Here $\left[\frac{a}{p}\right]_q$ refers to the $q$-th power rational reciprocity symbol, with $q,p$ prime. That is

$\left[\frac{a}{p}\right]_q=1$ if $q\nmid p-1$ or $a^{\frac{p-1}{q}}=1\bmod p$

$\left[\frac{a}{p}\right]_q=-1$ if $q|p-1$ and $a^{\frac{p-1}{q}}\neq 1\bmod p$.

Note that when $q=2$, this corresponds to the Legendre symbol $\left(\frac{a}{p}\right)$. And further, we can utilize quadratic reciprocity and the fact that $\left(\frac{p}{a}\right)=-1$ for infinitely many primes $p$ to show that $\left(\frac{a}{p}\right)=-1$.

Due to the lack of rational* power-reciprocity laws, reciprocity cannot be used to show this for a general $q$. My question is, is there another way we can show it to be true that

$\left[\frac{a}{p}\right]_q=-1$ for infinitely many primes $p$, $p\equiv 1\pmod q$?

*there is power reciprocity laws for rings of integers in cyclotomic fields (https://en.wikipedia.org/wiki/Eisenstein_reciprocity)

  • $\begingroup$ Wouldn't $p|a^{\frac{p-1}{q}}+\zeta_q$, for some $\zeta_q^p=1\mod p$, instead of $a^{q^k}+1$? I think this would make your approach harder as $\zeta_q$ changes. Or am I misunderstanding your approach? Why are you raising to the power of $q^k$? $\endgroup$
    – Tejas Rao
    May 21, 2021 at 20:24
  • $\begingroup$ Yeah, my idea doesn’t work. $\endgroup$ May 21, 2021 at 20:47
  • 3
    $\begingroup$ Assuming $a\not \in \Bbb{Q}^{*q}$, use Chebotarev theorem in $\Bbb{Q}(\zeta_q,a^{1/q})$ looking at the automorphism $\sigma$ sending $\zeta_q$ to itself and $a^{1/q}$ to $\zeta_q a^{1/q}$. @ThomasAndrews $\endgroup$
    – reuns
    May 21, 2021 at 21:10

1 Answer 1


For $q$ prime if $a\not \in \Bbb{Q}^{*q}$ then $x^q-a\in \Bbb{Q}[x]$ is irreducible (if $q$ is odd then $x^q-a$ is Eisenstein at some prime factor of $a$, the remaining case is $q=2,a <0$),

so that $q(q-1) \ | \ [\Bbb{Q}(\zeta_q,a^{1/q}):\Bbb{Q}]$ and $\sigma:(\zeta_q,a^{1/q})\to (\zeta_q,\zeta_q a^{1/q})$ is in $Gal(\Bbb{Q}(\zeta_q,a^{1/q})/\Bbb{Q})$.

By Chebotarev theorem (is there a more elementary way?) there are infinitely many primes $p$ and a prime ideal $\mathfrak{p}\subset O_{\Bbb{Q}(\zeta_q,a^{1/q})}$ above it such that $\forall b\in O_{\Bbb{Q}(\zeta_q,a^{1/q})}, \sigma(b)-b^p\in \mathfrak{p}$.

$\sigma$ generates the Galois group of $O_{\Bbb{Q}(\zeta_q,a^{1/q})}/\mathfrak{p}$, for $p\nmid aq$, that it fixes the reduction of $\zeta_q$ implies that $q| p-1$ and that $\sigma$ doesn't fix $a^{1/q}$ implies that $a\bmod p$ isn't a $q$-th power ie. $a^{(p-1)/q}\not \equiv 1\bmod p$.

  • 1
    $\begingroup$ Thank you! On a side note, I see you everywhere on math.stackexchange reuns, good work :) $\endgroup$
    – Tejas Rao
    May 22, 2021 at 0:51

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