Let $X$ be a smooth projective $k$-variety. I want to prove that if we have a $k$-morphism :
$$ f : \operatorname{Spec}(R) \rightarrow X$$
where $R$ is a DVR, which is compatible with the following diagram :
$\require{AMScd}$ \begin{CD} \operatorname{Spec}(k(X)) @>{}>> X\\ @VVV @VVV\\ \operatorname{Spec}(R) @>{}>> \operatorname{Spec}(k) \end{CD}
(where $k(X)$ is the field of rational functions of $X$)
Then $f$ map the closed point of $\operatorname{Spec}(R)$ to a closed point of $X$. I tried several things including some dimension/irreducible closed set arguments, but it didn't work. (Moreover, if $R$ was a $k$-algebra of finite type, it would be direct by the Nullstellensatz, but why it would be so?)
Can someone could help me?
Thank you!
Edit : I think maybe I have the answer. We consider $\overline{\{ f(x) \}}$ which is a closed irreducible subset of $X$, hence of dimension $0$ or $1$. If its dimension is $0$, then it's $\{f(x)\}$ and we have then $f(x)$ is a closed point of $X$. Otherwise, $\overline{\{ f(x) \}}$ has dimension $1$ and is then equal to $X$. But if we denote by $\mu$ the generic point of $\operatorname{Spec}(R)$, then by the uniqueness of the generic point of $X$, we have $f(x) = f(\mu)$. Then, I think it's ask for the valuation associated to $R$ to bo non trivial to make it impossible. Right ?
This is not true, here is a simple counterexample:
Let $k$ be any field, let $X=\mathbb{P}^1_k$ and let $x \in \mathbb{P}^1(k)$ be a point. Then the local ring $R=\mathcal{O}_{X,x}$ is a DVR with field of fractions $\operatorname{Frac} R$ isomorphic to $K(x)$, and moreover the following diagram commutes \begin{CD} \operatorname{Spec}(k(X)) @>{}>> X\\ @VVV @VVV\\ \operatorname{Spec}(R) @>{}>> \operatorname{Spec}(k). \end{CD}
However, the map $\operatorname{Spec} R \to X$ coming from the fact that $R$ is a local ring of $X$, has the generic point $\operatorname{Spec} k(X) \to X$ in its image.