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How to prove that only four dimensional division algebra (noncommutative) over $\Bbb{Q}$ is rational quaternions?

After a bit of internet research, I am very sure about the above statement, if not please surprise me!

How to proceed at such statement? Just letting $D$ be a $4$ dimensional division algebra, is not taking me forward.

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    $\begingroup$ Presumably you’re talking about central division algebras:; those whose center is $\Bbb Q$. Otherwise, any quartic field extension of $\Bbb Q$ satisfies. $\endgroup$
    – Lubin
    Sep 29, 2015 at 18:11

2 Answers 2

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It's not true. It would be true if $\mathbb Q$ were replaced by $\mathbb R$ in your statement, and the rational quaternions replaced with the real quaternions. There is an infinite family of pairwise non-isomorphic quaternion algebras over $\mathbb Q$ which are simple central divison algebras of dimension $4$ over $\mathbb Q$, and of which the rational quaternions are one example.

Moreover any field extension of $\mathbb Q$ of degree $4$ is a commutative division algebra over $\mathbb Q$ and there are infinitely many of those as well.

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  • $\begingroup$ Fine answer, can be improved only by giving an example. $\endgroup$
    – Lubin
    Sep 29, 2015 at 18:09
  • $\begingroup$ So, you are taking essential a subfield of the real quaternions, sort of like $\mathbb Q[i\sqrt{a},j\sqrt{b}]$, where $a,b$ are positive non-square integers? Also what happens when you take the algebraic reals as your original field? Is that enough to assure uniqueness? What if you take only the real surds (containing the rationals and closed under square roots for positive values.) $\endgroup$ Sep 29, 2015 at 18:09
  • $\begingroup$ Well, @ThomasAndrews, as to your first question, I tried a “random” construction of a different quaternion algebra over $\Bbb Q$, and I went down in flames. As to your second question, you’re asking about the Brauer group of some kind of real $2$-extension of $\Bbb Q$. $\endgroup$
    – Lubin
    Sep 30, 2015 at 1:06
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I think I will fill out Bruno’s excellent answer with an example and a few facts.

For this discussion, “$\infty$” is a prime, and the completion of $\Bbb Q$ at infinity is $\Bbb R$. The completion of $\Bbb Q$ at an ordinary prime $p$ is the $p$-adic number field $\Bbb Q_p$. Any central division algebra over $\Bbb Q$ of rank $4$ must be “ramified” at evenly many primes. The rank-$4$ division algebra $\Bbb A$ is “unramified at $p$” if $\Bbb A\otimes_{\Bbb Q}\Bbb Q_p$ is isomorphic to the ring of $2$-by-$2$ matrices over $\Bbb Q_p$, and “ramified at $p$” if that tensor product is still a rank-four central division algebra over $\Bbb Q_p$.

If you haven’t seen tensor product $\otimes$ before, don’t panic. It just means using the same formulas for multiplication of the basis elements, but taking the scalars to be in $\Bbb Q_p$ instead of $\Bbb Q$. The regular quaternion division algebra $\Bbb H$ that you know is ramified only at $\infty$ and $2$. This means that if you use $\Bbb R$-coefficients, you still get a division algebra (as you know), but you also get a division algebra if you allow $\Bbb Q_2$-coefficients instead. For any other prime $p$, if you allow $\Bbb Q_p$-coefficients, the resulting noncommutative algebra will be isomorphic to the ring of $2$-by-$2$-matrices over $\Bbb Q_p$.

Oh yeah, you say, what about a rank-four division algebra ramified only at $3$ and $5$? I’ll give you the constants, but I’m afraid I’ll have to leave it up to you to check that my claims are all right. The basis is $\{1,r,s,t\}$, where $r$ behaves like a $\sqrt3$, $s$ like a $\sqrt5$, and $t$ behaves like a $\sqrt{-15}$. Here’s the rules for multiplying the units:

$r^2=3$, $s^2=5$, $t^2=-15$, $rs=t=-sr$, $rt=3s=-tr$, and $ts=15r=-st\,$.

The important thing is the “reduced norm form”, which is what you get when you start with $a+br+cs+dt$, which generates a quadratic subfield of the algebra, and look at its field-theoretic norm (the constant term in its characteristic polynomial). This is $$ Q(a,b,c,d)=a^2-3b^2-5c^2+15d^2\,, $$ and you can check, just by looking at integer values of $a,b,c,d$ modulo $9$, that there are no nontrivial rational zeros $(a,b,c,d)$ of $Q$. It is this fact that guarantees that the quaternion algebra defined by the the relations above among the basis elements is indeed a division algebra. I don’t have the courage to verify that this algebra is unramified (“splits”) at all other primes than $3$ and $5$, but I’m pretty sure that this is the case.

Oh, yes, and one other fact about central division algebras over $\Bbb Q$ that you would not have expected just from knowing about $\Bbb H$ as an $\Bbb R$-division algebra: for every $n>1$, there are central division algebras over $\Bbb Q$ of rank $n^2$.

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  • $\begingroup$ Thank you Lubin for giving a thorough answer. I'm short on time these days, otherwise I would have happily upgraded my answer as you suggested. But, in any case, I couldn't have done better than this! $\endgroup$ Oct 1, 2015 at 19:14
  • $\begingroup$ @BrunoJoyal, the retired old guy usually has lots of time. $\endgroup$
    – Lubin
    Oct 1, 2015 at 22:18
  • $\begingroup$ And lots of wisdom! :) $\endgroup$ Oct 2, 2015 at 0:25

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