Lemma 10.40.8. Let $R$ be a ring and let $M$ be an $R$-module. If $M$ is a finitely presented $R$-module, then $\text{Supp}(M)$ is a closed subset of $\mathop{\mathrm{Spec}}(R)$ whose complement is quasi-compact.

Proof. Choose a presentation

$R^{\oplus m} \longrightarrow R^{\oplus n} \longrightarrow M \to 0$

Let $A \in \text{Mat}(n \times m, R)$ be the matrix of the first map. By Nakayama's Lemma 10.20.1 we see that

$M_{\mathfrak p} \not= 0 \Leftrightarrow M \otimes \kappa (\mathfrak p) \not= 0 \Leftrightarrow \text{rank}(A \bmod \mathfrak p) < n.$

Hence, if $I$ is the ideal of $R$ generated by the $n \times n$ minors of $A$, then $\text{Supp}(M) = V(I)$. Since $I$ is finitely generated, say $I = (f_1, \ldots , f_ t)$, we see that $\mathop{\mathrm{Spec}}(R) \setminus V(I)$ is a finite union of the standard opens $D(f_ i)$, hence quasi-compact. $\square$

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