Cardinalities of bases of a free $R$ module are same?

Question

Let $R$ be a ring with no zero divisiors such that for all $r,s\in R$ there exist $a,b\in R$ not both zero with $ar+bs=0$.

• If $R=K\oplus L$ then $K=0$ or $L=0$.
• if $R$ has an idendity then any two bases of free $R$ module $F$ have the same cardinality.

I am done with first statement, I guess that I will use first part to show the second one.

I need help in second one.

Thanks.

Answer 1

Update: after thinking about it for awhile, I remembered I already gave an elementary answer here. Basically, it says that the Ore condition (given here) allows you to do row reduction on matrices, generating a contradiction.

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Here is one elementary-ish solution. If you've studied noncommutative algebra, hopefully you've read about the problem of which domains can be embedded in division rings.

One main observation is that the condition about $ar+bs=0$ is actually just the left Ore condition in disguise. This means that $R$ embeds densely in a "division ring of left quotients $D$" almost identically as one does in the commutative case.

Basic algebra says that the $R$ linear homomorphism of $R^m$ to $R^n$ take the form of $m\times n$ matrices (operating on the right of row vectors from $R^m$.) If we have an isomorphism of $R^n$ with $R^m$, that amounts to an $m\times n$ matrix $A$ and an $n\times m$ matrix $B$ such that $AB=I_m$ and $BA=I_n$. But look: those matrices are also matrices over the division ring $D$ that you embedded $R$ into, and $A$ and $B$ are telling us there is a $D$ linear transformation of $D^m$ to $D^n$, which is impossible!

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There are a couple of non-elementary approaches that are worth mentioning which are closely related.

Since $D\otimes_R R^n\cong D^n$ as left $D$ modules, the isomorphism $R^m\cong R^n$ of left $R$ would change into an isomorphism $D^m\cong D^n$ of left $D$ vector spaces. Since dimension is well-defined for vector spaces over division rings, we have $m=n$.

The other closely related idea is that of uniform dimension.

Definition: A left $R$ module $M$ is said to have finite uniform dimension $k$ if there exists a set of uniform submodules of $M$ $N_1\ldots N_k$ such that $\oplus_{i=1}^k N_i$ is an essential submodule of $M$.

Lemma: Finite uniform dimension is well-defined. That is, if you have two such sums of submodules described in the definition of finite length and essential in $M$, then they have the same length.

The Ore condition on a domain forces any pair of nonzero submodules of $_RR$ to have nonzero intersection. Since $_RR$ is obviously dense in itself, $_RR$ has uniform dimension $1$, and $R^n$ has uniform dimension $n$ as a left $R$ module. It takes work to establish the lemma above, but after that is done, then it is obvious that $R^n\cong R^m$ implies $n=m$, since the uniform dimensions of isomorphic modules must match.

Answer 2

For bases of infinite cardinality it is clear.

Use induction in the other cases, so that if $F$ and $F'$ are bases with cardinality $m $ resp. $n$, then induct upon $max(m,n)$. The base case is clear from part 1 of the exercise. So suppose it is true for m-1, and let us prove it for m. Then say $F=R^m$ and $F'=R^n$ and let $\phi:R^m \rightarrow R^n$ be your isomorphism, and suppose wlog $m \geq n$. Then, you can quotient out by $R$ to get a map $\phi':R^{m-1} \rightarrow R^{n-1}$ and here you can show that this is an isomorphism, if you choose $\phi'$ "carefully enough".

I have left this argument vague by intent, but the ideas are there. You should work this out in more detail, and make it more precise with what I mean by quotiening out as I do. Why does $R^m/R \cong R^{m-1}$ hold? Why can I say that I can choose $\phi'$ to be an iso?