Lemma 10.40.5 (00L2)—The Stacks project

Lemma 10.40.5. 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 $M_{\mathfrak p} \not= 0$. Choose an element $m \in M$ whose image in $M_\mathfrak p$ is nonzero. Then the annihilator of $m$ is contained in $\mathfrak p$ by construction of the localization $M_\mathfrak p$. Hence a fortiori $I = \text{Ann}(M)$ must be contained in $\mathfrak p$.

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. $\square$

Comments (3)

Comment #5511 by Manuel Hoff on September 18, 2020 at 06:11

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.

Comment #5707 by Johan on November 15, 2020 at 22:23

Thanks very much! Your argument is given in this commit.

Comment #9929 by Tony on January 10, 2025 at 08:54

I don't see how Lemma 10.9.9 can be used in the second part of the proof. However, after giving up on it I found that there is an elementary alternative: Since $x_i/1 = 0/1 \in M_\mathfrak{p}$ there must exist an $f_i\notin\mathfrak{p}$ such that $f_ix_i = 0$ . Now choose $f = f_1\cdots f_r$ , and note that on one hand $f\notin\mathfrak{p}$ since $\mathfrak{p}$ is prime, and on the other hand $f\in \mathrm{Ann}(M)$ . Thus $\mathrm{Ann}(M)\not\subseteq\mathfrak{p}$ , which is equivalent to the desired result $\mathfrak{p}\notin V(\mathrm{Ann}(M))$ .

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