Local property of split exact sequence In the module category of a ring $A$, is a short exact sequence split if and only if the localization of this sequence is split for every prime ideal?
Thanks!
 A: Without extra finiteness assumptions, this is not true in general.
Even for $A=\mathbb{Z}$, there are infinitely generated $A$-modules $M$ that are locally free (in the sense that $M_\mathfrak{p}$ is a free $A_\mathfrak{p}$-module for every prime ideal $\mathfrak{p}$) but not projective. Then if $0\to N\to P\to M\to0$ is a (necessarily non-split) short exact sequence with $P$ projective, then for every prime ideal $\mathfrak{p}$, $0\to N_\mathfrak{p}\to P_\mathfrak{p}\to M_\mathfrak{p}\to0$ is a split short exact sequence, since $M_\mathfrak{p}$ is projective.
For non-Noetherian rings, there are counterexamples with $M$ finitely generated, since there can be finitely generated flat $A$-modules $M$ that are not projective. Letting $0\to N\to P\to M\to0$ be a (necessarily non-split) short exact sequence where $P$ is projective, then for any prime ideal $\mathfrak{p}$, $M_\mathfrak{p}$ is projective, since every finitely generated flat module for a local ring is projective, and so $0\to N_\mathfrak{p}\to P_\mathfrak{p}\to M_\mathfrak{p}\to0$ is a split short exact sequence.
However, as alluded to in comments, one finiteness condition that gives a positive answer is where the short exact sequence $0\to N\to X\to M\to0$ has $M$ finitely presented.
I'll expand on the proof sketched in the comments.
If $A\subseteq B$ is a flat ring extension (e.g., $B=A_\mathfrak{p}$), then for a finitely generated projective $A$-module $P$, the natural map
$$\text{Hom}_A(P,N)\otimes_AB\to\text{Hom}_B(P\otimes_AB,N\otimes_AB)$$
is an isomorphism.
Take a projective $A$-module resolution
$$\dots\to P_2\to P_1\to P_0\to M\to0$$
with $P_1$ and $P_0$ finitely generated. The natural maps
$$\text{Hom}_A(P_i,N)\otimes_AB\to\text{Hom}_B(P_i\otimes_AB,N\otimes_AB)$$
are isomorphisms for $i=0,1$, and taking homology in degree $1$ it follows that the natural map
$$\text{Ext}^1_A(M,N)\otimes_AB\to\text{Ext}^1_B(M\otimes_AB,N\otimes_AB)$$
is injective (even an isomorphism if $P_2$ is also finitely generated).
Taking the class $\zeta$ of $\text{Ext}^1_A(M,N)$ representing the original short exact sequence, it follows that if every localization of the sequence is split (so the image of $\zeta$ in $\text{Ext}^1_{A_\mathfrak{p}}(M_\mathfrak{p},N_\mathfrak{p})$ is zero for every $\mathfrak{p}$), then $\zeta_\mathfrak{p}$ is zero in $\text{Ext}^1_A(M,N)_\mathfrak{p}$ for every $\mathfrak{p}$.
But if $\zeta\neq0$ then the annihilator of $\zeta$ is a proper ideal of $A$, contained in some maximal ideal $\mathfrak{m}$, and so $\zeta_\mathfrak{m}\neq0$.   
A: Take $A$ to be a product of two fields, e.g. $\mathbb{Z}/6$. All localizations at prime ideals are fields, but there are non-split short exact sequences of $A$-modules.
