How to evaluate a section of a sheaf at a point?

Question

This is a really simple question, and I'm a bit embarrassed to even have to ask it.

If $\mathcal{F}$ is a sheaf on a scheme $X$ and $s \in \mathcal{F}(U)$ is a local section, what does it mean to evaluate $s$ at a point $x \in X$?

On the one hand, I would expect the answer to be "the image of $s$ in the stalk $\mathcal{F}_x$ at $x$". For concreteness say $X = \text{Spec}(A)$ is affine and $\mathcal{F} = \widetilde{M}$ came from a module on $A$. Am I right in thinking that for $m \in M = \Gamma(X,\widetilde{M})$ a global section, the "evaluation" $m(\mathfrak{p})$ is just the image of $m$ under the map $M \to M_\mathfrak{p}$? In particular, does this section take values in the localization $M_\mathfrak{p}$?

On the other hand, in the special case $\mathcal{F} = \mathcal{O}_X$, then $s \in A$ is a regular function and I've heard people say that $s(\mathfrak{p})$ actually doesn't take values in $A_\mathfrak{p}$... Instead it takes values in the residue field $\kappa(\mathfrak{p}) = A_\mathfrak{p} / \mathfrak{p}$. Is this also correct?

How do we square these two interpretations, which seem to suggest some ambiguity in the target of a local section? Is it because $\mathcal{O}_X$ is a sheaf of rings rather than a sheaf of modules? That seems unlikely, in light of this recent answer (which inspired this question) which says that sections of this sheaf of modules (this line bundle) actually takes values in the residue field...

Thanks!

Answer 1

Without being too formal, my understanding has always been that the stalk $s_x$ is something akin to a Taylor expansion of $f$ at $x$.

In order to get a proper Taylor expansion, at least when $x$ is a smooth point, you need to move to the formal completion in the $\mathfrak{p}_x$-adic topology of the local ring at $x$ which is non-canonically isomorphic to a power series ring.

Whereas the value of $s$ at $x$ is what you get when you tensor with the residue ring $\kappa(\mathfrak{p})$, as explained in hm2020's answer.

This is actually a straightforward extension of a more elementary notion. Forget schemes for a minute and think of complex-valued function on a complex variety $V$. If $f$ is such a function defined at $P\in V$ what you expect for the evaluation of $f$ at $P$ is just the value of $f$ at $P$, namely $$ \mathrm{ev}_P(f)=f(P)\in\mathbb{C}. $$ Now note that $\mathbb{C}=\mathcal{O}_P/\mathfrak{m}_P$ is the residue field at $P$.

Answer 2

Q: "If $\mathcal{F}$ is a sheaf on a scheme $X$ and $s \in \mathcal{F}(U)$ is a local section, what does it mean to evaluate $s$ at a point $x \in X$? .. On the other hand, in the special case $\mathcal{F} = \mathcal{O}_X$, then $s \in A$ is a regular function and I've heard people say that $s(\mathfrak{p})$ actually doesn't take values in $A_\mathfrak{p}$... Instead it takes values in the residue field $\kappa(\mathfrak{p}) = A_\mathfrak{p} / \mathfrak{p}$. Is this also correct?"

A: If $S:=Spec(A)$ is an affine scheme and $\mathcal{E}$ a sheaf of modules (corresponding to the $A$-module $E$) with $x \in S$ a point, let $s\in \Gamma(S,\mathcal{E})\cong E$ be a global section. Let $\mathfrak{p}$ be the prime ideal associated to the point $x$. Usually (there are many notations) when we write $s(x)$ we mean the image of the stalk $s_{\mathfrak{p}} \in E_{\mathfrak{p}}$ in the fiber $E\otimes_{A_{\mathfrak{p}}} \kappa(\mathfrak{p})$ where $\kappa(\mathfrak{p})$ is the residue field at $x$. There is a canonical map

$$ev_x: E \rightarrow E_{\mathfrak{p}} \otimes_{A_{\mathfrak{p}}} \kappa(\mathfrak{p})$$

and by definition $s(x):=ev_X(s)$. The element $s_{\mathfrak{p}} \in E_{\mathfrak{p}}$ is the germ of $s$ in the stalk of $E$ at $x$.

The thread you refer to

Definition of base-point-free sheaf

is imprecise - it says the section $s$ takes values in $\kappa(x)$ and this is not correct notation. One should write $\mathcal{E}(x)$ for the fiber (and not $\kappa(x)$ when the sheaf is invertible).

An explicit example: Let $C:=\mathbb{P}^1_k$ with $k$ a field of characteristic zero and let $L(d):=\mathcal{O}(d)$ with $d\in \mathbb{Z}$. Let $U:=D(x_0) \subseteq C$. It follows

$$L(d)(U) \cong k[t]x_0^d$$

is the free rank one $k[t]$-module on the basis element $x_0^d$ where $t:=x_1/x_0$ is an independent variable over $k$. If $x:=(t)$ is a point in $U$ it follows the stalk at $x$ is the $k[t]_{(t)}$-module

$$L(d)_x \cong k[t]_{(t)}x_0^d,$$

and the fiber $L(d)(x)$ is the one dimensional $k$-vector space

$$L(d)(x) \cong kx_0^d.$$

We may take powers of the maximal ideal $(t) \subseteq k[t]_{(t)}$

and we get

$$L(d)_x/m_x^{l+1}L(d)_x \cong k[t]_{(t)}/(t^{l+1})x_0^d.$$

In the module $L(d)_x/m_x^{l+1}L(d)_x$ you may Taylor expand a section. Hence you may view the canonical map

$$T^l: L(d)(U) \rightarrow L(d)_x/m_x^{l+1}L(d)_x$$

as the "Taylor expansion of order $l$" of a section $s \in L(d)(U)$.

The reason you should not forget about the element $x_0^d$, is that this basis element gives the transition function $g$ defining the invertible sheaf $L(d)$: On the open set $D(x_0)\cap D(x_1)$ we have the equality

$$ x_1^d=(x_1/x_0)^dx_0^d=t^dx_0^d$$

and this gives $L(d)$ the transition function $g(t):=t^d$.