# Do all exact sequences $0 \rightarrow A \rightarrow A \oplus B \rightarrow B \rightarrow 0$ split for finitely generated abelian groups?

Suppose $A$ and $B$ are finitely generated Abelian groups. Are all exact sequences of the form $0 \rightarrow A \rightarrow A \oplus B \rightarrow B \rightarrow 0$ split?

If not, is there an example?

• What's wrong with the canonical inclusion $B \to A \oplus B$ ? May 20 '14 at 15:44
• It isn't stated what the maps are, so it is not clear that that would give a splitting.
– RP_
May 20 '14 at 15:58
• @Alberto: that $0\to N\to G\to Q\to 0$ splits means that the map $Q\to G$ lifts the projection $G\to Q$. This can fail with the canonical inclusion.
– YCor
May 20 '14 at 16:07
• See Steven Landsburg's answer to the following question: mathoverflow.net/questions/157938/… May 20 '14 at 16:14
• Obviously in the the question homomorphisms are arbitrary, not the canonical ones. The answer is "yes, they're all split" by Landsburg's answer, see the link given by Dag Oskar Madsen.
– YCor
May 20 '14 at 16:37

This is true more generally for finitely generated modules over a noetherian ring. Your question is equivalent to asking whether the sequence $$0\rightarrow \operatorname{Hom}(B,A)\rightarrow \operatorname{Hom}(A\oplus B,A)\rightarrow \operatorname{Hom}(A,A)$$ is surjective on the right. To prove this, it suffices to localize and then complete at an arbitrary prime $$P$$, so we can assume we're working over a complete local ring where $$P$$ is the maximal ideal. Now it suffices to check surjectivity after modding out an arbitrary power $$P^n$$, which allows us to assume that all the modules are of finite length. Surjectivity follows because the lengths of the left-hand and right-hand modules add up to the length of the module in the middle.

Edited to add: The comments above (which I read after I posted this) remind me that I've posted this same argument before. If people think this instance should be deleted, I'm fine with that.

I thought it worth adding a reference to this:

Miyata, Takehiko, Note on direct summands of modules, J. Math. Kyoto Univ. 7 1967 65–69.

In the paper, the question of the OP is attributed to Matsumura and the solution to Toda. Then a short argument is given for this case.

The author generalizes this result to the following (which is a quote from MathSciNet):

"Let $$R$$ be a commutative, noetherian ring and let $$A$$ be an $$R$$-algebra of finite type. Moreover, let $$M$$ be a finitely generated $$A$$-module and let $$N$$ be a submodule of $$M$$. Using the usual tools of homological algebra and noetherian ring theory, the author establishes the following pair of results.

Theorem 1: If $$M$$ is isomorphic to $$N\oplus M/N$$, then $$N$$ is a direct summand of $$M$$.

Theorem 2: If $$0\to N\otimes_R T\to M\otimes_R T$$ is exact for all $$A$$-modules $$T$$ (i.e., $$N$$ is pure), then $$N$$ is a direct summand of $$M$$."

• Note: I think "$R$-algebra of finite type" means "finitely generated $R$-algebra".
– YCor
Jul 16 '19 at 19:23
• As does Wikipedia. Jul 17 '19 at 7:44