Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

So, in chapter 2, section 2 of Hartshorne, (prop 2.3), he describes how if $\varphi : A\rightarrow B$ is a homomorphism of rings, then you get a morphism of (affine schemes):

$\newcommand{\Spec}{\textrm{Spec }}$ $\newcommand{\oO}{\mathcal{O}}$ $\newcommand{\mf}[1]{\mathfrak{#1}}$ $(f,f^\sharp) : (\Spec B, \oO_{\Spec B})\rightarrow (\Spec A, \oO_{\Spec A})$

where $f:\Spec B\rightarrow\Spec A$ is given by $f(\mf{p}) = \varphi^{-1}(\mf{p})$. Here, for each $\mf{p}\in\Spec B$, $\varphi$ also gives us a map of localizations $\varphi_\mf{p} : A_{\varphi^{-1}(\mf{p})}\rightarrow B_\mf{p}$.

Also, $f^\sharp : \oO_{\Spec A}\rightarrow f_*\oO_{\Spec B}$ is given by sending $\sigma\in\oO_{\Spec A}(V)$ (for each open $V$), to the function $[\mf{p}\mapsto \varphi_\mf{p}(\sigma(f(\mf{p})))] \in \oO_{\Spec B}(f^{-1}(V)) = f_*\oO_{\Spec B}(V)$

So, my question is... What are the stalks of the direct image sheaf $f_*\oO_{\Spec B}$?

This is clearly a sheaf on $\Spec A$, so there should be a stalk for each $\mf{p}\in\Spec A$. There are two cases.

Firstly, suppose $\mf{p}\in\Spec A$ is in the image of $f$, then by definition, the stalk of $(f_*\oO_{\Spec B})_{\mf{p}}$ is the direct limit:

$\lim_{U\supset f^{-1}(\mf{p})}\oO_{\Spec B}(U)$

But this is not quite a stalk of $\oO_{\Spec B}$ (since $f$ may not be injective). However, Hartshorne seems to suggest that this is actually just $(\oO_{\Spec B})_{\mf{q}}$ which is just the localization of $B$ at $\mf{q}$, where $\mf{q}$ is any point of $f^{-1}(\mf{p})$. I don't really see why this must be true. (Especially since he seems to suggest that all the localizations at $\mf{q}$ are the same, for any $\mf{q}\in f^{-1}(\mf{p})$.

Secondly, suppose $\mf{p}\in\Spec A$ is not in the image of $f$. Then what? I can imagine that if there is some neighborhood $V$ of $\mf{p}$ such that $f^{-1}(V)$ is empty, then the stalk would be zero. But suppose there is no such $V$? Then What? (Alternatively, must there always exist such a $V$ in this case?)

Thanks guys

  • will
share|improve this question
1  
Maybe the reason you think Hartshorne suggests this is the way he states the "local homomorphism" condition in the definition of a local-ringed space? The group homomorphism he is talking about is not just the localization at P of the f# map. –  Sean Rostami Jun 2 '11 at 22:23

2 Answers 2

up vote 8 down vote accepted

I'm not sure what it is that you read in Hartshorne that suggested that $(f_*\mathcal O_{\mathrm{Spec} B})_{\mathfrak p}$ is equal to $(\mathcal O_{\mathrm{Spec} B})_{\mathfrak q}$, since this is not true.

My suggestion is that you consider two illustrative cases:

  1. Let $A = k$ (a field) and $B = k\times k$, with $A \to B$ the diagonal morphism. In this case Spec $A$ is a single point, and so there is only stalk to consider.

  2. Let $A = k[t]$ (again, $k$ is a field) and $B = k[t,t^{-1}]$, with $A \to B$ being the inclusion. In this case, the map Spec $B \to $ Spec $A$ coicides with the identity at all point of Spec $A$ other than the point $t = 0$, so the interesting case is the stalk of the pushforward at $t = 0$ (this is a case with empty fibre).

In each case you can compute the stalk you asked about directly from the definition, and I recommend that you try to do so.

