Proposition 2.3 Hartshorne

Question

The following is the proof of proposition 2.3 from Hartshorne. I have two questions:

1) Why is $\varphi_\mathfrak{p}$ a local homomorphism?

2) Why is $f^\#$ on a stalk $\varphi_\mathfrak{p}$?

Answer 1

Let me write $\mathfrak{P}$ for the prime ideal $\varphi^{-1}(\mathfrak{p})$ of $A$. If $\alpha:A\to A_\mathfrak{P}$ and $\beta:B\to B_\mathfrak{p}$ are the localization homomorphisms, then the map $\varphi_\mathfrak{p}:A_\mathfrak{P}\to B_\mathfrak{p}$ is the unique homomorphism such that $\varphi_\mathfrak{p}\circ\alpha=\beta\circ\varphi$ (its existence and uniqueness comes by applying the universal property of the localization $\alpha:A\to A_\mathfrak{P}$ to the map $\varphi\circ\beta:A\to B_\mathfrak{p}$). Explicitly, using the usual description of elements of localizations as fractions subject to an equivalence relation, we have $\varphi_\mathfrak{p}(a/s)=\varphi(a)/\varphi(s)$, where $a\in A$ and $s\in A\setminus\mathfrak{P}$. If $a/s\in\mathfrak{P}A_\mathfrak{P}$ (the maximal ideal of the local ring $A_\mathfrak{P}$), then $a\in\mathfrak{P}=\varphi^{-1}(\mathfrak{p})$, so $\varphi(a)\in\varphi(\varphi^{-1}(\mathfrak{p}))\subseteq\mathfrak{p}$. Thus $\varphi_\mathfrak{p}(a/s)=\varphi(a)/\varphi(s)$ is in $\mathfrak{p}B_\mathfrak{p}$ (the maximal ideal of the local ring $B_\mathfrak{p}$). This shows that $\varphi_\mathfrak{p}$ is local.

Let $X=\mathrm{Spec}(B)$ and $Y=\mathrm{Spec}(A)$. The homomorphism $\varphi$ induces a map of sets $f:X\to Y$ given by $f(\mathfrak{p})=\varphi^{-1}(\mathfrak{p})$, and the definition of the Zariski topology shows that this is continuous. To get a morphism of locally ringed spaces, we need a map of sheaves $f^\#:\mathscr{O}_Y\to f_*\mathscr{O}_X$ such that, for each $x\in X$, the induced map of stalks $f^\#_x:\mathscr{O}_{Y,f(x)}\to\mathscr{O}_{X,x}$ is local.

Hartshorne's definition of the structure sheaf $\mathscr{O}_Y$ of $Y=\mathrm{Spec}(A)$ is not the usual one, though it is more or less equivalent to it. In the usual approach, one introduces the notion of a sheaf on a basis of open sets and proves that such a sheaf extends uniquely to a sheaf on the space, defines a presheaf on the basis of standard open sets in $Y$ by $\mathscr{O}_Y(D(f))=A_f$ for $f\in A$, verifies that this is in fact a sheaf (this is where the real work is for this approach), and then extends to a sheaf on $Y$. Hartshorne's approach gives an explicit (but not terribly useful) definition of $\mathscr{O}_Y(V)$ for each open set $V\subseteq Y$ which clearly gives a sheaf $\mathscr{O}_Y$, and then proves (and this is where the real work is in Hartshorne's approach) that there is a canonical isomorphism $\mathscr{O}_Y(D(f))\simeq A_f$ for each $f\in A$.

I am going to write $\mathscr{O}_Y(D(f))=A_f$ and $\mathscr{O}_X(D(g))=B_g$ for $f\in A$ and $g\in B$, and you can either think of these as actual equalities (taking the first approach) or as canonical isomorphisms (taking Hartshorne's approach).

