An exchange property for bases of a free module over a division ring $R$

Question

Let $M$ be a module over a division ring $R$ and let $A$ and $B$ be bases of $M$.

Then $\forall a \in A \ \ \exists b \in B: \ \ (A \setminus \{ a \}) \cup \{ b \}$ is a basis of $M$.

Here's how a possible proof goes:

If $a \in B$, then $b = a$ serves.

Assume $a \notin B$, and let $S = A \setminus \{ a \}$. If $B \subseteq \langle S \rangle$, then every $b \in B$ is a linear combination of elements of $S$, and since $\langle B \rangle = M$, every $m \in M$ is a linear combination of elements of $S$, so, $\langle S \rangle = M$, and $a$ is a linear combination of elements of $S$, causing a cotnradiction to the fact that $A$ is linearly independent. Therefore $B$ is not contained in $\langle S \rangle$, and there is $b \in B$ that is not a linear combination of elements $S$. It can be easily proved that then $A' = S \cup \{b \}$ is linearly independent.

However, it still needs to be proven that this $A'$ is a also a generating (or spanning) set for $B$. Here's where I need a hand.

P.S. A proof cannot use the fact that a division ring satisfies IBN property.

Answer 1

You just need to prove that $a$ is in the span of $A'$.

Since $b$ is in the span of $A$, we can write it as $b=ar+v$, where $v$ is in the span of $S$. By the choice of $b$, we have $r\ne0$. Thus $a=br^{-1}+vr^{-1}$.

Answer 2

Every module over a division ring is free and all bases of such a module $M$ have the same number of elements. So in the finite dimensional case, say with $\dim M = d$, any $d$-element set of linearly independent elements is a basis and, therefore, is spanning. In your proof we have $A \setminus \{a\} \cup \{b\}$ is a linearly independent set with dimension many elements, and so it is a basis of $M$.

This exchange property is known as the basis exchange property in matroid theory. One can find a brief discussion of matroids defined over division rings at the end of Section 6.1 in Oxley's famous textbook Matroid Theory.