Splitting of quaternion algebras 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! 
 A: 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 $1\pmod 8$.
