On proving that $\bigoplus_{i=1}^{\infty}\mathbb{Z}_{n_i}$ is not a free $\mathbb{Z}$-module

Question

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?

Hint: What happens if you multiply an element in $\mathbb{Z}_n$ by $n$? What are the zero divisors of direct sums of $\mathbb{Z}$? — Severin Schraven, Nov 14 '18 at 14:43
Note that the direct sum $\bigoplus_{i=1}^{\infty}\mathbb{Z}_{n_i}$ is a free $\Bbb{Z}$-module if $n_i\leq1$ for all $i$. Do you consider $0$ a natural number? — Servaes, Nov 14 '18 at 14:48
@SeverinSchraven : Yes, thank you, I had read the answer below before your comment; would've been sufficient to solve it though. — Christopher.L, Nov 14 '18 at 15:44

score 3 · Accepted Answer · answered Nov 14 '18 at 14:42

3

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.

answered Nov 14 '18 at 14:42

user3482749

6,750

Ok, I had no idea what torsion was, so I had to read about that; Basically the analog to zero divisors for rings? Ok, it seems to be what everyone else is saying right, so: Let $\psi:\bigoplus_{i=1}^{\infty}\mathbb{Z}{n_i}\to\mathbb{Z}^{\oplus\Lambda}$ be the assumed isom., then for each $x\in \pi^{-1}(Z_k)\subset \bigoplus{i=1}^{\infty}\mathbb{Z}_{n_i}$, we have $0=\psi(0)=\psi(kx)=k\psi(x)\Rightarrow \psi(x)=0$, since $\mathbb{Z}^{\oplus\Lambda}$ has no torsion (aka torsion free?), and in this case it's directly analog to the fact that $\mathbb{Z}$ is a domain? – Christopher.L Nov 14 '18 at 15:31
Or, well, not each $x\in\pi^{-1}(Z_k)$, but rather each $x$ of the form $(0,0,\dots,x_i,0,\dots)$, where $x_i\in \mathbb{Z}_k$ i guess. – Christopher.L Nov 14 '18 at 15:34
Yes, that's correct. – user3482749 Nov 14 '18 at 15:34

score 1 · Answer 2 · answered Nov 14 '18 at 14:41

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 $M$.

score 1 · Answer 3 · answered Nov 14 '18 at 14:42

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 $n_i\neq0$ 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 $n_i\leq1$ for all $i$.

On proving that $\bigoplus_{i=1}^{\infty}\mathbb{Z}_{n_i}$ is not a free $\mathbb{Z}$-module

3 Answers3