I found in my notes that, for $k$ a field,

$M_n(k)$ is a direct sum of its column left ideals, so it is a semisimple $M_n(k)$-module.

I don't understand the first clause, though I feel it must be some obvious linear algebra.

I mean, what are the column left ideals of $M_n(k)$?

I think of the spaces generated by each element of the canonical basis for $M_n(k)$, i.e. The matrices with a 1-entry and all the other entries zero, but they're not column ideals.


1 Answer 1


Let's do $n=2$. The column left ideals in this case are $$I_1=\left\{\pmatrix{*&0\\*&0}\right\}$$ and $$I_2=\left\{\pmatrix{0&*\\0&*}\right\}$$ in what I hope is an obvious notation. Then $M_2(k)=I_1\oplus I_2$ and moreover $I_1=M_2(k)A$ for any nonzero $A\in I_1$ (and similarly for $I_2$) so that $I_1$ is a simple module etc.

  • $\begingroup$ Oh thanks, now that it is written out it's clearer. And that's also a nice way to see the simplicity of $I_1$. $\endgroup$
    – davideleo
    Sep 15, 2019 at 18:17

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