There exists no $M_n(\mathbb{C})$ module of dimension coprime to $n$

Question

Let $m,n\in\mathbb{N}:\gcd(m,n)=1$ prove there is no $M_n(\mathbb{C})$-module $V$ s.t. $\dim_{\mathbb{C}}V=m$

I think that since $V$ is an $M_n(\mathbb{C})$-module we should have that $\dim_{\mathbb{C}}V\mid n$ and thus we are done but is that really correct? If so how can I prove it?

Answer 1

The ring $A = M_n(\mathbb C)$ is simple, which implies that there is a unique simple left $A$-module $M$, and every left $A$-module is a direct sum of copies of $M$. In this case, one can take $M = \mathbb C^n$ with the natural action of $A$.

So yes, $\dim_{\mathbb C} V$ is divisible by $\dim_{\mathbb C} M = n$.

Answer 2

Clearly we can assume instead it’s a continuous morphism of algebras $\Phi: M_n(\mathbb{C}) \rightarrow M_m(\mathbb{C})$.

Then $\det \circ \Phi: GL_n(\mathbb{C}) \rightarrow \mathbb{C}^{\times}$ is a continuous group homomorphism, so it can be written as $f \circ \det$, where $f: \mathbb{C}^{\times} \rightarrow \mathbb{C}^{\times}$ is a continuous group homomorphism.

We know that for each $z$, $f(z^n)=\det\, \Phi(zI_n)=z^m$. It follows that $n|m$.