\times \mathbb{A}^1$ with those of $X$.

Seth
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1 Answers1

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Your questions can all be answered by reducing to the case where $X = \operatorname{Spec} R$ is affine.

We have $Y = \operatorname{Spec} R[t]$, with the projection map $\pi$ induced by the inclusion $R\to R[t]$. First of all, notice that if $K$ is the fraction field of $R$, then $K(t)$ is the fraction field of $R[t]$. And in general, localizations of $R$ inject into corresponding localizations of $R[t]$.

What are the codimension $1$ points of $Y$, i.e. the height $1$ primes of $R[t]$? Well, if $P \subset R[t]$ is a height $1$ prime, then either $P\cap R = Q$ is a height $1$ prime of $R$, or $P\cap R = (0)$.

And what are the corresponding localizations in each case? In the first case, $R_Q \subset R[t]_P = (R_Q)[t]$. In the second, $K=R_{(0)} \subset R[t]_P = K[t]_{P'}$, where $P'$ is the (maximal) ideal of $K[t]$ generated by $P$.

In other words, we have, in the first case, points that just correspond to the codimension-1 points of $X$ (and are in fact the generic points of the fibers over these points), and, in the second case, basically the closed points of $\mathbb{A}^1_{K(X)}$ (though since $K(X)$ is not algebraically closed, be a little careful of thinking this way, as Hartshorne is always working over algebraically closed fields).

Andrew Dudzik
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  • Hey, thanks for the answer! You've already been a big help on some of my other questions. Its getting late for me but I will be sure to read this carefully tomorrow! – Seth Oct 21 '14 at 00:56
  • Hey, I read your answer but I am not really understanding, so I will have to spend some more time on this when I have the chance. It's been a long day. – Seth Oct 21 '14 at 23:30
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    @Seth Understandable. Hopefully the basic point is somewhat clear: most basic properties of integral schemes can be reduced to basic properties of some integral domain. I am basically asserting that your questions come down to classifying the height $1$ primes of a polynomial ring over an integral domain. – Andrew Dudzik Oct 22 '14 at 00:05
  • @Slade I was wondering if you could elaborate a little on your claim:

    "if $\mathfrak{p} \subset R[t]$ is prime, then either $\mathfrak{p} \cap R = Q$ is a height $1$ prime of $R$, or $\mathfrak{p} \cap R=(0)$"

    I'm really not seeing why this is true at all. Is this some application of the going-up theorem?

     – Luke Jan 08 '18 at 07:59
  • @Seth may also be able to shed some light on this. I also just realized this is a 3 year old question, so hopefully you're still around here. – Luke Jan 08 '18 at 08:00
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    @Luke I meant to say that $P$ is a height 1 prime of $R[t]$, but maybe that was clear. Anyway, the point is that the height cannot go up—if $Q=P\cap R$ had height $>1$, then $Q[t]$ would have height $>1$ in $R[t]$. But $Q[t] \subset P$. – Andrew Dudzik Jan 08 '18 at 08:40
  • Ahh of course, I see! So the dimension can only go down, so the result must be dimension 0 or 1. Since it's an integral domain, the only height 0 prime is the trivial ideal. Thank you! Seems obvious now you spell it out, appreciated :) – Luke Jan 08 '18 at 09:02
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    About the formulas for the localizations: the first one looks wrong to me, and the second one is unclear. I think one can use the formula $R[T]{\mathfrak{p}[T]}\cong R{\mathfrak{p}}[T]{(\mathfrak{p}R{\mathfrak{p}})[T]}$ to compute this. –  Jul 02 '18 at 13:10