# A ring that has exactly 7 left ideals (T. Y. Lam)

Exercise 3.25 in Lam's First Course states:

Let $R$ be a ring that has exactly seven nonzero left ideals. Prove that one of them is an ideal (i.e. left and right) and provide an example of such a case.

I also checked the solution book and the proof starts like this:

Assume the contrary, that $R$ does not have proper ideals. Then $R$ is a simple ring. Also, because it has a finite number of left ideals, it must be left artinian. So, from Wedderburn-Artin, we have that $R=\mathbb{M}_n D$, for a division ring $D$ and some $n \geq 2$.

If $n \geq 3$, we can construct more than 7 left ideals (using column-like matrices), so $n$ must be 2.

Now let $\alpha:$ { lines through the origin in $D^2$, as a right $D$-vector space} $\rightarrow$ {nontrivial left ideals of $R$} defined by taking annihilators.

Then $\alpha$ is bijective, so $\#D=6$, a contradiction.

Could any of you please describe more specifically what does $\alpha$ do? I didn't understand how it is defined, let alone it being bijective.

Thank you.

-
if $v\in D^2$, $v\neq 0$, then $\alpha(vD)=\{A\in\mathbb{M}_2D;Av=0\}$. –  user8268 Mar 16 '12 at 13:54
Thanks. That clears things up a little... I'll come back if there's anything left. –  AdrianM Mar 18 '12 at 14:39
@user8268 It seems your answer helped the OP. Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. –  Julian Kuelshammer Jun 9 '13 at 18:59

A line through the origin is given by $vD$ for some $0\neq v\in D^2$. We define (by taking annnihilators) as $$\alpha(vD):=\{A\in \mathbb{M}_2D|Av=0\}.$$