The set of all finite rank operators forms a simple ring and contained in every nonzero ideal of $Hom_D(V,V)$? Let $V$ be an infinite dimensional vector space over a division ring $D$.
If $F$ is the set of all finite rank operators, that is $F=\{ \theta \in Hom_D(V,V) | dim(Im(\theta))<\infty \}$.
How to show $F$ is a simple ring, and $F$ is contained in every nonzero ideal of $Hom_D(V,V)$?
 A: In these arguments, there is a bit of clerical work one needs to do to actually write the arguments down in detail. I leave those to you and just explain the general ideas.

Lemma 1: If $f$ is a nonzero element of $End(V_D)$ with finite dimensional image, then the ideal $(f)$ generated by $f$ contains all other such maps (and zero.)
Proof start: If you are trying to generate the linear map $g$ from $f$ and the image of $g$ has dimension less or equal to the image of $f$, then things are easy. This assumption allows you to fabricate a linear map $r_1$ such that $Im(r_1f)=Im(g)$, and then by taking a basis of $Im(g)$ and looking at preimages through $r_1f$, one fabricates a second linear map $r_2$ such that $r_1fr_2=g$.
So the question is what to do when the $Im(g)$ has dimension greater than that of $Im(f)$. Since $f$ is nonzero, you can create an $r_3$ such that $Im(r_3f)\nsubseteq Im(f)$, and then $Im(f+r_3f)$ has dimension greater than $Im(f)$. You can repeat this many times to make a map in $(f)$ whose image has arbitrarily large (finite) dimension. Then you go back to the first paragraph.

I hope it is also obvious that if you have a map $h$ with infinite dimensional image, you can find an $r_4$ such that $Im(r_4h)<\infty$? You can just project onto finitely many coordintes, for example. At this point, you have enough information to thoroughly prove that the ideal consisting of transformations with finite dimensional images is the minimum nonzero ideal of the ring.

Lemma 2: $End(V_D)$ is a von Neumann regular ring.

You may or may not already know this... and it is a good exercise either way. I can provide more help if necessary.  And finally,

Lemma 3: Let $I$ be an ideal of a von Neumann regular ring $R$, and let $K$ be an ideal of $I$ (meaning that $K$ is an additive subgroup of $I$ such that $IK\subseteq K$ and $KI\subseteq K$.) Then $K\lhd R$.

This is the easiest of the three lemmas. The property that for every $a$ there exists an $x$ such that $axa=a$ is the way to go. Now you'll know that $F$ has to be simple: because if it wasn't, there'd be another ideal of $End(V_D)$ contained in $F$, but we know that there aren't any (other than $\{0\}$).
