Setup:
I've just started self-studying commutative algebra via video lectures from India Institute of Technology, and this question came up in the section on modules. It seems easy in a way, but I guess I'm missing something trivial. This is rather directly after introducing free modules, which I still haven't completely grasped. The definition of a free module used here is:
An $A$-module $M$ is said to be free if $M=\bigoplus_{\lambda\in\Lambda} M_\lambda$, where $\forall \lambda\in\Lambda \quad M_\lambda\cong A$.
Question 1: Would someone help me show (and to understand clearly) that $\bigoplus_{i=1}^{\infty}\mathbb{Z}_{n_i}$, where $n_i\in\{2,3,\dots\}$, is not a free $\mathbb{Z}$-module (where $\mathbb{Z}_{n_i}=\mathbb{Z}/n_i\mathbb{Z})$ ?
The lecturer gave the impression that this should be straight forward, so preferably I want to do this by showing that there cannot be an isomorphism
$\varphi: \mathbb{Z}^{\bigoplus\Lambda}\to\bigoplus_{i=1}^{\infty}\mathbb{Z}_{n_i}$,
where $\mathbb{Z}^{\bigoplus\Lambda}=\mathbb{Z}\oplus\mathbb{Z}\oplus…/"|\Lambda|\text{ times }" /…\oplus\mathbb{Z}$.
Question 2: Is there an even simpler answer than my potential one below?
Thoughts (Attempt):
The potential problem is likely that each $Z_{n_i}$ is a finite $Z$-module. By assuming $\bigoplus_{i=1}^{\infty}\mathbb{Z}_{n_i}$ is free, and projecting down, $\pi:\bigoplus_{i=1}^{\infty}\mathbb{Z}_{n_i}\to\mathbb{Z}_{n_i}$, we exhibit finite submodules $\pi^{-1}(Z_{n_i})\subset \bigoplus_{i=1}^{\infty}\mathbb{Z}_{n_i}$. Now, I found out that submodules of free modules over a PID are free, and thus since $\mathbb{Z}$ is a PID, we have that each submodule $\pi^{-1}(Z_{n_i})$ must be a free $\mathbb{Z}$-module. By the definition above there would then exist an isomorphism $\mathbb{Z}^{\bigoplus\Lambda}\to\pi^{-1}(Z_{n_i})$, which cannot be the case, since this function can not be injective since $\pi^{-1}(Z_{n_i})$ is finite. //
However, I cannot remember that we have yet mentioned that submodules of free modules over a PID are free. Is there thus an even easier way of getting an answer to Question 1?
Best Answer
Easy answer: isomorphisms preserve torsion elements in the underlying abelian group, but $$\bigoplus\limits_{i=1}^\infty\mathbb{Z}_{n_i}$$ contains a bunch of torsion elements (for example, if $n_1 \neq 1$, then $(1,0,0,0,\ldots)$ (otherwise move on to the first term such that $n_i \neq 1$: if there isn't one, then your module is trivial, which is free, so you should just throw that in your hypotheses)) has order $n_i$, so is torsion), but free $\mathbb{Z}$-modules have no torsion elements, so no such isomorphism can exist.