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!
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Sign up to join this communityIn 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!
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 $\zeta_\mathfrak{m}\neq0$.
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.