# For a fixed $a$, can we say $\left[\frac{a}{p}\right]_q=-1$ for infinitely many $p$?

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)

• 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 $q^k$? May 21, 2021 at 20:24
• Yeah, my idea doesn’t work. May 21, 2021 at 20:47
• 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 May 21, 2021 at 21:10

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. $$a^{(p-1)/q}\not \equiv 1\bmod p$$.