Is $\oplus_{i=1}^{\infty} \mathbb{Z}_{n_i}$ , where $n_i \in \mathbb{N} $ a free $\mathbb{Z} $
module?
This question is from my modules assignment and I think I need help while attempting this:
If it is a free $\mathbb{Z} $ module then there exists a basis set say { $x_1,…, x_n,… $} where each element is an infinite tuple.
Either I have to prove existence of such tuple or get some contradiction assuming that tuple exists.
Edit: I think $(1,0,0,…) , (0,1,0,0,…) , (0,0,1,0, \cdots),\cdots$ is such a basis and hence the module is free $\mathbb{Z}$ module. I took all the entries of direct sum as $\mathbb{Z}$ mistakingly.
Attempt for $\mathbb{Z}_{n_i}$'s : I think if I still define basis as (1,0,0,…) ,(0,1,0,0,0…),(0,0,1,0,0,…) this will still generate $\oplus_{i=1}^{\infty}\mathbb{Z}_{n_i}$ but it will not be linearly independent as if it is Linearly independent then there exists $a_i$'s such that $(a_1,…,a_n,….)=(0,…,0,…)$ then $a_i's$ could be 0 mod $n_i'$ but not in $\mathbb{Z}$. So, It is not Linearly independent.
So, is my proof fine?
Best Answer
A free $\mathbb{Z}$-module cannot have torsion elements, that is, elements $x$ such that $mx=0$ for some integer $m\ne0$, because $\mathbb{Z}$ has no torsion element and a free module is a direct sum of copies of $\mathbb{Z}$.
Your direct sum has torsion elements as soon as $n_i>0$ for some $i$.
It should be also noted that proving that the set you considered is not a basis doesn't prove that the module isn't free. For instance, $\mathbb{Z}$ is free and $\{2,3\}$ is a generating system that's not linearly independent. In other words, if you prove that some set is a basis, then the module is free; if you prove that some (generating) set is not a basis, you proved nothing.