Lemma 10.39.5. Let $R$ be a ring and let $M$ be an $R$-module. If $M$ is finite, then $\text{Supp}(M)$ is closed. More precisely, if $I = \text{Ann}(M)$ is the annihilator of $M$, then $V(I) = \text{Supp}(M)$.

Proof. We will show that $V(I) = \text{Supp}(M)$.

Suppose $\mathfrak p \in \text{Supp}(M)$. Then $M_{\mathfrak p} \not= 0$. Hence by Nakayama's Lemma 10.19.1 we have $M \otimes _ R \kappa (\mathfrak p) \not= 0$. Hence $I \subset \mathfrak p$.

Conversely, suppose that $\mathfrak p \not\in \text{Supp}(M)$. Then $M_{\mathfrak p} = 0$. Let $x_1, \ldots , x_ r \in M$ be generators. By Lemma 10.9.9 there exists an $f \in R$, $f\not\in \mathfrak p$ such that $x_ i/1 = 0$ in $M_ f$. Hence $f^{n_ i} x_ i = 0$ for some $n_ i \geq 1$. Hence $f^ nM = 0$ for $n = \max \{ n_ i\}$ as desired. $\square$

Comment #5511 by Manuel Hoff on

I don't see how Nakayama's Lemma helps in the first part of the proof. As I understand it, the argument is as follows:

If $m \in M$ with nonzero image in $M_{\mathfrak{p}}$, then it follows directly from the definitions that $\text{Ann}(m) \subseteq \mathfrak{p}$. But now we certainly have $\text{Ann}(M) \subseteq \text{Ann}(m)$.

In particular this direction does not need any finiteness condition.

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