$\textbf{1.}\,\,\,\,\,\,\,\,$ Let $(A,\mathfrak m_A)$ be a one dimensional local domain and let $B$ be its integral closure in the fraction field $L=\textrm{Frac}\,A$. Assume that $B$ is finitely generated over $A$. I would like to understand why the discrete valuations rings of $L$ dominating $A$ $are$ the localizations $B_{\mathfrak m}$ of $B$ at its maximal ideals.

I was able to show that such a localization is a DVR of $L$. But, just to conclude this first step, how does one prove that $\mathfrak m_A\subset \mathfrak mB_{\mathfrak m}$? We have \begin{equation} A\hookrightarrow B\hookrightarrow B_{\mathfrak m} \,\,\,\,\,\,\,\,\,x\mapsto x/1 \end{equation} but how do we know that $x\in\mathfrak m_A$ goes to something in $\mathfrak m$ via the first map? And, for showing the "converse" (from DVR to localization) I have no ideas.

$\textbf{2.}\,\,\,\,\,\,\,\,$ Afterwards, I would like to understand the following: suppose we have an integral variety $X$ with normalization $\pi:\tilde X \to X$. Let us take a (closed) codimension one subvariety $V\subset X$. The questions is: why do we have a correspondence between the DVR's of $L=K(X)$ dominating $A=\mathscr O_{X,V}$ and the (closed) subvarieties $Z\subset\tilde X$ mapping onto $V$? I feel like domination should translate $\pi(Z)=V$, but can't see it neatly.

Thank you!


(1).1. Let $P=\mathfrak mB_\mathfrak m\cap A$. Then $A/P\to B_\mathfrak m/\mathfrak mB_\mathfrak m$ is injective and integral. As the RHS is a field, this implies that $A/P$ is a field, hence $P=\mathfrak m_A$ and $\mathfrak m_A\subseteq \mathfrak mB_\mathfrak m$.

(1).2. Let $R$ be a DVR of $L$ dominating $A$. Any element $b\in B$ is integral over $R$ and belongs to $\mathrm{Frac}(R)$, so $b\in R$ because $R$ is integrally closed. Hence $B\subseteq R$. Let $Q=\mathfrak m_R\cap B$. Then $Q$ is a prime ideal of $B$ containing $\mathfrak m$, hence non-zero. As $\dim B=1$, $Q$ is maximal and $B_Q$ is a DVR of $L$ dominated by $R$, hence $B_Q=R$.

(2). One can apply (1) to $O_{X,V}$. The subvariety $Z$ gives rize to a DVR equal to $O_{\tilde{X}, Z}$. As $Z$ surjects to $V$, its generic point goes to the generic point of $V$, hence $O_{X,V}\to O_{\tilde{X}, Z}$ is a homomorphism of local rings. Conversely, a DVR dominating $O_{X,V}$ gives rize to a point $\xi$ of $\tilde{X}$ lying over the generic point of $V$ by 1.2. Let $Z$ be the Zariski closure of $\xi$. As $\pi$ is a finite map hence closed, $\pi(Z)$ is closed and dense in $V$, hence equals to $V$.

  • $\begingroup$ (1).2: does the assertion "$R$ is a DVR of $L$" include "$R$ dominates $A$"? (I ask this because DVR's are maximal local rings w.r.t. domination.) And to show $B\subset R$, could it be enough to say that $R\supset A$ and $R$ is integrally closed? (I ask this because I don't understand "any element $b\in B$ is integral over $R$": do we have $R\subset B$?) Afterwards, can I imitate (1).1 to see that $Q$ is maximal? ($B/Q\to R/\mathfrak m_R$ injective and integral). The rest is very clear, thank you QiL! $\endgroup$ – Brenin Sep 14 '12 at 9:03
  • $\begingroup$ @atricolf:(1).2: non, a DVR of L doesn't necessarily dominate $A$ (example: $L=\mathbb Q$ and $A=\mathbb Z$ localized at $p$). In the proof of $B\subseteq R$, I meant $b$, viewed as an element of $L$, is integral over $R$ (because integral over $A\subseteq R$). For the proof of $Q$ maximal, non $B/Q\to R/m_R$ is not, a priori, integral. We really need the assumption on $\dim A$ here. Otherwise it is false, $Q$ can be of any positive height. $\endgroup$ – user18119 Sep 14 '12 at 11:14

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