Recall that we call a sequence of $R-$modules $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ pure exact, if for every module $D$ the induced sequence $0\rightarrow A\otimes D\rightarrow B\otimes D\rightarrow C\otimes D\rightarrow 0$ is again exact.
So we know that every split exact sequence is pure exact and in excercise 3.31 from Rotman's homological algebra we are to show that the converse is not true in general.
We consider $P$ as the set of all primes in $\mathbb{Z}$ and the abelian group $M=\prod_{q\in P} \mathbb{Z}/(q)$. Now this exercise has 3 statements:
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$\oplus_{q\in P}\mathbb{Z}/(q)$ is the torsion subgroup of $M$, i.e. $tM=\oplus_{q\in P}\mathbb{Z}/(q)$
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$M/tM$ is a divisible abelian group (hence an injective $\mathbb{Z}$ module)
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The sequence $0\rightarrow tM\rightarrow M\rightarrow M/tM\rightarrow 0$ is pure exact (because $M/tM$ is torsion free) but does not split
So i solved the first two but i am not sure how to prove the last statement. I tried to assume it does split but i can't find a contradiction. I used the splitting lemma to obtain a surjective map $M\rightarrow tM$, which fixes $tM$ and i guess that such a map doesn't exist? I also thought about using statement $2.$ somehow but i don't really now how to apply it.
Best Answer
The third point follows from the second by the fact that there aren’t not null divisible elements in $M$. Indeed, if an element in $M$ is not null in the $q$-th coordinate, then it can’t be divided by $q$. On the other hand, if an injection $tM\subset M$ existed, than every element of $tM$ would be divisile for every $q\in P$. So we conclude that a similar injection couldn’t exist, in particular the sequence can’t split.