1
$\begingroup$

If we have a division algebra $A$, is it just a simple module over itself? Given a submodule $B$ of $A$ and $b \in B$, $\exists$ $b^{-1} \in B:bb^{-1}=1 \in B$, and so $ B = A$.

Is this argument correct?

$\endgroup$
1
$\begingroup$

It is not written 100% correctly, but the idea is fine.

The inaccuracies are that you need to say $b\in B\setminus\{0\}$, and the second thing is that there is no reason to say $b^{-1}$ is in $B$.

If $b\in B\setminus \{0\}$, then $1\in bb^{-1}\in B$ suffices to show $B=A$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.