Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.