The Stacks project

17.26 Rank and determinant

Let $(X, \mathcal{O}_ X)$ be a ringed space. Consider the category $\textit{Vect}(X)$ of finite locally free $\mathcal{O}_ X$-modules. This is an exact category (see Injectives, Remark 19.9.6) whose admissible epimorphisms are surjections and whose admissible monomorphisms are kernels of surjections. Moreover, there is a set of isomorphism classes of objects of $\textit{Vect}(X)$ (proof omitted). Thus we can form the zeroth Grothendieck $K$-group $K_0(\textit{Vect}(X))$. Explicitly, in this case $K_0(\textit{Vect}(X))$ is the abelian group generated by $[\mathcal{E}]$ for $\mathcal{E}$ a finite locally free $\mathcal{O}_ X$-module, subject to the relations

whenever there is a short exact sequence $0 \to \mathcal{E}' \to \mathcal{E} \to \mathcal{E}'' \to 0$ of finite locally free $\mathcal{O}_ X$-modules.

Ranks. Assume all stalks $\mathcal{O}_{X, x}$ are nonzero rings. Given a finite locally free $\mathcal{O}_ X$-module $\mathcal{E}$, the rank is a locally constant function

See Lemma 17.14.4. By definition of locally free modules the function $\text{rank}_\mathcal {E}$ is locally constant. If $0 \to \mathcal{E}' \to \mathcal{E} \to \mathcal{E}'' \to 0$ is a short exact sequence of finite locally free $\mathcal{O}_ X$-modules, then $\text{rank}_\mathcal {E} = \text{rank}_{\mathcal{E}'} + \text{rank}_{\mathcal{E}''}$, Thus the rank defines a homomorphism

Determinants. Given a finite locally free $\mathcal{O}_ X$-module $\mathcal{E}$ we obtain a disjoint union decomposition

with $X_ i$ open and closed, such that $\mathcal{E}$ is finite locally free of rank $i$ on $X_ i$ (this is exactly the same as saying the $\text{rank}_\mathcal {E}$ is locally constant). In this case we define $\det (\mathcal{E})$ as the invertible sheaf on $X$ which is equal to $\wedge ^ i(\mathcal{E}|_{X_ i})$ on $X_ i$ for all $i \geq 0$. Since the decomposition above is disjoint, there are no glueing conditions to check. By Lemma 17.26.1 below this defines a homomorphism

of abelian groups. The elements of $\mathop{\mathrm{Pic}}\nolimits (X)$ we get in this manner are locally free of rank $1$ (see below the lemma for a generalization).

Let $(X, \mathcal{O}_ X)$ be a ringed space. Instead of looking at finite locally free $\mathcal{O}_ X$-modules we can look at those $\mathcal{O}_ X$-modules $\mathcal{F}$ which are locally on $X$ a direct summand of a finite free $\mathcal{O}_ X$-module. This is the same thing as asking $\mathcal{F}$ to be a flat $\mathcal{O}_ X$-module of finite presentation, see Lemma 17.18.3. If all the stalks $\mathcal{O}_{X, x}$ are local, then such a module $\mathcal{F}$ is finite locally free, see Lemma 17.14.6. In general however this will not be the case; for example $X$ could be a point and $\Gamma (X, \mathcal{O}_ X)$ could be the product $A \times B$ of two nonzero rings and $\mathcal{F}$ could correspond to $A \times 0$. Thus for such a module the rank function is undefined. However, it turns out we can still define $\det (\mathcal{F})$ and this will be an invertible $\mathcal{O}_ X$-module in the sense of Definition 17.25.1 (not necessarily locally free of rank $1$). Our construction will agree with the one above in the case that $\mathcal{F}$ is finite locally free. We urge the reader to skip the rest of this section.