I am sure this must have been answered somewhere but I can't find them, so I shall try my luck here.
Let $A$ be a Noetherian integral domain and $M$ a finitely generated $A$-module. Then there exists $0 \neq f\in A$ such that $M_f$ is a free $A_f$-module.
I have tried reading several sources but it boils down to the proof that we may write $M$ as a composition series, say
$$0\subseteq M_1\subseteq\cdots\subseteq M_{n-1}\subseteq M$$
where each
$$M_i/M_{i-1}\cong A/\frak p_i$$
for some prime $\frak p_i$. According to the remark in [Matsumura, page 185], it suffices to choose $a$ in the intersection of all those $\frak p_i$ which are not zero and we get the localisation of $M_i/M_{i-1}$ at $a$ a free $A_a$ module. How is this possible?
If we make things very simple say $M=Am$ with $Ann(m)=\frak p$, why would choosing $p\in\frak p$ would allow $M_p$ to be a free $A_p$ module?
And second question (a bit different, but related nonetheless) is that returning to the special case where $M=Am$, why should the set
$$U=\{{\frak{q}} \in Spec A: M_{\frak{q}}\text{ is a free }A_{\frak{q}}\text{ module}\}$$
be non empty? I am reasoning out this way that given an exact sequence
$$0\rightarrow {\frak{p}} \rightarrow A\rightarrow Am\rightarrow 0$$
if we localise at the zero ideal, we get
$$0\rightarrow {\frak{p}}_{(0)} \rightarrow K(A)\rightarrow Am\otimes_{A} K(A)\rightarrow 0$$
so by dimension counting, ${\frak{p}}_{(0)}$ should be a zero vector space, and hence $U$ should be non empty by taking the zero ideal, but here I am contradicting myself in some sense that if $\frak p$ is not zero, then its localisation is probably not zero. Can someone point out the mistake to me?
Thanks!
I don't know to what refers "How is this possible?", but $(A/\mathfrak p_i)_a\simeq A_a/\mathfrak p_iA_a$, and there are two cases: $\mathfrak p_i\ne 0$ and then $a\in \mathfrak p_i$, so $\mathfrak p_iA_a=A_a$ (which gives $(A/\mathfrak p_i)_a=0$), or $\mathfrak p_i=0$ and then we get $A_a$. (Now I hope that you know how to finish the proof. If not, then let me know.)
If $M=Am$ then $M\simeq A/\operatorname{Ann}(m)$ and as before $M_p\simeq(A/\mathfrak p)_p\simeq A_p/\mathfrak pA_p=0$. Moreover, $M_{\mathfrak q}\simeq A_{\mathfrak q}/\mathfrak pA_{\mathfrak q}$ and whether $\mathfrak p\nsubseteq\mathfrak q$ then $\mathfrak pA_{\mathfrak q}=A_{\mathfrak q}$. (For instance, if $\mathfrak p\ne 0$ you can take $\mathfrak q=0$. Otherwise, $\mathfrak p=0$ and thus $M\simeq A$.)