Can we classify those commutative rings $R$ such that $\operatorname{Supp}(M)=V(\operatorname{Ann}M)$ for every module $M$ over $R$ ? (https://en.wikipedia.org/wiki/Support_of_a_module)

[NOTE: For a subset $S$ of a commutative ring $R$ with unity, $V(S):=\{\mathfrak p : \mathfrak p$ is a prime ideal of $R$ and $S \subseteq \mathfrak p\}$ ]

(We know that if $M$ is a finitely generated module over a commutative ring with unity $R$ then $\operatorname{Supp}(M)=V(\operatorname{Ann} M)$. Is it true that over every commutative ring $R$, there is a module $M$ such that $\operatorname{Supp}(M)\ne V(\operatorname{Ann} M)$ ?)

  • $\begingroup$ If $R$ is local and the maximal ideal is its only prime, then every non-zero module satisfies your requirement, so the answer to your last question is negative. $\endgroup$
    – user26857
    Commented Jan 2, 2018 at 19:10
  • $\begingroup$ @user26857: Ah true, a zero dimensional local ring ... and anything in the converse direction ? $\endgroup$
    – user495643
    Commented Jan 2, 2018 at 19:15
  • $\begingroup$ If $R$ is local with $\dim R=1$, and $\bigcap_{k\ge 1}m^k=(0)$ (e.g., $R$ is Noetherian), then set $M=\bigoplus_{k\ge 1}R/m^k$. In this case $\operatorname{Supp}(M)\ne V(\operatorname{Ann} M)$. (Not a characterization, but this shows a different behavior for $\dim R>0$.) $\endgroup$
    – user26857
    Commented Jan 2, 2018 at 19:20
  • $\begingroup$ @user26857: I'm having difficulty determining what is Supp$(M)$ and $Ann M$ in your second example $\endgroup$
    – user495643
    Commented Jan 2, 2018 at 19:54
  • 2
    $\begingroup$ Titles should be informative and should help people searching quickly determine whether the question is relevant to what they are looking for. There's no reason to edit a post weeks later to remove information from its title. $\endgroup$ Commented Mar 3, 2018 at 7:33

1 Answer 1


This is true of a ring $R$ iff $\operatorname{Spec} R$ is a finite discrete space (or equivalently, the quotient of $R$ by its nilradical is a finite product of fields).

First, suppose $\operatorname{Supp}(M)=V(\operatorname{Ann} M)$ for every $R$-module $M$. We will first prove that every prime ideal in $R$ is maximal. Suppose $\mathfrak{p}\subset R$ is prime but not maximal. Let $S=R/\mathfrak{p}$; then $S$ is a domain but not a field. Let $a\in S$ be a nonzero element which is not a unit, and consider the $R$-module $M=\bigoplus_{n\in\mathbb{N}} S/(a^n)$. Let $I=\bigcap_n (a^n)\subset S$ and let $J$ be the inverse image of $I$ in $R$. It is clear that $I$ is the annihilator of $M$ as an $S$-module, so $J$ is the annihilator of $M$ as an $R$-module.

On the other hand, $I$ generates a proper ideal in the localization $S[a^{-1}]$. Indeed, since $a$ is not a zero divisor, the only way $I$ could fail to generate a proper ideal is if $a^n\in I$ for some $n$, which would imply $a^n\in (a^{n+1})$ so $a^n=a^{n+1}b$ for some $b\in S$. Again since $a$ is not a zero divisor, this implies $1=ab$ so $a$ is a unit, contrary to our choice of $a$.

Thus $I$ generates a proper ideal in $S[a^{-1}]$, and we can extend it to a maximal ideal. This maximal ideal pulls back to a prime ideal $\mathfrak{q}\subset R$ such that $J\subseteq\mathfrak{q}$ but $\bar{a}\not\in\mathfrak{q}$ where $\bar{a}\in R$ is an element whose image in $S$ is $a$. Since every element of $M$ is annihilated by some power of $\bar{a}$, $M_\mathfrak{q}=0$, so $\mathfrak{q}\not\in \operatorname{Supp}(M)$. But since $J\subseteq \mathfrak{q}$, $\mathfrak{q}\in V(J) =V(\operatorname{Ann M})$. This contradicts our assumption that $\operatorname{Supp}(M)=V(\operatorname{Ann} M)$ for all $R$-modules $M$.

