Definition 31.12.1. Let $X$ be an integral scheme with generic point $\eta $. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. The rank of $\mathcal{F}$ is $\dim _{\kappa (\eta )} \mathcal{F}_\eta $.
31.12 Ranks of modules
This section is the analogue of More on Algebra, Section 15.23 for quasi-coherent modules. Recall that if $X$ is an integral scheme with generic point $\eta $, then the local ring $\mathcal{O}_{X, \eta }$ is equal to the residue field $\kappa (\eta )$, see Morphisms, Lemma 29.50.5.
This does not conflict with the defition of the rank of a locally free module (in the case both definitions apply). It does not seem like a good idea to define the rank of a quasi-coherent module if the scheme $X$ is not irreducible or if the local ring $\mathcal{O}_{X, \eta }$ at the generic point is not a field1
Lemma 31.12.2. Let $X$ be an integral scheme. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Let $r$ be a cardinal. The following are equivalent
$\mathcal{F}$ has rank $r$,
for all $U \subset X$ nonempty affine open $\mathcal{F}(U)$ is an $\mathcal{O}_ X(U)$-module of rank $r$, and
for some $U \subset X$ nonempty affine open $\mathcal{F}(U)$ is an $\mathcal{O}_ X(U)$-module of rank $r$.
Proof. Omitted. $\square$
Lemma 31.12.3. Let $f : X \to Y$ be a dominant morphism of integral schemes. Let $\mathcal{G}$ be a quasi-coherent $\mathcal{O}_ Y$-module. Then the rank of $\mathcal{G}$ on $Y$ is equal to the rank of $f^*\mathcal{G}$ on $X$.
Proof. Use Lemma 31.12.2, Morphisms, Lemma 29.8.8, and More on Algebra, Lemma 15.23.5. $\square$
Lemma 31.12.4. Let $X$ be an integral scheme. Let $0 \to \mathcal{F} \to \mathcal{F}' \to \mathcal{F}'' \to 0$ be a short exact sequence of quasi-coherent $\mathcal{O}_ X$-modules. Then the rank of $\mathcal{F}'$ is the sum of the ranks of $\mathcal{F}$ and $\mathcal{F}''$.
Proof. Omitted. See More on Algebra, Lemma 15.23.3. $\square$
Lemma 31.12.5. Let $X$ be an integral scheme. Let $\mathcal{F}$ and $\mathcal{G}$ be quasi-coherent $\mathcal{O}_ X$-modules.
The rank of $\mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{G}$ is the product of the ranks of $\mathcal{F}$ and $\mathcal{G}$.
If $\mathcal{F}$ is of finite presentation, then the rank of $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G})$ is the product of the ranks of $\mathcal{F}$ and $\mathcal{G}$.
Proof. Omitted. Note that part (2) makes sense because $\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{F}, \mathcal{G})$ is quasi-coherent by item (10) in Schemes, Section 26.24. The algebraic version of this lemma is More on Algebra, Lemma 15.23.4. $\square$
Lemma 31.12.6. Let $X$ be an integral scheme. Let $\mathcal{F}$ be a finite type quasi-coherent $\mathcal{O}_ X$-module. Then
$\mathcal{F}$ has finite rank $r \geq 0$, and
there exists a nonempty open $U \subset X$ such that $\mathcal{F}|_ U$ is free of rank $r$.
Proof. See More on Algebra, Lemma 15.23.6 for the algebraic version. $\square$
[1] If $X$ is irreducible and $\mathcal{F}_\eta $ is free over $\mathcal{O}_{X, \eta }$ it would appear natural to use the rank of this free module. Beware that some references define rank using the length of $\mathcal{F}_\eta $.
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