$X$ 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.
Any advice and comments are applicated. Thanks in advance.
Best Answer
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 $1$.