1
$\begingroup$

This is exercise 28 from Noncommutative Algebra / Farb & Dennis, I.1:

(a) If $V$ is a vector space of countably infinite dimension over a field $k$, show that the set of finite rank operators (i.e., those elements of $End_k(V)$ whose image is finite dimensional) forms a two-sided ideal in $End_k(V)$; hence $End_k(V)$ is not simple, in contrast to the fact that finite endomorphism rings of finite dimensional vector spaces are simple.

(b) Use part (a) to construct a simple ring which is not semisimple.

My attempt at (a):

It seems like I can prove this without the assumption that $V$ is of countable dimension, which makes me suspect I have a mistake:

The finite rank opearators are an additive subgroup of $End_k(V)$ because if $\varphi(V)$ and $\psi(V)$ are spanned by $B:=\{b_1,\dotsc,b_m\}$ and $C:=\{c_1,\dotsc,c_n\}$ respectively, then $(\varphi + \psi)(V)$ is spanned by $B \cup C$.

Now let $\varphi$ be a finite rank operator and let $\psi \in End_k(V)$ such that $\varphi(V)$ is spanned by the finite $B \subset V$,then $(\varphi \circ \psi)(V)$ is finite dimensional as a subspace of $\varphi(V)$ and $(\psi \circ \varphi)(V)$ is finite dimensional because it is spanned by $\psi(B)$.

Is anything wrong here?

  • 1
    Looks good to me!2011-11-23

1 Answers 1

1

You're absolutely right: the countability of $\operatorname{dim}_k V$ is not used in the proof of part a).

However, it will be used in the proof of part b). (Hint: suppose that $V$ had uncountable dimension $\kappa$. Then the collection of all operators of at most countable rank forms a strictly larger two-sided ideal in $\operatorname{End}_kV$. Why is this a problem?)

  • 0
    @Kim: Everything you say is correct. (And you're welcome.)2011-11-23