I am reading the beginning of John Voight's draft of a book on Quaternion Algebras.
In definition 2.2.1 (page 20) he defines the quaternion algebra $(\frac{a,b}{F})$ by specifying a mutliplication table on a $4$ dimensional vector space over $F$.
In Lemma 2.2.5 (page 21) he says that "an $F$-algebra $B$ is a quaternion algebra if and only if there exists generators $i,j\in B$ such that $i^2=a,j^2=b,ij=-ji$.
I found two ways of interpreting the statement of the lemma, but both are problematic (the fact that $(\frac{a,b}{F})$ satisfies the relations is clear, I am talking about the reverse implications).
Interpretation 1: "as written" - but then, it is only clear that $B$ is a quotient of the presentation $F\{i,j\}/(i^2-a,j^2-b,ij+ij)$, and we don't know that $B$ has the universal property of the presentation (which seems to be used in the proof).
Interpretation 2: "it is actually meant that $B$ is $F\{i,j\}/(i^2-a,j^2-b,ij+ij)$", but then it is immediate that $B$ is a quaternion algebra since by the universal property of the presentation it surjects onto $(\frac{a,b}{F})$, and so it is enough to prove that $\dim_FB\leq 4$ which is clear using the relations (so why bother giving a more complicated proof?).
My question:
- Assuming "Interpretation 2" is correct, is what I say in the discussion of this interpretation correct?
- Assuming "Interpreatation 1" is correct, is the universal property of the presentation really used? How is it justified?
Note: In retrospect, I know that the statement of the lemma is correct because quaternion algebras are simple (so no proper quotients), but this seems not to be known to the reader at this point, and my question is about the basics of quaternion algebras as presented in this book.
You may interpret the statement either way and it would be correct. "Interpretation 1" is stronger, and you're right that it follows from the fact that quaternion algebras are simple (so no further quotients), but we don't know this so early in the book. Also, your short proof of the statement under "Interpretation 2" is correct, as far as I can tell.
Alas, the proof of the statement under "Interpretation 1" uses that the ideal of relations in the free algebra is invariant under $i \mapsto -i$, and we don't know that. (Well, it's true, but there is an argument missing.)
Rather than trying to fill something in, here is a hopefully satisfying replacement proof (and I've just updated the book draft), one that proves the stronger "Interpretation 1". Let $\alpha = t+xi+yj+zij=0$ and compute $$ 0 = i(\alpha i + i \alpha) = 2a(t+xi).$$ Since $2 \neq 0$ in $F$ and $a \neq 0$, we conclude $t+xi=0$. Repeating with $j,ij$, we similarly find $t+yj=t+zij=0$. Thus $$ \alpha - (t+xi)-(t+yi)-(t+zij) = -2t =0 $$ so $t=0$, thus $x=y=z=0$.
In particular, the algebra cannot be a further quotient. (This argument can be used in a similar way to show that a quaternion algebra is simple.)
I'm sorry for any confusion! You can always reach be directly by e-mail if you would prefer--I don't check this site regularly. I'm happy to receive by email any further comments or questions you may have.