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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:

$p\in\mathrm{Supp}\;M$ 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 $\mathrm{Ann}\;m_i\subset p$ if and only if $\mathrm{ann}\;M=\bigcap_{i=1}^n\;\operatorname{Ann}m_i\subset p$. And so $\mathrm{Supp}\;M=V(\operatorname{Ann}M)$.

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

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1 Answer 1

up vote 2 down vote accepted

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

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

Edit In the above example, take the direct sum on $n\ne 0$. One can also consider the $\mathbb Z$-module $\oplus_{n\ge 0} \mathbb Z/2^n\mathbb Z$ in which case the annihilator is $0$, but the support is only one prime ideal $2\mathbb Z$.

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