# Voight's Quaternion Algebras, Corollary 7.1.2

I am reading the proof that if $$B$$ is a central simple $$F$$-algebra of dimension $$4$$ ($$\operatorname{char}(F) \neq 2$$), then $$B$$ is a quaternion algebra in John Voight's book Quaternion Algebras (Corollary 7.1.2), and have a question on the proof. The author first reduces to the case that $$B$$ is not a division algebra, then lets $$I$$ be a nontrivial left ideal with minimal dimension $$m$$. The author then says the following:

Then there is a nonzero homomorphism $$B \to \operatorname{End}_F(I) \cong M_m(F)$$ which is injective, since $$B$$ is simple. By dimension, we cannot have $$m = 1$$; if $$m = 2$$, then $$B \cong M_2(F)$$ and we are done. So suppose $$m = 3$$. Then by minimality, every nontrivial left ideal of $$B$$ has dimension $$3$$. But for any $$\alpha \in B$$, we have that $$I \alpha$$ is a left ideal, so the left ideal $$I \cap I\alpha$$ is either $$\{0\}$$ or $$I$$; in either case, $$I \alpha \subset I$$ and $$I$$ is a right ideal as well. But this contradicts the fact that $$B$$ is simple.

I know that I'm missing something here, because I don't see how the contradiction relies on the assumption that $$m = 3$$. It appears to me that if $$m = 2$$, then we can play the same game: for any $$\alpha \in B$$, $$I\alpha$$ is a left ideal, and so $$I \cap I\alpha$$ is as well. But then $$I \cap I\alpha \subset I$$ and by minimality of the dimensions of $$I$$, we have that $$I \cap I \alpha = \{0\}$$ or $$I \cap I \alpha = I$$. But then $$I\alpha \subset I$$, so $$I$$ is a right ideal, contradiction. What is wrong here? How is Voight using the fact that every nontrivial left ideal of $$B$$ has dimension $$3$$ to get his contradiction?

There is a major glitch in this argument: if $$I \cap I \alpha = \{0\}$$, then this does not imply that $$I\alpha \subseteq I$$! This is just hogwash.
But what is true is that when $$m=3$$, we cannot have $$I \cap I \alpha =\{0\}$$, just by dimensions: these are both $$F$$-vector spaces of dimension $$3$$, so if their intersection is $$\{0\}$$ then they span a $$2\cdot 3 = 6$$-dimensional space, a contradiction. Then the rest of the argument proceeds as there.