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?


  • 3
    $\begingroup$ If $M$ is a finitely generated $A$-module, and $A \rightarrow B$ is a flat morphism of rings, then $\mathrm{Ext}^*_B(M \otimes_A B, N \otimes_A B) = \mathrm{Ext}^*_A(M , N ) \otimes_A B$ : apply this with $B$ equal to various localizations of $A$ to get your result, when the last term of the exact sequence is finitely generated. Without the finite generation hypothesis, your assertion is false. $\endgroup$
    – js21
    Jun 7, 2017 at 10:53
  • 5
    $\begingroup$ I don't really understand why this question should be off-topic. As the mistaken answer below shows, it is not so easy to come up with a counterexample, is it ? $\endgroup$
    – Niels
    Jun 8, 2017 at 6:36
  • $\begingroup$ @js21 I think you need some extra finiteness condition, such as $A$ Noetherian, for the statement about $\text{Ext}^*$? $\endgroup$ Jun 8, 2017 at 8:09
  • $\begingroup$ @Jeremy Rickard: You are right. The correct statement is that it is true (for $\mathrm{Ext}^1$) whenever $M$ is $(-2)$-pseudo-coherent. Using Lemma 10.72.1 from Stacks Project $087M$, this amounts to show that $\mathrm{Ext}^1_A(M , N \otimes_A B) = \mathrm{Ext}^1_A(M , N ) \otimes_A B$. By Lazard's theorem, $B$ is a cofiltered colimit of finite free $A$-modules. One concludes using the finiteness assumtion on $M$ $\endgroup$
    – js21
    Jun 8, 2017 at 9:25
  • $\begingroup$ @js21 Actually, even if $M$ is just finitely presented, the natural map $\text{Ext}^1_A(M,N)\otimes_AB\to\text{Ext}^1_B(M\otimes_AB,N\otimes_AB)$ is injective, which is enough for this question. $\endgroup$ Jun 8, 2017 at 9:56

2 Answers 2


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

  • 2
    $\begingroup$ It is not true that a flat module over a local ring is projective. Projective modules over local rings are free, but flat ones aren't... for example the rationals over the locazation of the integers at a prime. $\endgroup$ Jun 13, 2017 at 20:59
  • 1
    $\begingroup$ @CountDracula Quite right. But finitely generated flat modules over a local ring are projective, which is enough for a counterexample for a non-Noetherian ring. I'll correct it. $\endgroup$ Jun 13, 2017 at 21:50
  • $\begingroup$ @JeremyRickard thank you!the answer is very clear and correct ,I'll try to understand finiteness condition. $\endgroup$
    – Jian
    Jun 14, 2017 at 0:56
  • $\begingroup$ @CountDracula Sorry, I read your comment just before I went to bed, and forgot to say thank you for pointing out my silly mistake. $\endgroup$ Jun 14, 2017 at 12:48

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.

  • 6
    $\begingroup$ I don't think there are ... $\endgroup$ Jun 7, 2017 at 12:17
  • $\begingroup$ What is the meaning of there are non-split s.e.s of A-modules? $\endgroup$
    – Jian
    Jun 7, 2017 at 12:42
  • $\begingroup$ @JeremyRickard: Definitely $0 \to {\mathbb Z} \to {\mathbb Z} \to {\mathbb Z}/6 \to 0$ is non-split. $\endgroup$
    – Sasha
    Jun 7, 2017 at 20:17
  • 2
    $\begingroup$ @JeremyRickard: You are right, sorry. $\endgroup$
    – Sasha
    Jun 7, 2017 at 20:39
  • 8
    $\begingroup$ @FernandoMuro I'm sure we've all been there! Rickard's Law: The probabiity of posting something embarrassingly wrong is proportional to the amount of time you'll then be away from a computer and unable to correct it. $\endgroup$ Jun 8, 2017 at 13:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy