Exercise 5.6 b) of Chapter II of Hartshorne's Algebraic Geometry asks to prove that if $A$ is a Noetherian ring and $M$ a finitely generated $A$-module then $Supp(\tilde{M})=V(Ann(M))$. Where $\tilde{M}$ is the sheaf on the scheme $Spec(A)$ associated to the module $M$.
I have "proved" the statement without using the Noetherianity condition on $A$ (or without noticing I'm using it!). So I guess I'm missing something. I would like some help in understanding where my error lies!
Here follows my "proof" of the statement:
First inclusion $V(Ann(M))\subseteq Supp(\tilde{M})$
Let $p\in V(Ann(M))$, so $Ann(M)\subseteq p$. By contradictions I suppose $M_p=0$, in specific for each $m \in M$ there exists $a\in A - p$ such that $am=0$. Let $\{m_i\}$ be a finite set of generators for $M$ and $\{a_i\} \subseteq A$ such that $a_im_i=0$ for each $i$. On one hand $\prod a_i\in A-p$ because $p$ is a prime ideal, but on the other hand $(\prod a_i)m_j=0$ for each $j$, so $\prod a_i\in \bigcap Ann(m_i) = Ann(M) \subseteq p$. Contradiction.
Second inclusion $Supp(\tilde{M}) \subseteq V(Ann(M))$
Let $p \in Supp(\tilde{M})$, then $M_p\neq 0$ and there exists $m\in M$ such that it does not exist $a \in A - p$ such that $am=0$. Then $Ann(M)\bigcap (A-p)=\emptyset$ so $Ann(M)\subseteq p$, in specific $p\in V(Ann(M))$.
So, what's wrong? Thank you for your help!