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By Algebraic Geometry I from Görtz, Wedhorn page 60 $f^\sharp_x$ is the unique ring homomorphism which makes the diagram $A\to B \to B_{p_x}$, $A\to A_{p_{f(x)}}\to B_{p_x}$ commutative. The first arrow is $\varphi$ and the last $f^\sharp_x$. The two other arrows are the canonical homomorphisms. Can you please explain, why $f^\sharp_x$ makes the diagram commutative and why only $f^\sharp_x$ makes the diagram commutative?

Background: $f:Spec B \to Spec A$ is a morphism of affine schemes, $\varphi:=\Gamma(f):=f^\flat_{Spec A}$ and $p_x$ is the prime ideal of $B$ corresponding to a point $X\in Spec B$.

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For any morphism of locally ringed spaces $f:X\to Y$, if $x\in X$, $y=f(x)$, the "stalk map" $f_x^\sharp:\mathscr{O}_{Y,y}\to\mathscr{O}_{X,x}$ is (by definition!) the unique homomorphism such that, for any open subset $V\subseteq Y$ containing $y$, the composite $\mathscr{O}_Y(V)\to\mathscr{O}_X(f^{-1}(V))\to\mathscr{O}_{X,x}$ is equal to $\mathscr{O}_Y(V)\to\mathscr{O}_{Y,y}\xrightarrow{f_x^\sharp}\mathscr{O}_{X,x}$ (the uniqueness and existence follow from the universal property of colimits, and you should check this if you don't understand).

Now let's specialize to the case $X=\mathrm{Spec}(B)$ and $Y=\mathrm{Spec}(A)$, and $f:X\to Y$ is the morphism corresponding to a ring map $\varphi:A\to B$. Then $x$ is a prime $\mathfrak{q}\in\mathrm{Spec}(B)$, $y$ is a prime $\mathfrak{p}\in\mathrm{Spec}(A)$, and $\varphi^{-1}(\mathfrak{q})=\mathfrak{p}$. By the definition of the structure sheaf of the spectrum of a ring, $\mathscr{O}_{Y,y}=A_\mathfrak{p}$ and $\mathscr{O}_{X,x}=B_\mathfrak{q}$ (these equalities are really canonical isomorphisms of $A$ and $B$-algebras, respectively), and these identifications are such that $\mathscr{O}_X(X)=B\to\mathscr{O}_{X,x}=B_\mathfrak{q}$ is the localization map, and likewise for $\mathscr{O}_Y(Y)=A\to\mathscr{O}_{Y,y}=A_\mathfrak{p}$. As a specific instance of what I've said above about the map $f_x^\sharp$, with $V=Y$, so $f^{-1}(V)=X$, we have the the map $\varphi=f^\sharp_Y:\mathscr{O}_Y(Y)=A\to\mathscr{O}_X(X)=B\to\mathscr{O}_{X,x}=B_\mathfrak{q}$ is the same as the map $A\to A_\mathfrak{p}\xrightarrow{f_x^\sharp}B_\mathfrak{q}$. So $f_x^\sharp$ makes the diagram (that I haven't written down, but have just expressed as an equality of composites of ring maps) commutative by the definition of $f_x^\sharp$ (that I explained above, for any map of LRS) and by the identifications coming from the definition of the structure sheaf on the spectrum of a ring. The fact that there is only one ring map $A_\mathfrak{p}\to B_\mathfrak{q}$ making the (implicit) diagram commute is a consequence of the universal property of localization, which ensures that $A\to A_\mathfrak{p}$ is an epimorphism in the category of rings. Indeed, if $\psi:A_\mathfrak{p}\to B_\mathfrak{q}$ were another map making the (implicit) diagram commute, then pre-composing with $A\to A_\mathfrak{p}$ and using the commutativity, we get that

$$A\to A_\mathfrak{p}\xrightarrow{f_x^\sharp}B_\mathfrak{q}=A\to A_\mathfrak{p}\xrightarrow{\psi}B_\mathfrak{q}$$

so, since $A\to A_\mathfrak{p}$ is a ring epimorphism, $\psi=f_x^\sharp$.

EDIT: This is a response to the questions asked by the OP in the comments. First, the use that localizations are epimorphisms to show that $A_\mathfrak{p}\to B_\mathfrak{q}$ is the unique $A$-algebra map between these two rings, where the target is an $A$-algebra via $A\to B\to B_\mathfrak{q}$, is not superfluous. I'm not talking about maps of locally ringed space here, just maps of rings. If I had said that, for each $f\in A$, $f\notin\mathfrak{p}$, $A_f\to A_\mathfrak{p}\to B_\mathfrak{q}$ is the same as $A_f\to B_{\varphi(f)}\to B_\mathfrak{q}$, then I'd be using the universal property of colimits (because $A_\mathfrak{p}=\mathrm{colim}_{f\notin\mathfrak{p}}A_f$, but I didn't). The condition characterizing $f_x^\sharp$ for a general map of locally ringed spaces is coming from the universal property of colimits, but in the special case of rings, you only have to consider the open set $V=Y$ in the discussion above (because of the universal property of localization).

