Suppose $A \cong M\_n(\mathbb{D})$ where $A$ is a simple algebra over division ring $D$. We want to find an explicit isomorphism between $A$ and $M\_n(\mathbb{D})$. I read from Ivanyos et al. (2012) that if we have a rank 1 element $e \in A$ then the left action of $A$ on $Ae$ provides an isomorphism.


When I take a look at other literature, this seems to be a trivial issue, but I am not sure how the construction works. Could anyone explain how this works or refer me to some sources that details this construction.


Thank you!



Solving the explicit isomorphism problem
The element $e\in A$ being of rank one is equivalent to asking $Ae$ to have dimension $n$ over $D$. Now, there is a homomorphism $A\cong End\_D(Ae)$ taking $a\in A$ to the endomorphism $be\mapsto abe$, which is readily seen to be $D$-linear. The map is injective, since otherwise the kernel $\{a\in A:abe=0,\forall b\in A\}$ would be a nontrivial two-sided ideal of $A$, contradicting the assumption that $A$ is simple.


Choosing a $D$-basis of $Ae$ identifies $End\_D(Ae)$ with $M\_n(D)$.


