If $\mathfrak{p}_y$ the preimage of $\mathfrak{m}_x$ by $\mathcal{O}_Y(Y)\rightarrow\mathcal{O}_X(X)\rightarrow \mathcal{O}_{X,x}$, show that $f(x)=y$

Question

I'm trying to solve the following problem:

Let $(X,\mathcal{O}_X)$ be a locally ringed space and $Y$ an affine scheme. Let $f:X \rightarrow Y$ be a morphism of locally ringed spaces. Then, for any $x\in X$, we consider the ring morphisms

$$\mathcal{O}_Y(Y)\rightarrow\mathcal{O}_X(X)\rightarrow \mathcal{O}_{X,x}.$$

Let $\mathfrak{p}_y$ be the preimage of the maximal ideal $\mathfrak{m}_x$ in $\mathcal{O}_Y(Y)$ and $y$ the point corresponding to $\mathfrak{p}_y$. Show that $f(x)=y$.

This is what I tried: I didn't know how to attempt this for a general locally ringed space, so I assumed that $(X,\mathcal{O}_X)$ is a scheme. In particular, by restraining $f$ to an affine open containing the point $x$, I think I can assume that $(X,\mathcal{O}_X)$ is affine.

Now, since $f:X \rightarrow Y$ is a morphism of locally ringed spaces, we know that the induced map $f^\sharp_x:(f^{-1}\mathcal{O}_Y)_x=\mathcal{O}_{Y,f(x)} \rightarrow \mathcal{O}_{X,x}$ is local, i. e., that $f_x^\sharp(\mathfrak{m}_{f(x)})\subset \mathfrak{m}_x$, which implies $(f_x^\sharp)^{-1}(\mathfrak{m}_x)=\mathfrak{m}_{f(x)}$.

I also know that if $\phi:\mathcal{O}_Y(Y)\rightarrow\mathcal{O}_X(X)$ is the induced map on global sections, the diagram

$$\require{AMScd} \begin{CD} \mathcal{O}_Y(Y) @>{\phi}>> \mathcal{O}_X(X)\\ @VVV @VVV \\ \mathcal{O}_{Y,f(x)} @>{f^\sharp_x}>> \mathcal{O}_{X,x} \end{CD}$$

is commutative (theorem 11.11 here).

Putting all of this together, if we denote $i:\mathcal{O}_Y(Y)\rightarrow \mathcal{O}_{Y,f(x)}$ and $j:\mathcal{O}_X(X)\rightarrow \mathcal{O}_{X,x}$, we have that

$$\mathfrak{p}_y=(j \circ \phi)^{-1}(\mathfrak{m}_x)=(f_x^\sharp \circ i)^{-1}(\mathfrak{m}_x)=i^{-1} \circ (f_x^\sharp)^{-1}(\mathfrak{m}_x)=i^{-1} (\mathfrak{m}_{f(x)})=i^{-1}(\mathfrak{p}_{f(x)}\mathcal{O}_{Y,f(x)})=\mathfrak{p}_{f(x)}$$

Is that correct/the right idea? How do I generalize it to the case of a locally ringed space?

Answer 1

Yes, the diagram is the correct thing to use, your argument is fine, and you don't even have to do anything different for the case of $X$ a locally ringed space: the diagram in your post is valid for any morphism of locally ringed spaces, and the only place you use that $Y$ is an affine scheme is in the structure of the morphism $\mathcal{O}_Y(Y)\to\mathcal{O}_{Y,f(x)}$. (The importance of the local condition is so that $f_x^\sharp$ is a local map of local rings.)

If you're looking for an outside reference, see for instance Stacks 01HY.