Lemma 10.39.8. 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

Let $A \in \text{Mat}(n \times m, R)$ be the matrix of the first map. By Nakayama's Lemma 10.19.1 we see that

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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