Csdrhrt

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 $M=bigoplus_{lambdainLambda} M_lambda$, where $forall lambdainLambda quad M_lambdacong A$.

Question 1: Would someone help me show (and to understand clearly) that $bigoplus_{i=1}^{infty}mathbb{Z}_{n_i}$, where $n_iin{2,3,dots}$, is not a free $mathbb{Z}$-module (where $mathbb{Z}_{n_i}=mathbb{Z}/n_imathbb{Z})$ ?

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

$varphi: mathbb{Z}^{bigoplusLambda}tobigoplus_{i=1}^{infty}mathbb{Z}_{n_i}$,

where $mathbb{Z}^{bigoplusLambda}=mathbb{Z}oplusmathbb{Z}oplus.../"|Lambda|text{ times }" /...oplusmathbb{Z}$.

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, $pi:bigoplus_{i=1}^{infty}mathbb{Z}_{n_i}tomathbb{Z}_{n_i}$, 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 $mathbb{Z}^{bigoplusLambda}topi^{-1}(Z_{n_i})$, 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?

Easy answer: isomorphisms preserve torsion elements in the underlying abelian group, but $$bigopluslimits_{i=1}^inftymathbb{Z}_{n_i}$$ 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.

You can show this using the fact that submodules of free modules are free over PID, but there is a much simpler way: in a free $mathbb Z$-module the equation $n cdot x = 0$ for some fixed $n$ has only one solution ($x=0$), but in $M = bigoplus mathbb Z / n_i mathbb Z$ all elements of the summand $mathbb Z / n_i mathbb Z$ are solutions of $n_i cdot x = 0$, and the summand injects canonically into the sum, providing non-zero solutions in $M$.

Indeed you want to show that no such isomorphism $varphi$ exists. To do so, it suffices to note that a free module over $Bbb{Z}$ has precisely one element of finite order; the identity element. If $n_ineq0$ for some $i$, then the $i$-th basis vector (the element of $bigoplus_{i=1}^{infty}mathbb{Z}_{n_i}$ that projects down to $1$ on $Bbb{Z}_{n_i}$ and to $0$ on every other coordinate) is of order $n_i$. From this observation it is not hard to deduce that $bigoplus_{i=1}^{infty}mathbb{Z}_{n_i}$ is a free $Bbb{Z}$-module if and only if $n_ileq1$ for all $i$.