# Integral extension of a discrete valuation ring

$$X$$ and $$Y$$ are integral noether schemes over $$\mathbb{C}$$, and $$F:X\rightarrow Y$$ is a surjective morphism.
Let $$R$$ be any discrete valuation ring over $$\mathbb{C}$$ with its fraction field $$K$$, and $$f:\operatorname{Spec}K \rightarrow Y$$ is given.
I'm trying to prove the following Lemma, 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 $$g: \operatorname{Spec}K' \rightarrow X$$ of $$\mathbb{C}$$-schemes, such that the following diagram commutes: $$\require{AMScd}$$ $$\begin{CD} \operatorname {Spec}K' @>\displaystyle g>> X\\ @V \displaystyle V V\ @VV \displaystyle{F} V\\ \operatorname {Spec}K@>>\displaystyle f> Y \end{CD}$$

My try :
Let $$p\in Y$$ be the image of $$f$$. Since $$F$$ is surjective, there is a $$p'\in X$$ such that $$F(p')=p$$. Take the integral closure of $$\mathcal{O}_{p'}$$. By localizing it at a maximal ideal, I get a DVR $$R'$$ which dominates $$\mathcal{O}_{p'}$$.
But I don't think this is an integral extension of $$R$$ in general, and I'm stuck.
• $R$ can’t be of finite type, I think. The best you can have is “$R$ is a localization of a $\mathbb{C}$-algebra of finite type.” Oct 6, 2020 at 14:28
I’m afraid that this statement is false. Consider the case $$X=Y=\mathrm{Spec}\,\mathbb{C}(T)$$ and $$K=\mathbb{C}(T)$$, $$f$$ is the identity, $$R$$ is the ring of rational fractions defined at $$1$$, and $$X \rightarrow Y$$ is $$f \in \mathbb{C}(T) \rightarrow f(T^2)$$.
If there were $$R’$$, $$K’$$ as claimed, then the intersection of $$R’$$ and $$K(X)$$ yields a DVR $$R’’ \subset \mathbb{C}(T)$$ integral over $$R$$ (beware the embeddings though). But $$R’’$$ is normal, so $$R’’$$ is the integral closure in $$\mathbb{C}(T)$$ of the ring of fractions $$f(T^2)$$ with $$f(T)$$ defined at $$1$$.
But said integral closure is the ring of fractions defined at $$1$$ and $$-1$$, which isn’t local. Indeed, if there are polynomials $$a_0,\ldots,a_n$$ with $$a_n(1) \neq 0$$ satisfying $$\sum_{k=0}^n{a_k(T^2)f^k(T)}=0$$, where $$f$$ is a rational fraction, then the denominator of $$f$$ divides $$a_n(T^2)$$ so vanishes neither at $$-1$$ nor at $$1$$.