Fibre of a module sheaf at a point

Let $X$ be a locally ringed space, let $\mathcal{F}$ be an ${\mathcal{O}}_X$-module in the sense of Modules on a ringed space, and let $x\in X$. Write ${\mathcal{F}}_x$ for the stalk of $\mathcal{F}$ at $x$ (The stalk of a presheaf at a point); it is a module over the local ring ${\mathcal{O}}_{X,x}$, and the residue field is $\kappa (x)={\mathcal{O}}_{X,x}\mathrm{/}{\mathfrak{m}}_x$ with maximal ideal ${\mathfrak{m}}_x\subseteq {\mathcal{O}}_{X,x}$ (The residue field at a point of an affine scheme). The fibre of $\mathcal{F}$ at $x$ is the tensor product of modules over the ring ${\mathcal{O}}_{X,x}$ (The tensor product $M{\otimes }_{R}N$ from the additive group underlying the free $\mathbb{Z}$-module on $M\times N$, elementary tensors, and finite tensor sums) $\mathcal{F}(x):={\mathcal{F}}_x{\otimes }_{{\mathcal{O}}_{X,x}}\kappa (x)\mathrm{.}$ The tensor product carries the $\kappa (x)$-module structure induced by the scalar action on the second factor, so $\mathcal{F}(x)$ is a vector space over the residue field $\kappa (x)$.

Right exactness of the tensor product (Tensoring is right exact) applied to the exact sequence ${\mathfrak{m}}_x\hookrightarrow {\mathcal{O}}_{X,x}\twoheadrightarrow \kappa (x)$ identifies this vector space with the quotient of the stalk by the submodule ${\mathfrak{m}}_x{\mathcal{F}}_x$: $\mathcal{F}(x)\cong {\mathcal{F}}_x\mathrm{/}{\mathfrak{m}}_x{\mathcal{F}}_x,m\otimes \bar{\lambda }⟼\lambda m<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>{\mathfrak{m}}_x{\mathcal{F}}_x\mathrm{.}$

The fibre and the stalk are different objects: ${\mathcal{F}}_x$ is a module over the local ring ${\mathcal{O}}_{X,x}$ and need not be a vector space, while $\mathcal{F}(x)$ is a vector space over $\kappa (x)$ equipped with a canonical surjection ${\mathcal{F}}_x\twoheadrightarrow \mathcal{F}(x)$. For the zero module $\mathcal{F}=0$ one has ${\mathcal{F}}_x=0$ and $\mathcal{F}(x)=0$ at every point; for $\mathcal{F}={\mathcal{O}}_X$ one has ${\mathcal{O}}_X(x)=\kappa (x)$ at every point. The construction is functorial: a morphism $\mathcal{F}\to \mathcal{G}$ of ${\mathcal{O}}_X$-modules induces a $\kappa (x)$-linear map $\mathcal{F}(x)\to \mathcal{G}(x)$ for every $x\in X$, and restriction to an open $U\subseteq X$ gives $(\mathcal{F}{\mathrm{\mid }}_U)(x)=\mathcal{F}(x)$ for $x\in U$.