I guess your second question is why $f(x)=\varphi^{-1}(\mathfrak{q})$. So now you're working with an arbitrary morphism of LRS $f:\mathrm{Spec}(B)\to\mathrm{Spec}(A)$ and you want to prove that $f(x)=\varphi^{-1}(\mathfrak{q})$ if $x=\mathfrak{q}\in\mathrm{Spec}(B)$, i.e., the map of topological spaces is determined by the map on global sections for morphisms of spectra of rings. For this, see my answer here for a complete explanation of the result you're interested in (I don't want to rewrite the whole thing here).

EDIT: I guess I'll elaborate a bit. We have $f(\mathfrak{q})=\mathfrak{p}$ and we want to prove that $\varphi^{-1}(\mathfrak{q})=\mathfrak{p}$. We know what the map $f_x^\sharp:A_\mathfrak{p}\to B_\mathfrak{q}$ is, and we know that it is local because we're working with maps of local rings. We also know that this map $f_x^\sharp$ is compatible with the respective localization maps $A\to A_\mathfrak{p}$ and $B\to B_\mathfrak{q}$ (there's a commutative diagram here that I don't know how to make on MSE). My argument at the aforementioned link shows that $f(\mathfrak{q})$ is the inverse image in $A$ of $\mathfrak{q}B_\mathfrak{q}$ under $A\to B\to B_\mathfrak{q}$, which is $\varphi^{-1}(\mathfrak{q})$ (the inverse image of $\mathfrak{q}B_\mathfrak{q}$ in $B$ is $\mathfrak{q}$). Now if we pull back $\mathfrak{q}B_\mathfrak{q}$ to $A$ the other way around the diagram, which we already know by commutativity of the diagram gives us $f(\mathfrak{q})$, we get $\mathfrak{p}A_\mathfrak{p}$ in $A_\mathfrak{p}$ (since the map is local), and then it is a standard property of localization that the inverse image of the maximal ideal of $A_\mathfrak{p}$ in $A$ is $\mathfrak{p}$. So $\varphi^{-1}(\mathfrak{q})=f(\mathfrak{q})=\mathfrak{p}$, and we win.

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  • $\begingroup$ First, thanks a lot for your answer. I'm not so familiar with category theoretic arguments. For example I don't know the universal property of colimits. But I understood the definition of a direct limit (probably not in the most abstract category theoretic sense but in the sense we need here). Can you formulate that universal property explicitly without using category theory? Furthermore, epimorphism are surjective homomorphisms, right? But why should $A\to A_p$ be surjective? Let $A=\mathbb{Z}$ and $p=(3)$ then it isn't since $A$ is a proper subset of $A_p$. $\endgroup$ – principal-ideal-domain Jul 23 '14 at 19:12
  • $\begingroup$ In this case, you can take colimit to be direct limit. Any definition of a direct limit of rings, modules, etc., will have a universal property given that characterizes it. Epimorphism is a categorical term and it does not mean surjective. A ring map $g:R\to S$ is an epimorphism if for all rings $T$ and all ring maps $u,v:S\to T$, if $u\circ g=v\circ g$, then $u=v$. Using the universal properties of the objects and morphisms involved in this argument is the cleanest (and arguably the "right") way to do it, so if you do not understand the argument in terms of universal properties (not category $\endgroup$ – Keenan Kidwell Jul 23 '14 at 19:18
  • $\begingroup$ theory), then I suggest you review universal properties of localizations and direct limits. $\endgroup$ – Keenan Kidwell Jul 23 '14 at 19:18
  • $\begingroup$ Thanks again. That helps me so much (so each surjective ring map is a ring epimorphism but the converse is not true in general). Now I understood at least the last part of your answer completely. I'm going to work on the remaining part. $\endgroup$ – principal-ideal-domain Jul 23 '14 at 20:22
  • $\begingroup$ Dear @principal-ideal-domain, Right, surjective ring maps are epimorphisms, but the converse is false (localizations serve as an example to illustrate this). For the second part, how do you define $f_x^\sharp$ if not by the abstract characterization I've given of it? Perhaps you give an explicit description of the stalks $\mathscr{O}_{Y,y}$ and $\mathscr{O}_{X,x}$ as equivalence classes of pairs consisting of an open set and a section of $\mathscr{O}_X$ over that open set? Then you can "define" $f_x^\sharp$ via an explicit formula. If this is the way you do it, then as an exercise, verify $\endgroup$ – Keenan Kidwell Jul 23 '14 at 20:48

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