Local property of split exact sequence

Question

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!

Answer 1

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 $\zeta_\mathfrak{m}\neq0$.

Answer 2

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.