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Suppose we have $g_{2} = \sum_{t=0}^{p-1} \omega^{t^{2}}$, where $\omega$ denotes the p^th root of unity, and p is a prime equivalent to $1 \pmod{4}$. Let $q$ be a prime equivalent to $1 \pmod{4}$. How do I show that $g_{2} \in \mathbb{F}_{q} \iff $ $q$ is a quadratic residue mod ${p}$, without using the law of quadratic reciprocity. Similarly, how do I show that $p$ is a quadratic residue mod $q$ if and only if $q$ is a quadratic residue mod $p$, once again, without using the law of quadratic reciprocity?

user100101212
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1 Answers1

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Note that $g_2^2=\sum_{0 \leq k,l < p}{\omega^{k^2+l^2}}=\sum_{k \in \mathbb{F}_p}{c_k\omega^k}$, where $c_k=|\{(x,y) \in \mathbb{F}_p^2, x^2+y^2=k\}|$.

Now, let $i \in \mathbb{F}_p$ be a square root of $-1$. Then, $c_k=|\{(x,y) \in \mathbb{F}_p^2,\,(x+iy)(x-iy)=k\}|=|\{(x,y) \in \mathbb{F}_p^2,\,xy=k\}|$, so $c_k=p-1$ if $k \neq 0$ and $c_0=2p-1$. It follows $g_2^2=p$, thus $g_2 \in \mathbb{F}_q$ iff $p$ QR mod $q$.

Edit: I had misread the question, and indeed, I missed a whole part of it. If $q$ is a QR mod $p$, then the sequences $(t^2 \pmod{p})_{0 \leq t < p}$ and $(qt^2 \pmod{p})_{0 \leq t < p}$ are permutations one of the other. It follows $g_2^q=g_2$ hence $g_2 \in \mathbb{F}_q$.

However, we can see that if $q$ is not a QR mod $p$ and $g_2 \in \mathbb{F}_q$, then the reunion of two sequences is a permutation of $(t \pmod{p})_{0 \leq t < 2p}$, hence $g_2^q+g_2=0$, therefore $g_2=0$ and $p=g_2^2=0$, a contradiction.

So similarly $g_2 \in \mathbb{F}_q$ iff $q$ is a QR mod $p$.

Aphelli
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  • This is only one direction, rifht? – darij grinberg Nov 06 '19 at 20:16
  • No. The identity $g_2^2=p$ always holds in some finite extension of $\mathbb{F}_q$. So $p$ is a square in $\mathbb{F}_q$ iff the roots of $X^2-p$ are in $\mathbb{F}_q$ iff some root of $X^2-p$ is in $\mathbb{F}_q$. Then (2) implies $g_2 \in \mathbb{F}_q$, which implies (3). – Aphelli Nov 06 '19 at 20:23
  • Ah, you're right. – darij grinberg Nov 06 '19 at 20:33
  • Sorry, I am still not understanding. So if $(g_{2})^{2} = p$, how do I know that $g_{2} \in \mathbb{F}{q} \iff q$ is a $QR$ mod p? Then how do you deduce the law of quadratic reciprocity from that result. Essentially, I am not seeing how the conclusions follow from $g{2}^{2 } = p$, so if you could explain it a little slowly, I would appreciate it. – user100101212 Nov 06 '19 at 21:51
  • So I can see that q $ QR$ mod $p \implies g_{2} \in \mathbb{F}{q}$, but I am unclear on the converse. So we can say $g{2} \in \mathbb{F}{q} \implies g{2} ^{q} = g_{2} \implies \sum_{t=0}^{p-1} \omega^{qt^{2}} = \sum_{t=0}^{p-1} \omega^{t^{2}}$. Is this sufficient to claim that ${qt^{2} \pmod{p}}$ ia a permutation of ${t^{2} \pmod{p}$ and hence is a QR mod p. Lastly, how do I use these results to claim that $q$ is a QR $mod p$ if and only $p$ is a QR mod $q$. Thanks for the help. – user100101212 Nov 06 '19 at 22:37
  • The thing is that if $q$ is not a QR mod $p$, then ${qt^2}$ is the “complement” of ${t^2}$ so $g_2^q+g_2=2\sum_{t=0}^{p-1}{\omega^t}=0$, so either $g_2 \notin \mathbb{F}_q$ or $g_2=0$, but since $g_2^2=p$, only the first choice is possible. Note finally that since $g_2^2=p$, $g_2 \in \mathbb{F}_q$ iff $p$ QR mod $q$. But from the above we know $g_2 \in \mathbb{F}_q$ iff $q$ QR mod $p$, hence the quadratic reciprocity law. – Aphelli Nov 06 '19 at 22:44
  • Where did we use the condition that $p,q \equiv 1 \pmod{4}$ ? – user100101212 Nov 06 '19 at 22:53
  • The condition on $q$ isn’t used. The condition on $p$ is used when counting the $c_k$: else I think it’s $c_0=1$, $c_k=p+1$ and thus $g_2^2=-p$. Note that the QR law is the same regardless of $q$ mod $4$ as long as $4|p-1$. – Aphelli Nov 06 '19 at 22:59
  • got it. Thanks alot – user100101212 Nov 06 '19 at 23:02