1
$\begingroup$

Let $X^{\prime}$ and $X$ be integral noether schemes over $\mathbb{C}$, and $p:X^{\prime}\rightarrow X$ be a surjective morphism.
Let $R$ be any discrete valuation ring over $\mathbb{C}$ with its fraction field $K$, and $i:\operatorname{Spec}K \rightarrow X$ is given.
I'm trying to prove the following claim, but I have a problem.

There exists an integral extension $R\subset R'$, where $R'$ is another discrete valuation ring over $\mathbb{C}$ with fraction field $K'$, and a morphsim $j: \operatorname{Spec}K^{\prime} \rightarrow X$ of $\mathbb{C}$-schemes, such that the following diagram commutes: $\require{AMScd}$ \begin{CD} \operatorname {Spec}K^{\prime} @>\displaystyle j>> X^{\prime}\\ @V \displaystyle V V\ @VV \displaystyle{p} V\\ \operatorname {Spec}K@>>\displaystyle i> X \end{CD}

My try :
Let $x\in X$ be the image of $i$. Since $p$ is surjective, there is a closed point $x^{\prime}\in X^{\prime}$ such that $p(x^{\prime})=x$. Then,$k(x)\subset k(x^{\prime})$ and $K \subset K^{\prime}:=k(x^{\prime})\otimes K$ is a finite algebraic extension.
We get $R\subset K^{\prime}$. Set $R^{\prime}$ as the integral closure of $R\hookrightarrow K^{\prime}$.

But I'm not sure if $R^{\prime}$ is DVR, and I'm stuck.
Any advice and comments are applicated. Thanks in advance.

Improve this question
$\endgroup$
4
  • $\begingroup$ 1. This seems like it might be better suited for MSE and your attempt is a bit of a mess. $R'$ need not be a DVR (consider the integral closure of $k[t^2-1]_{(t^2-1)}$ in $k(t)$). 2. It's somewhat strange you picked noetherian instead of finite type. Are you really only assuming noetherian and not finite type? $\endgroup$
    – KReiser
    Nov 12, 2020 at 5:22
  • $\begingroup$ Thanks. Excuse me, but I should ask at MSE. If I assume $R$ and $X$ is of finite type over $k$, is this true? $\endgroup$
    – Aoki
    Nov 12, 2020 at 14:34
  • $\begingroup$ Yes, see for instance the strategy outlined in Hartshorne exercise II.4.11(a). $\endgroup$
    – KReiser
    Nov 13, 2020 at 11:54
  • $\begingroup$ Take a look at this question: math.stackexchange.com/questions/3691688/… $\endgroup$
    – G. Gallego
    Apr 29, 2021 at 10:37

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.