Thus every prime in $R$ is maximal. Now if $\mathfrak{p}$ is any maximal ideal, the module $R/\mathfrak{p}$ has support $\{\mathfrak{p}\}$. Thus for any subset $A\subseteq\operatorname{Spec} R$, the module $M=\bigoplus_{\mathfrak{p}\in A}R/\mathfrak{p}$ has support $A$. Since $\operatorname{Supp}(M)=V(\operatorname{Ann M})$ and $V(I)$ is a closed set for any ideal $I$, this implies every subset of $\operatorname{Spec} R$ is closed, so $\operatorname{Spec} R$ is discrete. Finally, since $\operatorname{Spec} R$ is quasicompact, this implies it is also finite.

Conversely, suppose $R$ is a ring such that $\operatorname{Spec} R$ is a finite discrete space. Then $R$ can be identified with the finite product $\prod_{\mathfrak{p}}R_\mathfrak{p}$ of its localizations at all its prime ideals. When we make this identification, each prime ideal $\mathfrak{p}$ corresponds to the elements of $\prod_{\mathfrak{p}}R_\mathfrak{p}$ whose $\mathfrak{p}$-coordinate is in the unique maximal ideal of $R_\mathfrak{p}$. Also, any $R$-module $M$ is naturally isomorphic to the direct sum $\bigoplus_{\mathfrak{p}}M_\mathfrak{p}$ of its localizations at all the prime ideals. The annihilator of $M$ is then the product $\prod_\mathfrak{p} \operatorname{Ann}_{R_\mathfrak{p}} M_\mathfrak{p}\subseteq\prod_{\mathfrak{p}}R_\mathfrak{p}$. Thus $\operatorname{Ann} M\subseteq\mathfrak{p}$ iff $\operatorname{Ann}_{R_\mathfrak{p}} M_\mathfrak{p}$ is contained in the unique maximal ideal of $R_\mathfrak{p}$. Since any proper ideal in $R_\mathfrak{p}$ is contained in the unique maximal ideal, this means $\operatorname{Ann} M\subseteq\mathfrak{p}$ iff $M_\mathfrak{p}\neq 0$. That is, $\mathfrak{p}\in V(\operatorname{Ann} M)$ iff $\mathfrak{p}\in \operatorname{Supp}(M)$, so $\operatorname{Supp}(M)=V(\operatorname{Ann} M)$.

  • $\begingroup$ How can you identify $R$ as the product at its localizations ? What is the isomorphism .? $\endgroup$
    – user495643
    Commented Jan 3, 2018 at 14:54
  • $\begingroup$ Just take the canonical map from $R$ to the product of its localizations. That map is an isomorphism by the sheaf property of the structure sheaf on $\operatorname{Spec} R$, using the open cover consisting of all the singletons. $\endgroup$ Commented Jan 3, 2018 at 17:13
  • $\begingroup$ Okay ... unfortunately I don't know anything about sheaf ... can you point out any easy way to see that the canonical map is injective ? $\endgroup$
    – user495643
    Commented Jan 3, 2018 at 17:27
  • $\begingroup$ I don't think there's any particularly simple proof. If you assume $R$ is reduced, then the intersection of all the prime ideals is $0$, so the Chinese remainder theorem gives an isomorphism $R\cong\prod_\mathfrak{p}R/\mathfrak{p}$, and then you can show $R/\mathfrak{p}$ is actually the same as $R_\mathfrak{p}$ (this is easy by direct computation in the ring $\prod_\mathfrak{p}R/\mathfrak{p}$ which is a product of fields). But the argument is messier if $R$ is not reduced. $\endgroup$ Commented Jan 3, 2018 at 18:09
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    $\begingroup$ This answer has been mostly copied to an arxiv paper, without acknowledgement. See meta.mathoverflow.net/questions/3798/… $\endgroup$
    – YCor
    Commented Jul 8, 2018 at 13:02

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