8
$\begingroup$

A rational (definite) quaternion algebra is an algebra of the form

$$ \mathcal{K} = \mathbb{Q} + \mathbb{Q}\alpha + \mathbb{Q}\beta + \mathbb{Q}\alpha \beta $$

with $\alpha^2,\beta^2 \in \mathbb{Q}$, $\alpha^2 < 0$, $\beta^2 < 0$, and $\beta \alpha = - \alpha \beta$.

For a place $v$ of $\mathbb{Q}$, we say that $\mathcal{K}$ splits at $v$ if $\mathcal{K} \otimes \mathbb{Q}_v \simeq M_2(\mathbb{Q}_v)$; otherwise, we say that it ramifies.

This comes up because the endomorphism ring of an elliptic curve over a finite field may be a quaternion algebra.

I have pretty much no intuition for these things. For instance, I'm just thinking about the rational quaternions, and I don't see how tensoring up with $\mathbb{Q}_v$ could introduce zero-divisors. Doesn't the existence of a multiplicative, positive-definite norm preclude this?

I would very much appreciate some examples of quaternion algebras (preferably, examples that arise as endomorphism of elliptic curves and an explanation as to why) which split/ramify at some places. I want to get a feel for what these things "look like."

Thanks!

$\endgroup$
  • 7
    $\begingroup$ Tensoring produces zero divisors for the same reason tensoring fields can produce zero divisors: ${\mathbf C}$ is a field but ${\mathbf C} \otimes_{\mathbf R} {\mathbf C}$ is not a field: its ring structure looks like ${\mathbf C}[X]/(X^2 + 1) = {\mathbf C} \times {\mathbf C}$. Likewise, although the real quaternions ${\mathbf H}({\mathbf R})$ form a division ring, if you tensor this with ${\mathbf C}$, the resulting 4-dimensional ${\mathbf C}$-algebra is not a division ring but in fact is ${\rm M}_2({\mathbf C})$. Do you know about that? $\endgroup$ – KCd Jun 24 '12 at 23:25
  • 4
    $\begingroup$ Positive-definiteness of the norm becomes pretty irrelevant as a constraint against zero divisors when you tensor with the $p$-adics, where positivity has no meaning. And just because $x^2 + y^2$ is positive definite on ${\mathbf R}^2$, that doesn't exactly make it positive definite on ${\mathbf C}^2$. I trust you can find lots of nonzero pairs $(x,y)$ in ${\mathbf C}^2$ such that $x^2 + y^2 = 0$. $\endgroup$ – KCd Jun 24 '12 at 23:27
  • 3
    $\begingroup$ Let $F$ be a field not of characteristic 2. For $a, b \in F^\times$, let $(a,b)_F = F + Fi + Fj + Fij$ where $i^2 = a$, $j^2 = b$, and $ji = -ij$. This is a division algebra unless $ax^2 + by^2 = 1$ has a solution $(x,y)$ in $F$, in which case it is isom. to ${\rm M}_2(F)$. (An equivalent condition is that $b = x^2 - ay^2$ for some $(x,y)$ in $F$, not the same $x$ and $y$ as before of course.) With this in mind, for an odd prime $p$ set $B_p = (-1,-p)_{\mathbf Q}$. This is a definite quaternion algebra and $B_p \otimes_{\mathbf Q} {\mathbf Q}_\ell = (-1,-p)_{{\mathbf Q}_\ell}$. $\endgroup$ – KCd Jun 24 '12 at 23:40
  • 2
    $\begingroup$ For this to be ${\rm M}_2({\mathbf Q}_\ell)$, it is necessary and sufficient that $-p = x^2 + y^2$ for some $x$ and $y$ in ${\mathbf Q}_\ell$. If $\ell$ is an odd prime other than $p$ there are such $x$ and $y$ in ${\mathbf Q}_\ell$ (but not in ${\mathbf Q}$, of course) by Hensel's lemma since we can solve $-p \equiv x^2 + y^2 \bmod \ell$ with either $x$ or $y$ nonzero mod $\ell$ and then lift $\ell$-adically. If $\ell = \infty$ there is no solution to $-p = x^2 + y^2$ in ${\mathbf Q}_\ell = {\mathbf R}$. As for the cases $\ell = p$ and $\ell = 2$, I'll leave it to you to see what happens. $\endgroup$ – KCd Jun 24 '12 at 23:44
  • 3
    $\begingroup$ For $a$ and $b$ in ${\mathbf Q}^\times$, the behavior of $(a,b)_{{\mathbf Q}_p}$ -- is it a matrix algebra or a division algebra -- is governed by whether the $p$-adic Hilbert symbol $(a,b)_p$ is $1$ or $-1$. If you learn about the Hilbert symbol and explicit formulas for it, such as in Serre's Course in Arithmetic, you can generate a huge number of examples yourself. $\endgroup$ – KCd Jun 24 '12 at 23:46
3
$\begingroup$

