# $\mathbb{Z}$-module $\prod\limits_{\text{p prime} } \mathbb{Z}/p\mathbb{Z}$

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?

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.
• 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. – amator2357 Feb 11 at 12:49