Let $F$ be a totally real number field having *at least* two different real embeddings $\sigma\_1 : F \hookrightarrow \mathbb{R}$ and $\sigma\_2 : F \hookrightarrow \mathbb{R}$.


Does a quaternion algebra $A = \left(\frac{a,b}{F}\right)$ over $F$ exist such that $A$ is not itself a matrix algebra, but which splits at exactly those two infinite primes $\sigma\_1$ and $\sigma\_2$?


If so, can one make it explicit (i.e. provide $a$, $b$ and $F$)?



Does a quaternion algebra exist over a number field that is split over some infinite real places, but not others?
Let $a$ be any totally negative element and pick $b$ to be an element such that $\sigma\_1(b),\sigma\_2(b)$ are positive, while $\sigma(b)$ is negative for all other $\sigma:F\to\mathbb R$. Such elements exist by suitable approximation theorems in number fields. For $A$ defined using these elements, after tensoring by $\mathbb R$ using the first element we get a matrix algebra, while using the other $\sigma$'s we recover Hamilton's quaternions.


It remains to guarantee $A$ itself is not split. If $[F:\mathbb Q]>2$ this is clear as there is some non-split infinite place. Otherwise, we can pick $a,b$ to satisfy suitable approximations at any finite place, so as to guarantee $A$ still has some non-split place.


