Is the length of a module over a simple artinian ring an invariant?

Question

If $R$ is a simple artinian ring, Wedderburn theory tells us that $R=Mat_n(D)$ for some $n\geq 1$ and division ring $D$ and also every $R$-module $M$ is a direct sum of finitely many copies of the simple $R$-module $D^n$.

Is the number of copies of $D^n$ is an invariant of $M$?

Many thanks.

Answer 1

and also every $R$-module $M$ is a direct sum of finitely many copies of the simple $R$-module $D^n$.

This is false. Every finitely generated module is a sum of finitely many copies, though. There are certainly modules for semisimple (+Artinian) rings which aren't a finite sum of simple $R$ modules, even if $R$ itself is finite.

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Is the number of copies of $D^n$ is an invariant of $M$?

If you are asking if the number of copies is well-defined for finitely generated modules, then yes.

The Krull-Schmidt theorem says that when the number of copies is finite, the number is uniquely determined.

Answer 2

If $R$ is simple artinian with unique irreducible left $R$-module $I$, and if $D := \text{End}_R(I)$ denotes the associated division ring, then Wedderburn's theorem tells us that $I\otimes_{D^{\text{op}}} -: D^{\text{op}}\text{-Mod}\to R\text{-Mod}$ is an equivalence of categories with inverse $\text{Hom}_R(I,-): R\text{-Mod}\to D^{\text{op}}\text{-Mod}$. In particular, for any $R$-module $M$ the canonical map $I\otimes_{D^{\text{op}}} \text{Hom}_R(I,M)\to M$ is an isomorphism, so that any $R$-module is a direct sum of copies of $I$, and given any other isomorphism $M\cong I^{(\kappa)}\cong I\otimes_{D^{\text{op}}} (D^{\text{op}})^{(\kappa)}$, we have $(D^{\text{op}})^{(\kappa)}\cong\text{Hom}_R(I,M)$ as $D^{\text{op}}$-modules, so the cardinality of $\kappa$ is uniquely determined as the rank of $\text{Hom}_R(I,M)$ over $D^\text{op}$.