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.

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 $ax^2+by^2-abz^2$ 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. For $73x^2+5y^2=365z^2$ 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 $73$) 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.

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 $41$. Then a quaternion algebra ramified at $41$ 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 $a=41$, $b=3$.

The form has two positive eigenvalues and one negative eigenvalue, clearly not definite. The form In the $41$-adic numbers, for $41x^2+3y^2=123z^2$ to have a solution in $\mathbb z_{41}$, $y$ must be a multiple of $41$. Without loss of generality, then, $x$ and $z$ are not multiples of $41$ (otherwise we divide all three by $41$ and check again), so we have $x^2-3z^2=0$ mod $41$. But $3$ is not a quadratic residue mod $41$ so this is impossible.

This form, in $\mathbb Q(\sqrt[4]{2})$, remains ramified in the four primes lying over $41$ but is still split at the infinite primes.

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.