Added: If $f: X \to Y$ and $\mathcal F$ is a sheaf on $X$, then for any $x \in X$ there is a canonical map of stalks $(f_*\mathcal F)_{f(x)} \to \mathcal F_x,$ given as follows: if $V$ is a n.h. of $f(x)$, then $f^{-1}(V)$ is a n.h. of $x$, and by definition $f_*\mathcal F(V) = \mathcal F(f^{-1}(V)).$ If $V$ runs over all n.h.s of $f(x)$, then $f^{-1}(V)$ will range over some (but typically not all) n.h.s of $x$, and so there will be an induced map $(f_*\mathcal F)_{f(x)} \to \mathcal F_x$, but this will typically not be an isomorphism (exactly because $f^{-1}(V)$ typically doesn't range over all n.h.s of $x$, but just certain ones). In the case of a morphism $f:X \to Y$ of ringed spaces, the given map $\mathcal O_Y \to f_*\mathcal O_X$ then induces maps of stalks $(\mathcal O_Y)_{f(x)} \to (f_*\mathcal O_X)_{f(x)}$ (by functoriality of the construction of stalks) and $(f_*\mathcal O_X)_{f(x)} \to (\mathcal O_X)_x$ (via the above construction). Their composite is the morphism $(\mathcal O_Y)_{f(x)} \to (\mathcal O_X)_x$ that Hartshorne uses when he makes the definition of a morphism of locally ringed spaces.

share|improve this answer
    
Alright, so in (1), I guess the direct limit in the definition of the stalk is just the direct limit of a single group, which is just $O_B(B) = k\times k$. And in (2), since the primes of B are just the primes of A except for $(t)$, the direct limit just limits over all nonempty open sets of Spec B, so if we only consider the open sets $D(f)$ for $f\in B$, for which $O_B(D(f)) = B_f$, so it seems like the direct limit (ie the stalk) is just $k(t)$? –  Will Chen Jun 5 '11 at 0:50
    
Also, I thought that $(f_*\oO_{\Spec B})_\mf{p} = (\oO_{\Spec B})_\mf{q}$ because of the way he defined the morphism of sheaves $f^\sharp$ and the local homomorphisms $\varphi_\mf{p} : A_{\varphi^{-1}}(\mf{p})\rightarrow B_\mf{p}$. Ie, he said "The induced maps $f^\sharp$ on the stalks are just the local homomorphisms $\varphi_\mf{p}$", but these local homomorphisms are only defined for $\mf{p}\in\Spec B$, whereas they should be defined for all $\mf{q}\in\Spec A$, so it seemed like he was saying that as $\mf{p}$ ranges over $\Spec B$, $f(\mf{p})$ ranges over all of $\Spec A$, which is false.. –  Will Chen Jun 5 '11 at 1:24
    
so I guess these stalks in general are just the direct limit of localizations, where the maps are just inclusions. (at least for integral domains) –  Will Chen Jun 5 '11 at 1:32
    
Dear oxeimon, Yes, your computations of the stalks in your first comment are correct. As for "the induced map on stalks" remark of Hartshorne, the point is that if $\mathfrak p \mapsto \mathfrak q,$ and if $V$ is a n.h. of $\mathfrak p$, then $f^{-1}(V)$ will be a n.h. of $\mathfrak q$, so there is an induced map on stalks $\mathcal O_{\mathrm{Spec} B,\mathfrak p} \to \mathcal O_{\mathrm{Spec} A,\mathfrak q},$ which is the map Hartshorne is discussing. (See my updated answer for slightly more detail, although it may be superfluous at this point.) Regards, Matthew –  Emerton Jun 5 '11 at 20:24
    
$\newcommand{\fF}{\mathcal{F}}$ Sorry to come back to this, but after rereading Hartshorne's definition of a morphism of locally ringed spaces, that even though as $V$ ranges over all open nbhd's of $f(P)$, $f^{−1}(V)$ ranges over a subset of the nbhd's of $P$, he still claims that $\lim_V \oO_X(f^{−1}(V)) = \oO_{X,P}$. (In your addendum to your original response, you said in general this limit, which you wrote as $(f_∗\fF)_{f(x)}$ is not ismorphic to $\fF_x$) –  Will Chen Jul 28 '11 at 5:28

You consider $B$ as an $A$-module. Then $f_*O_X$ ($X = Spec B$) is the quasi-coherent $O_Y$-module ($Y = Spec A$) corresponding to the $A$-module $B$. Thus for a prime ideal $y \in Y$, $(f_*O_X)_y$ is the localization of the $A$-module $B$ in $y$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.