Show that $D \cong {\rm End}_A(D^n)$ where $D$ is a division algebra and $A\cong M_n(D)$

Question

Define $A$ to be a finite dimensional simple algebra over a field $k$. $D$ is a $k$-division algebra (not necessarily commutative) such that $A\cong M_n(D)$ for some integer $n$. Let $L$ be a minimal left ideal of $A$. Then $L\cong D^n$.

I want to show that $D\cong {\rm End}_A(D^n)$. Do I do this by mapping the matrix of $f\in {\rm End}_A(D^n)$ to its determinant?

Many thanks!

Answer 1

EDIT: The below is wrong if $D$ is not commutative. Not sure what I was thinking 5.5 years ago, so not sure what the thought process was/whether it's salvagable.

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Note that $$\text{End}_D(D^n)=\text{Mat}_n(D)=A$$

Taking the center on both sides yields

$$Z(\text{End}_D(D^n))\cong D$$

But, the center of $\text{End}_D(D^n)$ is just the matrices which are $\text{Mat}_n(D)$-linear. So, by tracking this through the isomorphism $A\cong \text{Mat}_n(D)$, you see this turns into the statement $Z(\text{End}_D(D^n))\cong \text{End}_A(D^n)$.

Answer 2

There is a natural ring homomorphism $\theta:D\to End(_RV)$ given by right multiplication by elements of $D$. That is:

$\theta(d)=\text{right multiplication map by $d$ on $V$}$,

i.e. $(\theta(d))(v)=v\cdot d$

This is immediately injective since $D$ is a division ring. So it remains to show that it is surjective. This involves little more than some elementary linear algebra.

Let $f\in End(_RV)$ be some element, and (writing it as acting on the right of column vectors) suppose

$$\begin{bmatrix}1 \\ 0 \\ \vdots \\ 0\end{bmatrix}f=\begin{bmatrix}d \\ \ast \\ \vdots \\ \ast\end{bmatrix}$$

for some element $d\in D$.

Then

$$ \begin{bmatrix}a_1 \\ a_2 \\ \vdots \\ a_n\end{bmatrix}f=\begin{bmatrix}a_1 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots \\ a_n & 0 & \cdots & 0\end{bmatrix}\begin{bmatrix}1 \\ 0 \\ \vdots \\ 0\end{bmatrix}f=\begin{bmatrix}a_1 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots \\ a_n & 0 & \cdots & 0\end{bmatrix}\begin{bmatrix}d \\ \ast \\ \vdots \\ \ast\end{bmatrix}=\begin{bmatrix}a_1d \\ a_2d \\ \vdots \\ a_nd\end{bmatrix}=\begin{bmatrix}a_1 \\ a_2 \\ \vdots \\ a_n\end{bmatrix}\theta(d) $$

This shows that $f=\theta(d)$, proving surjectivity.

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This is a special case of the double centralizer theorem for Artinian rings. You can find many versions of the proof online and in textbooks.