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This is a refined version of a question I asked days ago and have no answers yet. I am completely illietarte in algebra so, please, don't kill but explain.

The question is all in the title: is there a complete decomposition of an arbitrary PID module into a direct sum of indecomposable submodules? I understand that in general uniqueness and the cardinality of the set of indecomposable submodules are disputable. I just need the existence. Thank you.

Edit Thanks for all comments. Maybe I am close to understanding where my problem is. I think what I can always do is the following (I put it for the case of $\mathbb{Z}^\infty$ for simplicity). For every $x\in\mathbb{Z}^\infty$ such that $\mbox{gcd}(\{x_n\})=1$ the cyclic submodule $\mathbb{Z}x$ is maximal. Two such submodules either coincide or intersect trivially. Moreover, there exists by Zorn a minimal set $\{x_i\}_{i\in I}$ such that $$ \mathbb{Z}^\infty=\sum_{i\in I}\mathbb{Z}x_i. $$ However, this sum is still not direct, because there may be a proper submodule $\mathbb{Z}y$ with $y\in\mathbb{Z}x_j$ such that $$ \mathbb{Z}y\in\sum_{i\in I\setminus\{j\}}\mathbb{Z}x_i. $$ For a general PID module you may not have maximal cyclic submodules but infinitely growing sums of cyclic submodules, but the principle is apparently the same. This is too bad :(

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    $\begingroup$ have you read this page? people.brandeis.edu/~igusa/Math101b/Chap5.pdf $\endgroup$
    – roy smith
    Apr 16, 2016 at 18:54
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    $\begingroup$ Recall that $\mathbf{Z}$-modules are abelian groups. Consider the abelian group $A=\mathbf{Z}^X$ with $X$ infinite countable. Every nontrivial subgroup of $A$ has a nontrivial homomorphism onto $\mathbf{Z}$, but $\mathrm{Hom}(A,\mathbf{Z})$ is countable. So $A$ cannot be an infinite direct sum of nonzero submodules. Probably it can't be a finite direct sum of indecomposable subgroups (I don't have an argument right now). $\endgroup$
    – YCor
    Apr 16, 2016 at 19:00
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    $\begingroup$ @Ycor: That's correct. For details, see the second paragraph of my answer here. $\endgroup$
    – Eric Wofsey
    Apr 16, 2016 at 19:14
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    $\begingroup$ There are written many books over abelian groups (I can recommend Fuchs: Infinite abelian groups I, II). This suggests that there is no simple / complete answer on the structure of such groups in general. $\endgroup$
    – Todd Leason
    Apr 16, 2016 at 20:09
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    $\begingroup$ Another excellent book on the subject is Irving Kaplinsky. Infinite abelian groups. University of Michigan Press, Ann Arbor, 1954. $\endgroup$
    – Peter May
    Apr 16, 2016 at 20:15

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