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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 $F$)?

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    $\begingroup$ You can always make a quaternion algebra which ramifies precisely at any finite even number of places. $\endgroup$
    – Kimball
    Commented Jun 17, 2021 at 13:28
  • $\begingroup$ Hey, yes, thank you, I have recently come across this theorem, and that does fix it! $\endgroup$
    – Doryan Temmerman
    Commented Jun 18, 2021 at 13:13

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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.

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  • $\begingroup$ Thanks a lot for this quick and clear answer. I was still wondering about the $[F:Q] = 2$ case. What would be the suitable conditions on $a$, $b$ for it to not be split at some finite place? Say we take $F = Q(\sqrt 2)$, can I write down conditions for $a$ and $b$ so that it is not split at the prime 3? What would that look like? $\endgroup$
    – Doryan Temmerman
    Commented Jun 15, 2021 at 14:32
  • $\begingroup$ @DoryanTemmerman You can find an explicit condition here - we want to pick $a,b$ so that one is a uniformizer, and another is a special kind of unit. In your specific example, you may take $a,b\in\mathbb Z[\sqrt{2}]$ such that $a\equiv -3\pmod 9,b\equiv\sqrt{2}\pmod 3$. $\endgroup$
    – Wojowu
    Commented Jun 15, 2021 at 14:51
  • $\begingroup$ I have tried taking $a = -3$ and $b = 3 + \sqrt 2$, so that they satisfy the condition of your first answer and the congruences just above. Plugging this into GAP I get a split algebra however... Also, I see where the $a \equiv -3 \mod 9$ comes from (so that $a$ becomes a uniformizing parameter), but does $b \equiv \sqrt 2 \mod 3$ correspond to a non-square unit of $\mathcal{O}_{\mathbb{Q}_3(\sqrt 2)}$? $\endgroup$
    – Doryan Temmerman
    Commented Jun 15, 2021 at 16:16
  • $\begingroup$ @DoryanTemmerman Woops, that's my bad, $\sqrt{2}$ is a square. $b\equiv 1+\sqrt{2}$ should work though. $\endgroup$
    – Wojowu
    Commented Jun 15, 2021 at 16:46
  • $\begingroup$ I agree that $1 + \sqrt{2}$ is a unit and should not be a square in $\mathcal{O}$, however when I try the constants $a = -3$ and $b = 4 + \sqrt{2}$, GAP still tells me that the algebra $\left( \frac{a,b}{\mathbb{Q}(\sqrt{2})} \right)$ is not a division ring. $\endgroup$
    – Doryan Temmerman
    Commented Jun 15, 2021 at 19:06

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