Let $R$ be a ring and $M$ an $R$-module.

If we suppose $M$ finitely generated then (following Matsumura) if we write $M=Rm_1+\cdots+Rm_n$ we have:

$p\in\operatorname{Supp}M$ if and only if $M_p\neq0$ if and only if there exists an $i$ such that $m_i\neq0$ in $M_p$ if and only if there exists an $i$ such that $\operatorname{Ann}m_i\subset p$ if and only if $\operatorname{ann}M=\bigcap_{i=1}^n\operatorname{Ann}m_i\subset p$. And so $\operatorname{Supp}M=V(\operatorname{Ann}M)$.

If $M$ is not finitely generated where does this proof fail?

It seems to me that if we write $M=\langle m_i\rangle_{i\in I}$ nothing will change.

And if this proof fails could you give me an example of a module such that $\operatorname{Supp}M\neq V(\operatorname{Ann}M)$?


The problem is, when there are infinitely many $i$, the condition $\cap_i \operatorname{Ann}(m_i)\subseteq p$ doesn't imply in general that $p$ contains an $\operatorname{Ann}(m_i)$. So the support of $M$ is always contained in $V(\operatorname{Ann} M)$, but this can be a strict inclusion.

Example: $R=\mathbb Z$, $M=\bigoplus_{n\ge 2} \mathbb Z/n\mathbb Z$. Then $\operatorname{Ann} M=0$, but $M\otimes \mathbb Q=0$. So the zero ideal belongs to $V(\operatorname{Ann} M)$, but is not in the support of $M$. The latter is in fact the set of the maximal ideals of $\mathbb Z$.

Edit. One can also consider the $\mathbb Z$-module $\bigoplus_{n\ge 1} \mathbb Z/2^n\mathbb Z$ in which case the annihilator is $0$, but the support is only one prime ideal $2\mathbb Z$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.