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Recently, someone stated that every short exact sequence (of, say, modules) of the form $$0 → M → M \oplus N → N → 0$$ splits. I think this is false in general because the arrow $M → M \oplus N$ might not be the natural inclusion. (The difference pointwise isomorphic/diagramwise isomorphic.)

Maybe one can realize an arrow $M\oplus N → N$ with a kernel isomorphic to $M$, but not being an (internal) direct summand of $M \oplus N$? I tried something along the lines of $ℚ$ and $ℚ/ℤ$, but I have been unsuccessful so far.

Can anyone provide me with a counterexample?

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  • $\begingroup$ Something is confusing me, by definition that sequence splits, I do not see anything to be proven there. Maybe this can help mathworld.wolfram.com/SplitExactSequence.html $\endgroup$
    – Luigi M
    Commented Dec 27, 2014 at 10:42
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    $\begingroup$ My guess is that you're allowing the maps to not be the usual ones. Is that right? $\endgroup$
    – Hoot
    Commented Dec 27, 2014 at 10:46
  • $\begingroup$ @Hoot et al. Yes, of course. I added the explanation. There’s a difference between diagrams being pointwise isomorphic and being isomorphic, well, as diagrams. $\endgroup$
    – k.stm
    Commented Dec 27, 2014 at 10:48
  • $\begingroup$ I realized my question already has an (exactly the same) answer already within this question. $\endgroup$
    – k.stm
    Commented Dec 27, 2014 at 11:36

1 Answer 1

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It took me a while, but I came up with a counterexample myself: $$0 → ℤ \overset α \longrightarrow ℤ \oplus \bigoplus_ℕ ℤ/2ℤ \overset β \longrightarrow \bigoplus_ℕ ℤ/2ℤ → 0,$$ with

  • $α$ being multiplication by $2$ (postcomposed by the natural inclusion)
  • $β$ being given on the left summand as the projection $ℤ → ℤ/2ℤ$ (postcomposed by the inclusion into the first summand) and on the right summand as a right shift.

More concretely, set $M = ℤ$ and $N = \bigoplus_ℕ ℤ/2ℤ$

\begin{align*} α&\colon M → M \oplus N,~k ↦ (2k,0)\\ β&\colon M \oplus N → N,~ (k,(κ_1,κ_2,…)) ↦ ([k]_2,κ_1,κ_2,…) \end{align*} Now $\ker β = 2ℤ \oplus 0 = \operatorname{img} α$, because for all $x = (k,κ) ∈ M \oplus N$ with $κ = (κ_1,κ_2,…)$ $$ β(x) = 0 ⇔ ([k]_2,κ_1,κ_2,…) = 0 ⇔ k ∈ 2ℤ~\text{and}~κ = 0.$$

The sequence doesn’t split, though, because the element $x = (1,0,0,…)$ in $N$ has order $2$, but there is no element of order $2$ in $β^{-1}(x) = (1,0) + \ker β = (1,0) + (2ℤ\oplus 0)$, so there cannot be any linear map $s \colon N → M \oplus N$ with $(β∘s) (x) = x$ (as $β∘s$ strictly decreases the order of $x$).

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  • $\begingroup$ Is there an example with finitely generated modules? If $0 \longrightarrow M \longrightarrow M \oplus N \longrightarrow N \longrightarrow 0$ is exact and $M$ and $N$ are finitely generated, then the sequence splits? $\endgroup$ Commented Nov 18, 2017 at 23:29
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    $\begingroup$ My question in the case $R$ is a $k$- algebra with finite dimension over $k$. If $M$ and $N$ are finitely generated over an arbitrary ring, I don't know. $\endgroup$ Commented Nov 18, 2017 at 23:45
  • $\begingroup$ @WernerGermánBusch Let $R = \prod_ℕ ℤ$. Then an anologous construction of an exact sequence $$0 → ℤ →ℤ \oplus \prod_ℕ ℤ/2ℤ → \prod_ℕ ℤ/2ℤ → 0$$ should work. Both $M = ℤ$ and $N = \prod_ℕ ℤ/2ℤ$ are finitely generated over $R$. I don’t know what happens if $R$ is a finite dimensional algebra over a field. $\endgroup$
    – k.stm
    Commented Nov 20, 2017 at 18:11
  • $\begingroup$ I was about to say that if $A$ is a finite dimensional algebra over a field, then it's true, using the $Hom(-,A)$ funtor and using dimensions. $\endgroup$ Commented Nov 21, 2017 at 4:18
  • $\begingroup$ Is there a counterexample where the first injective map is the canonical inclusion map ? $\endgroup$
    – uno
    Commented Apr 3, 2020 at 3:44

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