0
$\begingroup$

Let $R$ be a ring with identity and $M$ be an $R$-right module. Then $M$ is called free over $X \subseteq M$ if for every module $N$ with mapping $\alpha : X \to N$ we can extend it uniquely to a homomorphism $\overline \alpha : M \to N$. For a ring $R$ denote by $R_R$ the module structure if $R$ is regarded as a module over itself. Now I am looking for a slick and short proof of the following

A right $R$-module $M$ is free (over some set $X$) if and only if it is a direct sum of copies of $R_R$.

By some "high-level" theorems, like that $X$ is a minimal generating set for $M$ and the elements of $X$ are lineary independent in $M$ (i.e. every element from $M$ could be uniquely written as linear combination of finitely many elements from $X$) we get an isomorphism by mapping onto the "coordinates" and the direct product of a ring is module isomorphic to the direct sum of $R_R$, thus showing that every free modules is isomorphic to such a direct sum. But I hoped there might be some easier argument, without supposing linear independence of the $X$ and that $M = \langle X \rangle$.

My attempt. If $x \in X$ consider the $R$-right module $xR$ and the mapping $\beta : X \to \{x\}$, then we have a unique $\overline \beta : M \to xR$ which happens to be surjective as $\overline \beta(xr) = \beta(xr) = xr$. Now $M = xR \oplus \mbox{ker}(\overline \beta)$ as $m = rx + (m - rx)$ with $rx = \overline\beta(m)$. Now I wanted to show that $\mbox{ker}(\overline \beta)$ is free over $X \setminus \{x\}$ and hence proceed inductively. But then I realised that this is just possible if $X$ is countable or has some other structure which permits such inductive reasoning. So I guess such a reasoning is not valid in general, right? If this proof scheme does not work, are there any other short proofs of this fact?

$\endgroup$
2
  • $\begingroup$ I'm not sure I understand the end of your second paragraph. What do you mean by "without supposing linear independance of the $X$"? It seems that without this assumption, you can't prove that $M$ is free. $\endgroup$ Feb 11, 2016 at 14:00
  • $\begingroup$ Without knowing that as a theorem, like for example it is cited here: ma.utexas.edu/users/voloch/Homework/garrett.pdf but if it might be necessary then of course it might be included in the proof, but at least I am looking for a short proof without citing many other theorems. $\endgroup$
    – StefanH
    Feb 11, 2016 at 14:03

1 Answer 1

3
$\begingroup$

It's not entirely clear to me what you are looking for exactly, but here is a short proof of your statement:

Consider the $R$-module $N= \bigoplus_{x \in X} R_x$, where each $R_x$ is just a copy of $R_R$. Denote the unit of $R_x$ by $1_x$. Then the map of sets $f: X \to N, x \mapsto 1_x$ gives a unique module homomorphism $M \to N$ extending $f$ (and which I will abusively denote by $f$ as well). Furthermore, the maps $g_x: R \to M, g_x(r) = xr$ sum up to give a homomorphism $g:N \to M$. I claim that $f \circ g = \mathrm{id}_N$ and $g \circ f = \mathrm{id}_M$. For the first equality, consider an element $a = (a_x)_{x \in X} \in N$. Then $f \circ g(a) = f(\sum_x x a_x) = \sum_x f(x) a_x = \sum_x a_x = a$ which proves the first identity. For the latter equality, we have $g \circ f (x) = g_x(1_x) = x$, and hence $g \circ f$ is the identity on $X$. Since M is free, and $g \circ f$ is a module homomorphism, this means that it must be equal to $\mathrm{id}_M$.

$\endgroup$
1
  • $\begingroup$ That is good, I am happy! $\endgroup$
    – StefanH
    Feb 11, 2016 at 15:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .