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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 $\operatorname{Supp}M\neq V(\operatorname{Ann}M)$?

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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 $2\mathbb Z$.

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