## Tag `056J`

Chapter 28: Morphisms of Schemes > Section 28.5: Supports of modules

Lemma 28.5.3. Let $\mathcal{F}$ be a finite type quasi-coherent module on a scheme $X$. Then

- The support of $\mathcal{F}$ is closed.
- For $x \in X$ we have $$ x \in \text{Supp}(\mathcal{F}) \Leftrightarrow \mathcal{F}_x \not = 0 \Leftrightarrow \mathcal{F}_x \otimes_{\mathcal{O}_{X, x}} \kappa(x) \not = 0. $$
- For any morphism of schemes $f : Y \to X$ the pullback $f^*\mathcal{F}$ is of finite type as well and we have $\text{Supp}(f^*\mathcal{F}) = f^{-1}(\text{Supp}(\mathcal{F}))$.

Proof.Part (1) is a reformulation of Modules, Lemma 17.9.6. You can also combine Lemma 28.5.1, Properties, Lemma 27.16.1, and Algebra, Lemma 10.39.5 to see this. The first equivalence in (2) is the definition of support, and the second equivalence follows from Nakayama's lemma, see Algebra, Lemma 10.19.1. Let $f : Y \to X$ be a morphism of schemes. Note that $f^*\mathcal{F}$ is of finite type by Modules, Lemma 17.9.2. For the final assertion, let $y \in Y$ with image $x \in X$. Recall that $$ (f^*\mathcal{F})_y = \mathcal{F}_x \otimes_{\mathcal{O}_{X, x}} \mathcal{O}_{Y, y}, $$ see Sheaves, Lemma 6.26.4. Hence $(f^*\mathcal{F})_y \otimes \kappa(y)$ is nonzero if and only if $\mathcal{F}_x \otimes \kappa(x)$ is nonzero. By (2) this implies $x \in \text{Supp}(\mathcal{F})$ if and only if $y \in \text{Supp}(f^*\mathcal{F})$, which is the content of assertion (3). $\square$

The code snippet corresponding to this tag is a part of the file `morphisms.tex` and is located in lines 645–663 (see updates for more information).

```
\begin{lemma}
\label{lemma-support-finite-type}
Let $\mathcal{F}$ be a finite type quasi-coherent module
on a scheme $X$. Then
\begin{enumerate}
\item The support of $\mathcal{F}$ is closed.
\item For $x \in X$ we have
$$
x \in \text{Supp}(\mathcal{F})
\Leftrightarrow
\mathcal{F}_x \not = 0
\Leftrightarrow
\mathcal{F}_x \otimes_{\mathcal{O}_{X, x}} \kappa(x) \not = 0.
$$
\item For any morphism of schemes $f : Y \to X$ the pullback
$f^*\mathcal{F}$ is of finite type as well and we have
$\text{Supp}(f^*\mathcal{F}) = f^{-1}(\text{Supp}(\mathcal{F}))$.
\end{enumerate}
\end{lemma}
\begin{proof}
Part (1) is a reformulation of
Modules, Lemma \ref{modules-lemma-support-finite-type-closed}.
You can also combine
Lemma \ref{lemma-support-affine-open},
Properties, Lemma \ref{properties-lemma-finite-type-module},
and
Algebra, Lemma \ref{algebra-lemma-support-closed}
to see this. The first equivalence in (2) is the definition
of support, and the second equivalence follows from
Nakayama's lemma, see
Algebra, Lemma \ref{algebra-lemma-NAK}.
Let $f : Y \to X$ be a morphism of schemes. Note that
$f^*\mathcal{F}$ is of finite type by
Modules, Lemma \ref{modules-lemma-pullback-finite-type}.
For the final assertion, let $y \in Y$ with image $x \in X$.
Recall that
$$
(f^*\mathcal{F})_y =
\mathcal{F}_x \otimes_{\mathcal{O}_{X, x}} \mathcal{O}_{Y, y},
$$
see
Sheaves, Lemma \ref{sheaves-lemma-stalk-pullback-modules}.
Hence $(f^*\mathcal{F})_y \otimes \kappa(y)$ is nonzero
if and only if $\mathcal{F}_x \otimes \kappa(x)$ is nonzero.
By (2) this implies $x \in \text{Supp}(\mathcal{F})$ if and only
if $y \in \text{Supp}(f^*\mathcal{F})$, which is the content of
assertion (3).
\end{proof}
```

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