Let $\pi=a+bi\in \mathbb{Z}[i]$ be a Gaussian prime with $a$ and $b$ nonzero, and $b$ even. For odd rational primes $p=\pi\bar\pi$ and $q\neq p$, define $\pi^{\frac{1}{2}\left(q-\left(\frac{-1}{q}\right)\right)} = \alpha + \beta i$, then: \begin{align} \left(\frac{p}{q}\right)=-1 &\iff q\vert\alpha,\;\mathrm{and}\\ \left(\frac{p}{q}\right)=1 &\iff q\vert\beta. \end{align}

I was able to prove this using expansions of $\alpha\beta$ for $q=3\;\mathrm{mod}\;4$, and $p\alpha\beta$ for $q=1\;\mathrm{mod}\;4$, which is straightforward but not particularly insightful. I have searched the literature for a proof that ties this result to the broader concepts of quadratic residues, but I have not been able to find any mention of it.

Specifically, my question is, is this a known result? If so, is there a reference that ties it into the broader context of quadratic residues and reciprocity laws?


1 Answer 1


We have the following generalization: Let $K$ be a quadratic extension of $\mathbb Q$, $\mathcal O_K$ the ring of integers, $q$ an odd prime of $\mathbb Q$ unramified in $K$, and $x \in \mathcal O_K$ prime to $q$. Let $e = \frac{ q-1}{2}$ if $q$ splits in $K$ or $e=\frac{q+1}{2}$ if $q$ is inert in $K$.

Then $x^{e} \in \mathbb Z + q \mathcal O_K $ if and only if $\left( \frac{\operatorname{Norm}(x) }{q} \right)=1$ and $\operatorname{tr} (x^e) \in q \mathbb Z$ if and only if $\left( \frac{\operatorname{Norm}(x) }{q} \right)=-1$.

Proof: The key identity here is that $$x^q = \operatorname{Frob}_q(x) = \begin{cases} x & \textrm{if }q\textrm{ split} \\ \overline{x}& \textrm{if }q\textrm{ inert} \end{cases}$$ which combined with the usual formula for the quadratic residue gives (mod $q$)

$$\left( \frac{ \operatorname{Norm}(x)}{q} \right) = \left( \frac{ x\overline{x}}{q} \right) = x^{ \frac{q-1}{2}} {\overline{x}}^{\frac{q-1}{2}} = x^{ \frac{q-1}{2}} {\overline{x}}^{\frac{q-1}{2}} \cdot \frac{ \operatorname{Frob}_q(x)}{x^q} =\begin{cases} x^{ - \frac{q-1}{2}} \overline{x}^{\frac{q-1}{2}} & \textrm{if }q\textrm{ split} \\ x^{ - \frac{q+1}{2}} \overline{x}^{\frac{q+1}{2}} & \textrm{if }q\textrm{ inert} \end{cases}$$ $$ = \frac{ \overline{x}^e}{ x^e} = \frac{ \overline{x^e}}{x^e}.$$

If the quadratic residue symbol is $1$ this gives $\overline{x^e}=x^e$ (mod $q$), so $x^e \in \mathbb Z + q \mathcal O_K$, while if the quadratic residue symbol is $-1$ his gives $\overline{x^e}=-x^e$ (mod $q$), so $\operatorname{tr}(x^e) = x^e + \overline{x^e}=0$ mod $q$.

The proof should probably clarify a little more what's going on here.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.