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.
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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?
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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?
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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!
Best Answer
(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.