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?


1 Answer 1


Sorry for not seeing this sooner!

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.

I added this to the errata and fixed the online version. Thanks, and sorry for the mistake!


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