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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.

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    if $v\in D^2$, $v\neq 0$, then $\alpha(vD)=\{A\in\mathbb{M}_2D;Av=0\}$.2012-03-16
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    Thanks. That clears things up a little... I'll come back if there's anything left.2012-03-18
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    @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](http://meta.math.stackexchange.com/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868).2013-06-09

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