## Explicit description of a quaternion algebra with a prescribed set of ramified places

Let $k$ be an algebraic number field. I understand that given a finite set of non-complex places $S\subset V(k)$ of even cardinality, there exists a unique quaternion algebra $Q$ over $k$ such that $Q$ ramifies at all $v\in S$ (i.e. $Q \otimes_k k_v$ is a division algebra) and $Q$ splits at all the other places. Is there a way to get an explicit description of $Q$ of the form $Q(a,b|k)$ for some parameters $a,b, \in k^{\times}$?

The reason for my question is to find an explicit example of a division quaternion algebra over the field $\mathbb{Q}(\sqrt[4]{2})$ that splits at all the archimedean places and is moreover of the form $Q_0 \otimes_{\mathbb{Q}(\sqrt{2})} \mathbb{Q}(\sqrt[4]{2})$ for a quaternion algebra $Q_0$ defined over $\mathbb{Q}(\sqrt{2})$.

Any hint on the general question or on the specific example is greatly appreciated. Thank you!

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The theory of the Hilbert symbol reduces your question to a bunch of congruence conditions on $a,b$. – David Hansen Jul 7 at 6:01

To have zero-divisors, the norm of an an element must be $0$. The norm of an element is a three-variable quadratic form. You need a quadratic form that has zeroes at the split primes but not at the non-split primes.

Wikipedia provides an explicit description of all quaternion algebras, which gives quadratic forms ore of type $ax^2+by^2-abz^2$. So the problem is to find a quadratic form of type $x^2+by^2-abz^2$ which fails to have roots at some specified set of primes. I'm not sure exactly how to do this.

It's much easier if the requirements are not exact. For simplicity, I'm going to find a quaternion algebra defined over $\mathbb Q$ and tensor up to $\mathbb Q(\sqrt[4]{2})$. Take a split prime, say $73$. Then a quaternion algebra ramified at $73$ will remain ramified in the extension. To ensure that it is split at $\infty$, the only thing we need to check is that the form is not definite, ensuring a solution in $\mathbb R$. Take $a=73$, $b=5$.

The form has two positive eigenvalues and one negative eigenvalue, clearly not definite. Foror $73x^2+5y^2=365z^2$ to have a solution in $\mathbb Z_{73}$, $y$ must be a multiple of $73$. Without loss of generality, then, $x$ and $z$ are not multiples of $73$ (otherwise we divide all three by $73$ and check again), so we have $x^2-5z^2=0$ mod $73$. But $5$ is not a quadratic residue mod $73$ so this is impossible.

This form, in $\mathbb Q(\sqrt[4]{2})$, remains ramified in the four primes lying over $73$ but is still split at the infinite primes.

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Thanks a lot, this example is very helpful. Do you mind telling me why you chose 41? How do you know it is completely split? I can see that it is split in $\Q(\sqrt{2})$, since $8$ is a quadratic residue mod 41. Is there a similar criterion for extensions of degree 4? – anonymous Jul 9 at 10:51
What you need are four different roots to the equation $x^4-2=0$, so four fourth roots of $2$. The two things you can check with simple congruence conditions are the existence of a square root of $2$ and a square root of $-1$. Then you look at all primes satisfying this congruence condition until you find one where $2$ is a fourth power. You can do this by computing the quartic character $2^{(p-1)/4}$, for instance. If I computed correctly then the first one I found was $41$. – Will Sawin Jul 9 at 17:38
Actually $41$ doesn't work. $73$ does though, since $2^18=1$ modulo $73$. – Will Sawin Jul 9 at 17:45
$2$ is a quartic residue mod $p \equiv 1 \bmod 4$ iff $p = 64a^2+b^2$. The first example is indeed $73$, with $(a,b) = (1,3)$. Since also $73 = 3^4 - 2^3$ we have $2^3 \equiv 3^4 \bmod 73$, so $2^4 = 2 \cdot 3^4$ and we get the explicit 4th root $2/3 \equiv 25$ of $2 \bmod 73$. – Noam D. Elkies Jul 10 at 5:48