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
  • 1
    $\begingroup$ 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. $\endgroup$ Jun 2, 2011 at 22:23

2 Answers 2


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.

  • $\begingroup$ 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 $k(t)$? $\endgroup$
    – Will Chen
    Jun 5, 2011 at 0:50
  • $\begingroup$ 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.. $\endgroup$
    – Will Chen
    Jun 5, 2011 at 1:24
  • $\begingroup$ 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) $\endgroup$
    – Will Chen
    Jun 5, 2011 at 1:32
  • $\begingroup$ 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 $\endgroup$
    – Emerton
    Jun 5, 2011 at 20:24
  • $\begingroup$ $\newcommand{\fF}{\mathcal{F}}$ 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 $\fF_x$) $\endgroup$
    – Will Chen
    Jul 28, 2011 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 $y$.

  • $\begingroup$ Wonderful: a crystal-clear, crisp and completely general answer. What a pity it took me seven years to come across this post :-) $\endgroup$ May 26, 2018 at 19:26

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