To show that $(Y,\mathscr{O}_Y)$ is a locally ringed space, one proves that there is a canonical isomorphism $\varinjlim_{\mathfrak{P}\in D(f)}\mathscr{O}_Y(D(f))\to A_\mathfrak{P}$ for every prime ideal $\mathfrak{P}$ of $A$. This amounts to a canonical isomorphism $\mathscr{O}_{Y,y}\simeq A_\mathfrak{P}$, and allows us to identify the map $\mathscr{O}_Y(D(f))\to\mathscr{O}_{Y,y}$ with the map $A_f\to A_\mathfrak{P}$ for any $f\in A$ with $f\notin\mathfrak{P}$ (here the point $y\in Y$ is really the prime ideal $\mathfrak{P}$—I just prefer not to type expressions like $\mathscr{O}_{Y,\mathfrak{P}}$). This step is pretty much the same in either approach.

So now we have to define $f^\#:\mathscr{O}_Y\to f_*\mathscr{O}_X$, and I can no longer use $f$ to denote an element of $A$ (I don't think my use of $f$ in this way in the preceding paragraphs will cause any confusion). While this can be done in an obvious way using Hartshorne's definition of structure sheaves (where the sections are actual functions on open subsets of the space), this description is not useful for identifying $f_x^\#$ with $\varphi_\mathfrak{p}$ (when $x\in X=\mathrm{Spec}(B)$ is the prime ideal $\mathfrak{p}$). Instead I am only going to describe the effect of $f^\#$ on sections over a standard open set $D(a)\subseteq Y=\mathrm{Spec}(A)$, where $a\in A$. That is, I want to give a ring map $f_{D(a)}^\#:\mathscr{O}_Y(D(a))\to(f_*\mathscr{O}_X)(D(a))=\mathscr{O}_X(f^{-1}(D(a)))$. You can verify using the definition of the set map $f:X\to Y$ that $f^{-1}(D(a))=D(\varphi(a))$. Thus we need a ring map $f_{D(a)}^\#:A_a\to B_{\varphi(a)}$. The natural candidate is the map $\varphi_a:A_a\to B_{\varphi(a)}$, uniquely determined by the universal property of the localization $A\to A_a$ applied to the map $A\to B\to B_{\varphi(a)}$. In the usual approach, one then uses the canonicity of these maps to ensure that we have a map of sheaves $\mathscr{O}_Y\to f_*\mathscr{O}_X$ on the basis of standard open sets in $Y$, and then one knows that this extends uniquely to a map of sheaves on $Y$.

Finally, if one carefully writes out the definitions of stalks as colimits over open sets, one arrives at the following useful characterization: for $x\in X$, the map $f_x^\#:\mathscr{O}_{Y,f(x)}\to\mathscr{O}_{X,x}$ is the unique homomorphism for which the composition

$$\mathscr{O}_Y(V)\to\mathscr{O}_{Y,f(x)}\xrightarrow{f_x^\#}\mathscr{O}_{X,x}$$

is equal to the composition

$$\mathscr{O}_Y(V)\to\mathscr{O}_X(f^{-1}(V))\to\mathscr{O}_{Y,f(x)}$$

for every open subset $V$ of $Y$ containing $f(x)$ (I don't know how to make commutative diagrams on MSE). Of course, it is enough to consider $V$ from some basis of open sets for $Y$, and we will use the basis of standard open sets in $Y$. Combining this characterization with the identifications made above, we see that, for $x\in X$ corresponding to the prime ideal $\mathfrak{p}$ of $B$, the map $f_x^\#:A_\mathfrak{P}\to B_\mathfrak{p}$ is the unique homomorphism such that, for every $a\in A$ with $a\notin\mathfrak{P}$, the composition

$$A_a\to A_\mathfrak{P}\xrightarrow{f_x^\#}B_\mathfrak{p}$$

equals the composition

$$A_a\to B_{\varphi(a)}\to B_\mathfrak{p}\text{.}$$

The map $\varphi_\mathfrak{p}:A_\mathfrak{P}\to B_\mathfrak{p}$ also has this property, so $f_x^\#=\varphi_\mathfrak{p}$.

It is probable one could shorten the reasoning some what by working explicitly with elements of the localizations and colimits involved here, but I personally understand things better when phrased in terms of universal properties, so I have emphasized these.

I should add that this answer has some overlap with my answer here: on the adjointness of the global section functor and the Spec functor. The latter answer is more general than this one.