Definition of the Fibre of a Sheaf

Question

I'm working through an exercise and have a few questions about the following construction. If anyone thinks they should be split into separate posts please let me know, but they seem related and I'm not sure that they each warrant their own question.

Let $X$ be an affine variety over an algebraically closed field $k$, $A=\Gamma(X,\mathcal{O}_X)$, $M$ an $A$-module of finite type, $\mathcal{F}=\widetilde{M}$ the associated sheaf on $X$, and $x\in X$.

Let $\mathfrak{m}_x=\{f\in A\mid f(x)=0\}$, and set $k(x)=A/\mathfrak{m}_x\cong k$. Then we define $\mathcal{F}(x)=M\otimes_Ak(x)$, which is a $k$-vector space.

My questions about this are:

1. Is the action of $k$ on $\mathcal{F}(x)$ given by factoring the action of $k(x)$ sending $r\cdot\sum(m_i\otimes_Aa)\mapsto\sum(m_i\otimes_A(r\cdot a))$ through the isomorphism $k(x)\cong k$?

2. Is there a canonical name for $\mathcal{F}(x)$? My textbook refers to this as the "sheaf fibre", but this question suggests that this refers instead to $\mathcal{F}_x=M\otimes_AA_{\mathfrak{m}_x}$.

3. Is there an intuitive picture behind $\mathcal{F}(x)$? I can imagine $\mathcal{F}_x$ as an analogue of the stalk of $x$ on $\mathcal{O}_X$ since $A_{\mathfrak{m}_x}\cong\mathcal{O}_{X,x}$, but I can't seem to grasp $\mathcal{F}(x)$ in the same way. I know that $\mathcal{F}(x)=\mathcal{F}_x/\mathfrak{m}_x\mathcal{F}_x$, but I'm not sure how I should interpret this.

Any help would be much appreciated.

Note: As pointed out by jgon, I have misread the linked question, and fibre agrees with the term used there.

Answer 1

Nicolas Hemelsoet has already written an answer, which is correct, but this answer will contain a different perspective and address the OPs questions in a different manner.

1. Yes. What you've written correctly describes the action of $k$. $k(x)\simeq k$, so since $\newcommand\calF{\mathcal{F}}\calF(x)=M\otimes_A k(x)$ is always a $k(x)$ module, it is also naturally a $k$-module via the isomorphism.

2. You've misread the question you've linked. In the question we see

   The fiber of the sheaf $\calF$ is $\calF(x):=\calF_x\otimes_{ \newcommand\calO{\mathcal{O}}\calO_x} k(x)$ where $k(x)$ is the residue field for some point $x\in X$.

Thus the question also says that the fiber is $\calF(x)$ rather than $\calF_x$. Thus fiber is correct. (Or fibre if you're British.)

3. The distinction between the stalk and the fiber of a sheaf is directly analagous to the distinction between the germ of a function at a point and the value of a function at a point. If we think of sections of $\calF$ as "functions" on $X$ (let's not worry too much about where these functions are valued, this is an analogy), $\calF_x$ consists of functions together with their local behavior near $x$ just as $\calO_{X,x}$ consists of germs of functions. Then $\calF(x)$ is the result of evaluating all of the germs at $x$.

In general, when we quotient by a maximal ideal, we're evaluating a "function" at a point, and this is still true here. $\calF(x) = \calF_x/\mathfrak{m}_x\calF_x$ is the collection of all possible values of sections of $\calF$ at $x$.

Answer 2

1 : I'm not sure I understand your notation, but since $\mathcal F$ is a $\mathcal O_X-$ module there is a map $R \times M \to M$ and tensoring everything by $k(x)$ gives the multiplication map $R \times M(x) \to M(x)$ (it factors through $k(x) \times M(x) \to M(x)$.

2 : "Fiber" is fine I think.

3 : The intuition is as follows : imagine for simplicity that $\mathcal F$ is locally free, i.e it's the sheaf of sections of a vector bundle $E$ (algebraically it means that $M$ is projective as a $A$-module). Then, the stalk at $x$ is the set of sections in a "infinitesimally small" neighbourhood of $x$. For example if $\mathcal F = \mathcal O_X$ (i.e $M = R$) then the stalk is the localization at the maximal ideal $\mathfrak m_x$.

There is a similar interpretation for the fiber of a sheaf, which is the set of section $\{x\} \to E$ ! So this is much smaller. The two examples are related by the canonical evaluation map $\mathcal F_x \to \mathcal F(x)$. For example, when $\mathcal F = \mathcal O_X$ the fiber is just the residual field $k(x) = R/\mathfrak m_x$. In fact if $M$ is locally free of rank $r$ then $\mathcal F(x) \cong k(x)^r$. I hope this helps.