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In page 134, proposition 6.6 Hartshorne mentions that type 2 is a point $x \in X $ x $ \mathbb A^1 $ of codimension one, whose image in $X$ is the generic point of $X$. I realized that this point $x$ corresponds to a prime ideal $\mathcal p$ of height one in Spec $A[t]$ where Spec $A$ is open in $X$ and Spec $A[t]$ is $\pi^{-1} $(Spec$A$). But then I am not able to understand the fact that $A[t]_ \mathcal p $ is a localisation of $K[t]$ at some maximal ideal, where $K$ is the functon field of $X$. Can anyone please explain this?

Also in page 137 while defining $f^*$, is it true that $v_P$ restricted to $K(Y)^*$ and $v_Q$ have the same valuation ring $\mathcal O_Q$?

Suhas
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1 Answers1

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For the first question:

Since $p$ maps to the generic point of $X$, we have $p\cap A = 0$. Thus, in the ring $A[t]_p$, all nonzero elements of $A$ are inverted. So if $S$ denotes the multiplicatively closed set $A[t]-p$, the natural inclusion $A[t]_p \rightarrow S^{-1}(K[t])$ is an isomorphism. Since $K[t]$ is a UFD, it follows that $A[t]_p$ is a UFD (cf. question 140584 on this site). We conclude that $A[t]_p$ is a DVR, which is what Hartshorne was trying to prove.

Now, your question was why is $A[t]_p$ actually isomorphic to $K[t]_{p'}$ for some $p'$ of height 1 in $K[t]$. To see this, let $f$ in $K[t]$ be an irreducible polynomial generating the unique nonzero prime ideal of $A[t]_p$. Clearing denominators, we can assume $f$ is in $p$ in $A[t]$. Let $p'$ denote the prime ideal generated by $f$ in $K[t]$. Then, $p'$ contracts to $p$ and again we have a natural injective map $A[t]_p \rightarrow K[t]_{p'}$ which again is an isomorphism (it's actually the same map as before).

Cass
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