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I am trying to establish if the $\mathbb{Z}$-module $\displaystyle\prod_{\text{p prime} } \mathbb{Z}/p\mathbb{Z}$ is torsion-free. So I think that the elements of $\displaystyle\prod_{\text{p prime} } \mathbb{Z}/p\mathbb{Z}$ are all infinite tuples where on the first position in any of such tuple we'd have an element from $\mathbb{Z}/2\mathbb{Z}$, on the second position an element from $\mathbb{Z}/3\mathbb{Z}$, etc. Is that correct? So my guess would be that this module is not torsion-free. For example, we have $2\{[0],[1],[0],\dots,[0]\}=\{[0],[0],[0],\dots,[0]\}$. Is that right? In general, assuming that the above is true, what would be that module's complete torsion part?

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Your example is almost right; the nonzero second coordinate $[1]$ is an element of $\Bbb{Z}/3\Bbb{Z}$, not of $\Bbb{Z}/2\Bbb{Z}$ so $$2([0],[1],[0],\ldots)=([0],[2],[0],\ldots).$$ You are on the right track though. Note that elements of a product are usually denoted with round braces, not curly ones. Also, since the product is infinite it is customary not to write a last coordinate, but in stead to end with some dots...

The torsion part is the submodule $\bigoplus_{p\text{ prime}}\Bbb{Z}/p\Bbb{Z}$ of tuples with finite support. I'll leave it for you to prove.

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  • $\begingroup$ Of course, I should've written $2([1],[0],[0],\dots)=([0],[0],[0],\dots)$. Thank you for pointing that out. OK, thank you! I'll try to prove that. $\endgroup$ – amator2357 Feb 11 at 12:49

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