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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!

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    $K \otimes \mathbb Q_v$ is a product of (more than one) copies of extensions of $\mathbb Q_v$, which explains the zero-divisors2012-06-24
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    Class field theory tells us that any division algebra ramifies at only finitely many places, so splitting is the norm.2012-06-24
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    Well, I trust the general theory, but I'd like to see explicit examples worked out. In terms of splitting of primes in a number field, one can hold out concrete examples; that's what I think I need.2012-06-24
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    I am also interested in examples...most books/papers just do the general theory and somehow the authors are able to just "see" cearly when such a thing splits. It is not so obvious to me...2012-06-24
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    Pick a prime $p$ such that $\left( \frac{\alpha^2}{p} \right) = 1$, for example.2012-06-24
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    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?2012-06-24
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    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 $x^2 + y^2 = 0$.2012-06-24
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    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 $B_p \otimes_{\mathbf Q} {\mathbf Q}_\ell = (-1,-p)_{{\mathbf Q}_\ell}$.2012-06-24
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    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.2012-06-24
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    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.2012-06-24

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