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.