## 17.5 Supports of modules and sections

Definition 17.5.1. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}_ X$-modules.

The *support of $\mathcal{F}$* is the set of points $x \in X$ such that $\mathcal{F}_ x \not= 0$.

We denote $\text{Supp}(\mathcal{F})$ the support of $\mathcal{F}$.

Let $s \in \Gamma (X, \mathcal{F})$ be a global section. The *support of $s$* is the set of points $x \in X$ such that the image $s_ x \in \mathcal{F}_ x$ of $s$ is not zero.

Of course the support of a local section is then defined also since a local section is a global section of the restriction of $\mathcal{F}$.

Lemma 17.5.2. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}_ X$-modules. Let $U \subset X$ open.

The support of $s \in \mathcal{F}(U)$ is closed in $U$.

The support of $fs$ is contained in the intersections of the supports of $f \in \mathcal{O}_ X(U)$ and $s \in \mathcal{F}(U)$.

The support of $s + s'$ is contained in the union of the supports of $s, s' \in \mathcal{F}(U)$.

The support of $\mathcal{F}$ is the union of the supports of all local sections of $\mathcal{F}$.

If $\varphi : \mathcal{F} \to \mathcal{G}$ is a morphism of $\mathcal{O}_ X$-modules, then the support of $\varphi (s)$ is contained in the support of $s \in \mathcal{F}(U)$.

**Proof.**
This is true because if $s_ x = 0$, then $s$ is zero in an open neighbourhood of $x$ by definition of stalks. Similarly for $f$. Details omitted.
$\square$

In general the support of a sheaf of modules is not closed. Namely, the sheaf could be an abelian sheaf on $\mathbf{R}$ (with the usual archimedean topology) which is the direct sum of infinitely many nonzero skyscraper sheaves each supported at a single point $p_ i$ of $\mathbf{R}$. Then the support would be the set of points $p_ i$ which may not be closed.

Another example is to consider the open immersion $j : U = (0 , \infty ) \to \mathbf{R} = X$, and the abelian sheaf $j_!\underline{\mathbf{Z}}_ U$. By Sheaves, Section 6.31 the support of this sheaf is exactly $U$.

Lemma 17.5.3. Let $X$ be a topological space. The support of a sheaf of rings is closed.

**Proof.**
This is true because (according to our conventions) a ring is $0$ if and only if $1 = 0$, and hence the support of a sheaf of rings is the support of the unit section.
$\square$

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