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Prove that if $M$ is a Noetherian module (over an arbitrary ring $R$), then $\operatorname{Supp}M$ is a closed Noetherian subspace of $\operatorname{Spec}R$.

This is the exercise 10 of chapter 6 in An Introduction to Commutative Algebra by Atiyah and McDonald. I have known that it is closed since $\operatorname{Supp}M=V(\operatorname{Ann}M)$. But why it is Noetherian?

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  • $\begingroup$ One should note that in general $\operatorname{Supp}M$ is not closed for an arbitrary module $M$. A sufficient condition for $\operatorname{Supp}M$ to be closed is that $M$ is a finitely generated $R$-module. $\endgroup$
    – Elías Guisado Villalgordo
    Commented Mar 25 at 10:25

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If $(m_i)_{1\le i\le n}$ is a set of generators of the noetherian module $M$, and $\mathfrak a_i=\operatorname{Ann}_{R}m_i$. Then $$\operatorname{Supp}M=\bigcup_{1\le i\le n}V(\mathfrak a_i)$$ so it is enough to prove the assertion for a cyclic noetherian module.

Now, $V(\mathfrak a_i))$ is homeomorphic to $\;\operatorname{Spec}(R/\mathfrak a_i)$, and $\;R/\mathfrak a_i\simeq Rm_i\;$ is a noetherian ring, so its spectrum is a noetherian space.

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