Absolutely irreducible/simple $A$-module iff Endomorphism ring consists of scalar matrices

abstract-algebradivision-algebrasmodulesnoncommutative-algebra

Let $A$ be a non-commutative $K$-algebra (where $K$ a field),
whose underlying $K$-vector space is finite dimensional.

Definition An $A$-module $M$ is said to be
absolutely irreducible or abs. simple if for every extension field $E$ of $K$,
the $M \otimes_K E$ is a irreducible $A \otimes_K E$-module.

Remark. The terminology simple and irreducible module means the same.

Theorem 2.3. (from these notes:
https://www.imsc.res.in/~amri/topics/modular.pdf )
An irreducible $A$-module $M$ is absolutely irreducible if
and only if every $A$-module endomorphism of $M$ is multiplication by a
scalar in the ground field, ie $\operatorname{End}_A(M) =K$.

I not completely understand the proof of the direction
$\operatorname{End}_A(M) =K$ implies
$M$ absolutely irreducible:

Proof. We know from Schur’s lemma that
$D := \operatorname{End}_A(M)$ is a division
ring. This division ring is clearly a finite dimensional vector space over
$K$ (in fact a subspace of $\operatorname{End}_K(M)=K^{d^2}$).
The image $B:=s(A)$ of $A$ in $\operatorname{End}_K(M)$ under canonical
map $s:A \to \operatorname{End}_K(M)$ induced by $A$-module structure is
a matrix algebra $M_n(D)$ over $D$. Why?
$M$ can be realised as a minimal left
ideal in $M_n(D)$. M is an absolutely irreducible $A$-module if and only
if it is an absolutely irreducible $B$-module.

If $\operatorname{End}_A(M) = K$, then $B = M_n(K)$, and $M \cong K^n$.
$B
\otimes_K E = M_n(E)$, and $M \otimes_K E = E^n$.
Thus $ M \otimes_K E$ is clearly an irreducible
$B \otimes_K E $ module. Therefore, $M$ is absolutely irreducible.

Several steps appear nebulous to me.

  1. Why is the image $B:=s(A)$ of $A$ in $\operatorname{End}_K(M)$ is
    a matrix algebra $M_n(D)$ over $D$? Wedderburn–Artin theorem implies that $B$ is a direct product of matrix algebras $\operatorname{Mat}_{n_i}(D_i)$, $D_i$ divisor rings. Why $B$ consists of only one such factor?

  2. How $M$ should be realised as a minimal left ideal in $M_n(D)$. Is there used tacitly certain structure result?

Best Answer

For your first question, to go from a product of $M_n(D_i)$ to a single one, observe that the representation of $B$ is both faithful and simple. If we had a product of matrices over division rings, then we would have nontrivial central idempotents which act non trivially by faithfulness, contradicting simplicity.

For your second question, I think we are tacitly using the structure theorem of modules over $M_n(D)$, that they are all direct sums of the standard module $D^n$. It doesn’t seem relevant that we can realise $M$ as a minimal left ideal, aside from potentially making things more concrete (thinking of $M$ as matrices supported on a single column).

Overall one can think of this proof as being entirely a fact about the representation theory of matrices over division algebras, and the first part is the reduction step to this case.

It’s also probably worth keeping in mind that as categories, modules over $D$ and modules over $M_n(D)$ are equivalent, given by tensoring with the standard binodule. So after this first reduction step, one can pretend $n=1$ to make things a bit more concrete.

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