$\mathbb{Z}/m\mathbb{Z}$ is free when considered as a module over itself, but not free over $\mathbb{Z}$.

Question

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.

1. 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?

2. 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?

3. 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!

Answer 1

(1) No, $R$ does not have to be a subset of $M$. The map $R \times M \rightarrow M$ does need to satisfy some properties, namely:

$\bullet$ For all $a \in R$ and all $x,y \in M$, $a(x+y) = ax + ay$.
$\bullet$ For all $a,b \in R$ and all $x \in M$, $a(bx) = (ab)x$.
$\bullet$ For all $x \in M$, $1 x = x$.

When you talk about "the map $\mathbb{Q} \times \mathbb{Z} \rightarrow \mathbb{Z}$", I'm confused: what map do you have in mind? If you are pointing out that the standard multiplication of an integer by a rational number does not necessarily land in the integers: yes, you're right. Thus you cannot endow $\mathbb{Z}$ with the structure of a $\mathbb{Q}$-module in that way. In fact, you cannot endow $\mathbb{Z}$ with the structure of a $\mathbb{Q}$-module in any way, but the proof of this may be a bit beyond your current level of understanding.

(2) The elements $2$ and $3$ are not linearly independent over $\mathbb{Z}$: $3 \cdot 2 + (-2) \cdot 3 = 0$.

(3) You were not wrong about this in (1) above. Indeed for any abelian group $(M,+)$, the map $\mathbb{Z} \times M \rightarrow M$ given by $(n,m) = m + \ldots m$ ($n$ times) if $n$ is positive, the inverse of that if $n$ is negative, and $0$ if $n = 0$, makes $(M,+)$ into a $\mathbb{Z}$-module. In fact this is the unique $\mathbb{Z}$-module structure on $M$.

Now: for $n > 1$, $M = \mathbb{Z}/n\mathbb{Z}$ cannot be a free $\mathbb{Z}$-module since there is no nonempty subset which is linearly independent over $\mathbb{Z}$: for any $x \in M$, $nx = 0$. This does not stop $M$ from being a free $\mathbb{Z}/n\mathbb{Z}$-module, since $n = 0$ in $\mathbb{Z}/n\mathbb{Z}$. And indeed $1$ is a basis for $M$ as a $\mathbb{Z}/n\mathbb{Z}$-module.

Answer 2

(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.

Answer 3

(1) What in the world are R,M?

(2) No, $\,\{2,3\}\,$ is not a basis since $\,(-3)2+2\cdot 3 = 0\,$ and $\,(-3), 2\neq 0\,$

(3) I can't understand your reference to (1) since I can make sense of it, but any abelian group is trivially a $\,\Bbb Z$-module, and then

$$(a)\;\;\;\forall\,x\in\Bbb Z_m:=Z/m\Bbb Z\;,\;\;m\cdot x=0\in\Bbb Z_m\;\;\wedge\;m\neq0\implies Z_m\;\text{is not a free}\;\Bbb Z-\text{module} ;$$

$$(b)\;\forall\,x\in\Bbb Z_m\;,\;\;x=x\cdot 1\pmod m\implies \Bbb Z_m\;\text{is a free (and cyclic!)}\;\Bbb Z_m-\text{module}$$