# What is the stalk map for a morphism of affine schemes?

$$\newcommand{\Spec}{\operatorname{Spec}} \newcommand{\O}{\mathscr{O}}$$ Let $$X=\Spec A$$, and $$Y=\Spec B$$, and suppose that $$f:X\rightarrow Y$$ is a morphism coming from the ring homomorphism $$\phi:B\rightarrow A$$. This induces sheaf morphism $$f^\sharp:\O_Y\rightarrow (f_*\O_X)$$ which is given by localizations of $$\phi$$ at elements in $$b\in B$$, and $$\phi(b)\in A$$. It is easy to check that the stalk map $$(\O_Y)_y\rightarrow (f_*\O_X)_y$$, is then given by the induced map $$B_\mathfrak p\rightarrow A_\mathfrak p$$, where we are taking $$A$$ to be a $$B$$ module, and $$\mathfrak p$$ is the prime ideal corresponding to the point $$y$$. However, what about the stalk map $$(\O_Y)_{f(x)}\rightarrow (\O_X)_x$$?

In particular, for arbitrary locally ringed spaces, let $$f:X\rightarrow Y$$ is a morphism of local rings, and $$f_{f(x)}^\sharp:(\O_Y)_{f(x)}\rightarrow (f_*\O_X)_{f(x)}$$ be the unique stalk map given on equivalence classes $$[U,s]_{f(x)}\in (\O_Y)_{f(x)}$$ by $$[U,s]_{f(x)}\rightarrow [U,f^\sharp_U(s)]_{f(x)}$$. Then I would define $$f_x:(\O_Y)_{f(x)}\rightarrow (\O_X)_x$$ as the composition of the map of the above map with the map $$(f_*)_x:[U,s]_{f(x)}\in (f_*\O_X)_{f(x)}\mapsto [f^{-1}(U),s]_x\in (\O_X)_x$$. In other words all that last map is changing is how restrictive our equivalence relation is.

What is the analogue of this in the case of affine schemes? If $$\mathfrak p=\phi^{-1}(\mathfrak q)$$ is there somehow a natural morphism $$A_\mathfrak p\rightarrow A_\mathfrak q$$ that would clearly agree with the above construction?

The morphism $$B_{\mathfrak{p}} \to A_{\mathfrak q}$$ that you're looking for comes from the composition $$B \to A \to A_{\mathfrak{q}}$$ and the universal property of localization. All you need to check to get an induced morphism is that if $$b \in B \setminus \mathfrak{p}$$, then $$b$$ gets mapped to a unit in $$A_{\mathfrak{q}}$$. But if $$b \in B \setminus \phi^{-1}(\mathfrak{q})$$, then $$\phi(b) \notin \mathfrak{q}$$, so $$\frac{\phi(b)}{1}$$ is a unit in $$A_{\mathfrak{q}}$$

• How do we see that this agrees with the abstract version of this morphism? Commented May 22 at 14:29

In total generality if you have spaces $$X,Y$$ and a continuous map $$f:X\to Y$$ and a given morphism of sheaves $$\phi:\mathscr{F}\to f_\ast\mathscr{G}$$ for some sheaves $$\mathscr{F},\mathscr{G}$$ on $$Y,X$$ respectively, we can get our hands on maps $$\mathscr{F}_{f(x)}\to\mathscr{G}_x$$ using formal nonsense only. I think it is instructive to see this; it makes the construction feel less improvised.

Firstly, there is a map $$\phi_{f(x)}:\mathscr{F}_{f(x)}\to(f_\ast\mathscr{G})_{f(x)}$$ and it remains to find some kind of canonical homomorphism $$(f_\ast\mathscr{G})_{f(x)}\to\mathscr{G}_x$$. But indeed, $$(f_\ast\mathscr{G})_{f(x)}\cong(f^\ast(f_\ast\mathscr{G}))_x$$ and there is a canonical counit $$f^\ast f_\ast\mathscr{G}\to\mathscr{G}$$ which restricts to a map of the form we want.

Less abstractly, if you think about this isomorphism and how the counit is actually defined, this homomorphism works as you say. If $$\xi\in(f_\ast\mathscr{G})_{f(x)}$$ then it is represented as the germ of some $$\alpha\in(f_\ast\mathscr{G})(V)=\mathscr{G}(f^{-1}V)$$ where $$V$$ is an open neighbourhood of $$f(x)$$ and thus $$f^{-1}V$$ an open neighbourhood of $$x$$; the germ of $$\alpha$$ in $$Y$$, $$\xi$$, is mapped to the germ of $$\alpha$$ in $$X$$ i.e. viewing it as a section of $$\mathscr{G}$$. Though it feels like this does nothing, this homomorphism need not in general inject or surject I don't think. It does under some point-set assumptions though; what you really need is that the neighbourhoods of form $$f^{-1}V$$ are cofinal and this happens if the spaces involved are LCH and $$f$$ is (quasi)proper - not a situation common to algebraic geometry.

If $$X,Y$$ are affine schemes, spectra of $$A,B$$ respectively and $$\mathscr{F},\mathscr{G}$$ their structure sheaves and $$(f,\phi)$$ a scheme morphism we know that, really, $$\phi$$ is induced by a homomorphism $$g:B\to A$$ and for a given point $$\mathfrak{p}$$ of $$A$$, we need to understand the map $$(f_\ast\mathscr{O}(X))_{g^{-1}\mathfrak{p}}\to\mathscr{O}(X)_\mathfrak{p}\cong A_\mathfrak{p}$$. The general element of the left hand side is the germ of some section of $$\mathscr{O}(X)(f^{-1}D(u))=\mathscr{O}(X)(D(g(u)))$$, where $$g^{-1}\mathfrak{p}\in D(u)$$ i.e. $$g(u)\notin\mathfrak{p}$$, and therefore is the germ of some $$a/g(u)$$ (replacing $$u$$ with some power of $$u$$ if necessary) where $$a\in A$$ but be mindful this is a germ at a point of $$Y$$, not $$X$$. This element $$[a/g(u)]$$ is identified (via the isomorphism and canonical counit) with, well, "the same thing" i.e. the germ of $$a/g(u)$$ at $$\mathfrak{p}$$. So really, the overall induced map $$\mathscr{O}(Y)_{f(\mathfrak{p})}\cong B_{g^{-1}\mathfrak{p}}\to A_\mathfrak{p}\cong\mathscr{O}(X)_{\mathfrak{p}}$$ on stalks is just $$[b/u]\mapsto[g(b)/g(u)]_{\text{ in Y}}\mapsto[g(b)/g(u)]_{\text{ in X}}$$. The first map is just by definition of $$\phi$$ the sheaf map associated to the ring map $$g$$. Leaving equivalence classes understood, we just write this as $$b/u\mapsto g(b)/g(u)$$.