# The support of a non finitely generated module

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 $$2\mathbb Z$$.