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I'm trying to prove:

If $M$ is a nontrivial finitely generated left module over $M_2 (\mathbb{C})$, then the accompanying $\mathbb{C}$-vector space structure (just restrict the action to the scaling matrices) is at least dimension 2

I've been attempting to prove the contrapositive: assuming I've got a finitely generated module where every $x,y\in M$ satisfies $x=z\cdot y$ for some $z\in \mathbb{C}$, and trying to prove that the module is trivial.

So far I haven't been able to show much. I think I've shown that every $A\in M_2(\mathbb{C})$ has an associated complex number $z_A$ such that $A$ acts on every element in $M$ in the same way that $z_A$ does (where $z_A$ is really acting as does $z_A I_2$, a scaling of the identity matrix in $M_2 (\mathbb{C})$). Combining this with $M$ being finitely generated seems to let me write every element of $M$ in the form $z\cdot m$ for one particular generator $m$ and varying $z\in \mathbb{C}$.

As usual when I don't know what to do, I'm kind of blindly chugging away at it and don't see where this is going, if anywhere.

Any suggestions or comments would be appreciated.

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up vote 1 down vote accepted

You have constructed a map $\phi:A\in M_2(\mathbb C)\to z_A\in\mathbb C$. Show next that

  • $\phi$ is a morphism of $\mathbb C$-algebras, and that

  • the ring $M_2(\mathbb C)$ is simple, that is, that it does not have any non-zero proper bilateral ideal.

Using these two facts, reach a contradiction.

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