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Let's say I have a two odd primes, $p, q$ and $K$ is the field $\mathbb{Q}(\zeta_{pq})$. Let's say $\alpha \in \mathcal{O}$ is an arbitrary element in the ring of integers of $K$, $\frak{b} \subset \mathcal{O}$ is a prime ideal of $\mathcal{O}$, and $\alpha \notin \frak{b}$. Not sure what it's called but I'd like to compute a law that would allow me to "flip" the numerator and denominator in the following residue symbol:

$\Big(\frac{\alpha}{\frak{b}}\Big)_{q} = \alpha^{\frac{N(\frak{b}) - 1}{q}} \text{mod } \frak{b}$.

In my personal studies I've found I've had no other options but to resort to these symbols. I'm rather unfamiliar with any other reciprocity laws aside from quadratic, cubic, and quartic and any references or suggestions to learn based on where I stand would be appreciated.

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    $\begingroup$ I suggest that you read F. Lemmermeyer's very nice book on reciprocity laws which certainly has the answer to your question (and he may answer himself). $\endgroup$
    – Henri Cohen
    Commented Jun 19, 2022 at 21:43
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    $\begingroup$ The numerator and denominator of the power residue symbol are not the same kind of object: the denominator is a prime ideal while the numerator is an element and the numerator is sensitive to changing it by a unit multiple, so you can't pretend the numerator is a principal ideal. If $\mathcal O_K$ is a PID then you could impose a convention on the choice of generator of the denominator (like the use of positive primes in quadratic reciprocity and "primary" primes in cubic reciprocity), but in general I don't think it's realistic to expect you can "flip" the top and bottom of the symbol. $\endgroup$
    – KConrad
    Commented Jun 20, 2022 at 21:36
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    $\begingroup$ Keep in mind that in Artin reciprocity, the idea of flipping terms in a residue symbol is abandoned. The version of classical quadratic reciprocity that matches the formulation of Artin reciprocity is not about a relation between $(p|q)$ and $(q|p)$, but about periodicity of $(a|p)$ in $p$: if $p$ and $q$ are odd primes not dividing $a$ and $p \equiv q \bmod 4|a|$, then $(a|p) = (a|q)$. This is the form of quadratic reciprocity found by Euler (before Legendre symbols were created by Legendre). See p. 48 of websites.math.leidenuniv.nl/algebra/artin.pdf for "Euler's reciprocity law". $\endgroup$
    – KConrad
    Commented Jun 20, 2022 at 21:42

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I am not sure what you mean by "to resort to these symbols". If you want to compute their values given $\alpha$ and ${\mathfrak b}$, just use the definition.

Explicit reciprocity laws for higher powers do exist, but have a couple of natural drawbacks. If you look at the computation of $(\frac ab)$ in integers, you will notice that the process of inverting and reducing essentially boils down to applying the Euclidean algorithm. This also works for $n = 3, 4, 5, 8$ and a few other exponents as well, but not in general. As KConrad explained in his comments, reciprocity laws in the sense of Legendre only apply to principal ideals (or, more generally, to ideals whose order in the class group is coprime to $n$).

The natural way of looking at reciprocity in number fields is the modularity property: you have $(\alpha/{\mathfrak a})_n = (\alpha/{\mathfrak b})_n$ if the ideals ${\mathfrak a}$ and ${\mathfrak b}$ lie in the same ray class determined by the abelian extension $K(\sqrt[n]{\alpha})$ of $K = {\mathbb Q}(\zeta_n)$. I have discussed the case $n = 2$ in detail in my recent book on quadratic number fields.

If you absolutely must use classical reciprocity you should learn about Hilbert symbols. In any case, you cannot avoid class field theory if you're interested in higher reciprocity laws.

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