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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?

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    $\begingroup$ What's wrong with the canonical inclusion $B \to A \oplus B$ ? $\endgroup$ May 20 '14 at 15:44
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    $\begingroup$ It isn't stated what the maps are, so it is not clear that that would give a splitting. $\endgroup$
    – RP_
    May 20 '14 at 15:58
  • $\begingroup$ @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. $\endgroup$
    – YCor
    May 20 '14 at 16:07
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    $\begingroup$ See Steven Landsburg's answer to the following question: mathoverflow.net/questions/157938/… $\endgroup$ May 20 '14 at 16:14
  • $\begingroup$ 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. $\endgroup$
    – YCor
    May 20 '14 at 16:37
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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.

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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 $M$."

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

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