Number of left ideals in a simple ring I'm puzzling over a few algebra questions:
1) Give an example of a simple ring with exactly $12$ non-zero proper left ideals.
For this one I have no idea, I am not good with coming up with examples at all.  Any hint would be welcomed.
2) Let $S$ be a ring with $11$ non-zero proper left ideals. Show that $S$ has at least one non-zero proper two-sided ideal.
So for the second one, we can suppose $S$ has no non-zero proper two-sided ideals and then since it has a finite number of left-ideals we can use Artin-Wedderburn to say $S=M(D,n)$ where $D$ is a division ring and $n\ge2$. I think though if $n>3$ there are more than $11$ left-ideals via column-ideals so $n=2$ or $3$ but this is where I'm stuck. I believe I need to use the correspondence between $D$ vector spaces and left ideals.  Would this help? In $D2$ there are $6$ subspaces so that's a contradiction so that's good and in $D3$ I want to say there are $10$ but I am not sure.
 A: You started along the right path to narrowing it down. I'll try to sum up the rest.  (Sorry for switching sides on you, but it's easier to type matrices in latex for the right ideals I have in mind. You get left ideals if you transpose everything, of course.)
If you consider $M_3(\Bbb F_2)$ for a moment, we can do some classification of minimal right ideals. Consider ideals of the following forms:
$$
I_{1,\alpha,\beta}=\{\begin{bmatrix}a&b&c\\\alpha a&\alpha b&\alpha c\\\beta a&\beta  b&\beta c\\\end{bmatrix}\mid a,b,c\in\Bbb F_2\}
$$
$$
I_{2,\alpha}=\{\begin{bmatrix}0&0&0\\ a&b&c\\ \alpha a&\alpha b&\alpha c\end{bmatrix}\mid a,b,c\in\Bbb F_2\}
$$
$$I_3=\{\begin{bmatrix}0&0&0\\0&0&0\\ a&b&c\end{bmatrix}\mid a,b,c\in\Bbb F_2\}$$
This accounts for seven distinct minimal ideals. In each case, each one is uniquely of the form $eR$ for some idempotent $e$, where $(1-e)R$ is a complementary (and necessarily maximal) right ideal. We can conclude that the complements are all distinct as well, bringing us at least 14 right ideals (too many!).
Since any other field would have even more right ideals, we can safely say that we only need to concentrate on $n=2$. There, the mimimal right ideals are also the maximal right ideals, and they are all the nontrivial right ideals, and they are all 2-dimensional.
Now for any field, the sets of the following forms are nontrivial right ideals:
$$I_\alpha=\{\begin{bmatrix}a&b\\\alpha a&\alpha b\end{bmatrix}\mid a,b\in\Bbb F\}$$
$$I'=\{\begin{bmatrix}0&0\\a&b\end{bmatrix}\mid a,b\in \Bbb F\}$$
We can reason that these are all the nontrivial ideals in the following way. Each matrix in a nontrivial right ideal must be nonzero but noninvertible, so their rows must be linearly dependent. Given a particular right ideal $K$, we check to see if it has elements which are nonzero in the top row. If it doesn't, it is $I'$. If it does, then take an element nonzero in the first row $r_1$, and find the scalar $\alpha$ which makes $\alpha r_1=r_2$. Thus $K$ has nontrivial intersection with $I_\alpha$, hence it is equal.
This analysis shows that $M_2(\Bbb F_q)$ has $q+1$ nontrivial right ideals. That means the matrix rings for $q\in\{2,3,4,5,7,8,9,11\}$ are an exhaustive list of the simple rings with no more than 12 proper right ideals, but at least one proper right ideal. Twelve happen in $M_2(\Bbb F_{11})$. And, finally, you can see that none of the possibilities allow 11 nontrivial right ideals.
