Let $A$ be a connected noetherian ring (not necessarily irreducible), $M$ be a finitely presented flat $A$-module. Then $M_{\mathfrak{p}}$ is a free $A_{\mathfrak{p}}$-module for each $\mathfrak{p} \in \operatorname{Spec}(A)$.
Is it also true that $M$ is locally free in the sense that we find a generating system $f_1,...,f_n$ of $A$ such that the $M_{f_i}$ are free? When trying to come up with a proof, I needed that $A$ is irreducible, so I wonder if there is a counterexample if $A$ is reducible?
It is a standard fact in commutative algebra that, for an arbitrary commutative ring $A$, an $A$-module $M$ is finitely presented flat if and only if $M$ is finitely genertated projective if and only if $M$ is locally free of finite rank (:= there are elements $f_1,\dotsc,f_n \in A$ generating the unit ideal such that $M_{f_i}$ is finite free over $A_{f_i}$ for each $i$). A proof can be found in books on commutative algebra, and also in the stacks project. It follows that for an arbitrary scheme $X$ the flat $\mathcal{O}_X$-modules of finite presentation coincide with the locally free $\mathcal{O}_X$-modules of finite rank.