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This question is a follow-up of this one.

Now I'm trying to prove that the torsion submodule of $\prod \mathbb{Z}_p$ is not a direct summand of $\prod \mathbb{Z}_p$.

I found this interesting because it's an example that over a PID, a non-finitely generated module may fail to be decomposed as torsion $\oplus$ something free, showing even that it can't be decomposed as torsion $\oplus$ anything.

Now, it is easily proven that the torsion submodule of $\prod \mathbb{Z}_p$ is $\bigoplus \mathbb{Z}_p$.

So if we could decompose $\prod \mathbb{Z}_p$ as torsion $\oplus$ something it would have to be $\prod \mathbb{Z}_p = \bigoplus \mathbb{Z}_p \oplus \frac{\prod \mathbb{Z}_p}{\bigoplus \mathbb{Z}_p}$ (this is correct, right?)

Now, using the question linked above, we know the right summand is divisible ($\iff$ injective), besides it is torsion-free ($\iff$ flat).

(I think) I've learned my lesson from my previous question, so I also tried proving that there is a member in $\prod \mathbb{Z}_p$ which can't be written as a sum of something in $\bigoplus \mathbb{Z}_p$ plus something in $\frac{\prod \mathbb{Z}_p}{\bigoplus \mathbb{Z}_p}$, but I couldn't advance any further in this direction either.

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3 Answers 3

up vote 8 down vote accepted

Let $G=\prod_p\mathbb Z_p$ and $H=\bigoplus_p\mathbb Z_p$ and regard $H$ as a subgroup of $G$. You want to know if $H$ is a direct factor of $G$. If it was there would be a subgroup $K$ of $G$ isomorphic to $G/H$. There isn't. You know that $G/H$ is divisible, but there are no nonzero elements $x\in G$ having the property that $x=p y$ for some $y\in G$ for all primes $p$.

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Ask yourself this question: if $x \in \prod_p \mathbb{Z}_p$ is such that $x \in n \prod_p \mathbb{Z}_p$ for each $n \in \mathbb{Z} \setminus \{0\}$, then what can we say about $x$?

You should be able to conclude that there is no embedding from $\frac{\prod_p\mathbb{Z}_p}{\oplus_p \mathbb{Z}_p}$ to $\prod_p \mathbb{Z}_p$

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My answer will be quite advanced, but it might interest you (or show you new ideas): a module $A$ is self-small, if the $Hom_R(A, -)$ functor commutes with arbitrary direct sums of $A$. There can be proved (with a little effort), that endomorphic images (direct summands for example) of self-small modules are self-small and infinite direct sum cannot be self-small. With (much) more effort one could prove, or find here, a condition on self-smalness of product, which shows, that $\prod_p \mathbb{Z}_p$ is self small. If $\oplus_p \mathbb{Z}_p$ was a direct summand, it would have to be self-small, a contradiction.

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