Let us look at the Hamiltonian quaternions $\mathcal{K}$ with $\alpha^2=\beta^2=-1$. We all know, by Sir William's reasoning, that this algebra ramifies at the infinite place $\mathbb{Q}_v=\mathbb{R}$. Let $p$ be an odd prime. The claim is that there exists a negative integer $-m$ such that it has a square root in $\mathcal{K}$ as well as $\mathbb{Q}_p$. It is easy to find such integers, because several negative integers have square roots in $\mathcal{K}$. This is because $$(ai+bj+ck)^2=-a^2-b^2-c^2$$ for any triple of rational integers $a,b,c$. It is known that for example all odd integers that are not congruent to $7$ modulo $8$ can be written as a sum of three squares. OTOH, any integer that is congruent to a quadratic residue modulo $p$ has a square root in $\mathbb{Q}_p$ by a Hensel lift of the modular square root. Because $p$ is odd, such integers cannot cover the entire residue class $7+8\mathbb{Z}$. The claim follows.

This implies that $\mathbb{Q}_p$ contains a maximal subfield of the quaternion algebra, and that in turn implies that this places splits. Another way of seeing this is that when $z\in\mathbb{Q}_p$ satisfies $z^2=-m$, and simultaneously we have $m=a^2+b^2+c^2$ for some integers $a,b,c$, then the element $$ z\cdot1+a\cdot i+b\cdot j+ c\cdot k\in \mathbb{Q}_p\otimes\mathcal{K} $$ has zero norm, and hence cannot be invertible.

Edit: See Keith Conrad's comment below for a simpler way of showing that Hamiltonian quaternions split at all odd primes $p$.

The prime $p=2$ OTOH ramifies (class field theory also tells that any division algebra must ramify at at least two places). We already suspect as much from the above calculation, because an odd integer $m$ has a square root in $\mathbb{Q}_2$, iff $m\equiv 1\pmod 8$ (so $\sqrt{-m}\in\mathbb{Q}_2$ only, if $m\equiv 7\pmod 8$). But this time we should study the norm of an element $$ q=a_0\cdot1+a_1\cdot i+a_2\cdot j+a_3\cdot k\in\mathbb{Q}_2\otimes\mathcal{K}. $$ The norm is, of course, $$ N(q)=a_0^2+a_1^2+a_2^2+a_3^2. $$ I want to prove that this never vanishes. This will prove that $\mathcal{K}$ ramifies at $p=2$.Without loss of generality (scaling) we can assume that all the coefficients are $2$-adic integers, and that least one of them is a $2$-adic unit. An easy case-by-case analysis then shows that $N(q)$ is not divisible by $8$. Basically this follows from the fact that the squares of all the odd integers are congruent to $1\pmod 8$.

$\endgroup$
  • 7
    $\begingroup$ If $F$ is a finite field of odd characteristic, every element of $F$ is a sum of two squares. More generally, if $a$ and $b$ are nonzero in $F$, the equation $ax^2 + by^2 = c$ always has a solution $(x,y)$ in $F$ for each $c$ in $F$ (rewrite as $ax^2 = c-by^2$ and each side has $(q+1)/2$ values as $x$ or $y$ vary, so by pigeonhole there is an overlapping common value; there are other explanations of this too). In particular, for any odd prime $p$ we can solve $-1 \equiv x^2 + y^2 \bmod p$ and either $x$ or $y$ is nonzero mod $p$. Then by Hensel we get $-1 = r^2 + s^2$ for some $p$-adic (contd) $\endgroup$ – KCd Jun 24 '12 at 23:21
  • $\begingroup$ integers $r$ and $s$. Thus $1 + ri + sj$ is a zero divisor: $(1+ri+sj)(1-ri-sj) = 1 + r^2 + s^2 = 0$. $\endgroup$ – KCd Jun 24 '12 at 23:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.