Csdrhrt

Let $p$ be a prime number, let $mathbb{Q}_p$ be the local field, by Hensel's lemma, we know it has $p-1$-th roots of unity, let $zeta$ be a fixed primitive $p-1$-th root of unity in $mathbb{Q}_p$.

Let $a,binmathbb{Q}_p^*$, let $(a,b)_{zeta}$ be the cyclic algebra defined by $mathbb{Q}_plangle x,y|x^{p-1}=a,y^{p-1}=b, xy=zeta yxrangle$.

Is there an explicit way to tell which class $[(a,b)_zeta]inmathrm{Br}(mathbb{Q}_p)=mathbb{Q}/mathbb{Z}$ is? (For example, when $a,binmathbb{Q}$, how does the two rational numbers determine an element in $mathbb{Q}/mathbb{Z}$?)

I should clarify going into this that I don't know any class field theory, so I might say some silly things. The number you're trying to compute is the Hasse invariant of the cyclic algebra, and it should also correspond to a Hilbert symbol as in the comments. I found a nice formula for a special case of the Hilbert symbol in Section 3.3 of these notes (Lemma 3.22). Specialized to this case it gives

$$(a, b)_{xi} = omega left( (-1)^{nu(a) nu(b)} frac{b^{nu(a)}}{a^{nu(b)}} bmod p right)$$

where $nu : mathbb{Q}_p to mathbb{Z}$ is the valuation and $omega : mathbb{F}_p^{times} to mu_{p-1}(mathbb{Q}_p)$ is the Teichmüller character.

You asked in the comments about $p = 5, a = 2, b = 3$. In this case $nu(a) = nu(b) = 0$ so the Hilbert symbol is trivial. In order for it to be nontrivial at least one of $nu(a)$ and $nu(b)$ must be nonzero. For example, sticking to $p = 5$, if we take $a = 2, b = 5$ then we get

$$(1, 5)_{xi} = omega left( frac{1}{2} bmod 5 right) = omega(3)$$

which is the unique $4^{th}$ root of unity in $mathbb{Z}_5$ congruent to $3 bmod 5$, and in particular primitive. Unfortunately I don't know exactly how this is identified with an element of $mathbb{Q}/mathbb{Z}$, although it must be either $frac{1}{4}$ or $frac{3}{4}$.