TITLE: $\mathbb{Z}/m\mathbb{Z}$ is free when considered as a module over itself, but not free over $\mathbb{Z}$.
QUESTION [8 upvotes]: I am struggling to explain the following statement to myself in a convincing way. $\mathbb{Z}/m\mathbb{Z}$ is free when considered as a module over itself, but is not free when considered as a $\mathbb{Z}$-module.
But first of all, there are a few related questions that I am in doubt and could not find any useful clues.

When we defined a module R$\times$M$\to$M with R a ring and M an Abelian group, does R has to be a subset of M? The definition does not say so. But if we think about it, for example, take R be $\mathbb{Q}$ and M be $\mathbb{Z}$, then the map $\mathbb{Q}\times\mathbb{Z}$ need not be in $\mathbb{Z}$. What is wrong with my reasoning?
There is a statements: "If module is free, not every generating set necessarily contain a basis. Consider, for example, the generating set {2,3} for the $\mathbb{Z}$-module $\mathbb{Z}$. The subset {2} $\subseteq$ $\mathbb{Z}$ is a linearly independent set that can not be extended to a basis." My questions are: the generating set {2,3} is linearly independent so it is a basis, correct? Then what does it mean by "can not be extended to a basis", can we just add the element 3 into {2} then it becomes a basis?
Consider again the original question "$\mathbb{Z}/m\mathbb{Z}$ is free when considered as a module over itself, but is not free when considered as a $\mathbb{Z}$-module". Suppose I was wrong in point 1. that R has to be a subset of M, clearly $\mathbb{Z}$ is not a subset of $\mathbb{Z}/m\mathbb{Z}$. But in this case, does the $\times$ operation in
$\mathbb{Z}\times\mathbb{Z}/m\mathbb{Z}\to\mathbb{Z}/m\mathbb{Z}$ always mean $\times$ (modulo m), then what is the point of changing the ring R since the operation is the same and the image is the same?

Thank you very much for all helps!

REPLY [2 votes]: (1) A (left) module over $R$ is an abelian group $M$ equipped with a mapping $R\times M\to M$ satisfying certain axioms (that you easily find).
You probably are confused by the fact that there is always a linear monomorphism from $F$ to $V$, if $V$ is a nontrivial vector space over the field $F$; but the $\{0\}$ vector space already provides a counterexample.
Other counterexamples in modules are easy to find: every abelian group is a $\mathbb{Z}$-module, with the obvious map. Of course there's no way to embed $\mathbb{Z}$ into $\mathbb{Z}/2\mathbb{Z}$.
(2) Again you're confused by vector spaces. In vector spaces it's true that every linearly independent set can be extended to a basis; in modules this is not true. Indeed the result in vector spaces heavily depends on the possibility of multiplying by the inverse of any nonzero scalar.
The set $\{2\}$ in $\mathbb{Z}$ is linearly independent, but it's easy to show that any set $S$ with more than one element in $\mathbb{Z}$ is linearly dependent: take $a,b\in S$ with $a\ne b$: then $\alpha a + \beta b = 0$, with $\alpha=b$ and $\beta=-b$ (and one of the coefficients is surely nonzero).
In $\mathbb{Z}/2\mathbb{Z}$ no subset is linearly independent: indeed $\{1+2\mathbb{Z}\}$ isn't because $2(1+2\mathbb{Z})=0+2\mathbb{Z}$ and $2\ne0$. Of course the same is true for any torsion abelian group, among which is $\mathbb{Z}/m\mathbb{Z}$.
(3) We've seen that $\mathbb{Z}/m\mathbb{Z}$ is not free as a $\mathbb{Z}$-module, because it has no linearly independent subset.
Conversely $R$ is a free $R$-module (with the obvious action): in fact $\{1\}$ is a linearly independent set which generates $R$ as $R$-module.
This should not surprise you. If you change the base ring almost anything can happen. A different phenomenon, but related, is what happens when you consider $\mathbb{C}$ as a vector space over $\mathbb{C}$ or over $\mathbb{R}$; in the first case it has dimension one, in the second case it has dimension two. Similarly, an abelian group $M$ can be a module over different rings and have different properties in the various cases.