---
abstract: 'We provide a simple proof of the general rational quartic reciprocity law due to Williams, Hardy and Friesen.'
address: 'Mörikeweg 1, 73489 Jagstzell, Germany'
author:
- Franz Lemmermeyer
title: Rational Quartic Reciprocity
---

In 1985, K. S. Williams, K. Hardy and C. Friesen [@11] published a reciprocity formula that comprised all known rational quartic reciprocity laws. Their proof consisted in a long and complicated manipulation of Jacobi symbols and was subsequently simplified (and generalized) by R. Evans [@3]. In this note we will give a proof of their reciprocity law which is not only considerably shorter but which also sheds some light on the raison d’être of rational quartic reciprocity laws. For a survey on rational reciprocity laws, see E. Lehmer [@7].

We want to prove the following

Let $m \equiv 1 \bmod 4$ be a prime, and let $A, B, C$ be integers such that $$\begin{array}{rl} A^2=m(B^2+C^2),    \qquad & 2 \mid B, \\
   (A,B) = (B,C) = (C,A) = 1,\qquad & A+B \equiv 1 \bmod 4.
\end{array}$$ Then, for every odd prime $p>0$ such that $(m/p) = + 1$, $$\label{E1}
\left(\frac{A+B\sqrt m}{p}\right) =  \left(\frac{p}{m} \right)_4 .$$

Let $k = {\mathbb Q}(\sqrt{m}\,)$; then $K = {\mathbb Q}(\sqrt{m}, \sqrt{A+B\sqrt{m}}\,)$ is a quartic cyclic extension of ${\mathbb Q}$ containing $k$, as can be verified quickly by noting that $$A^2-mB^2 = mC^2 = (\sqrt{m}C)^2 \quad \text{ and } 
   \quad \sqrt{m}C \in k \setminus {\mathbb Q}.$$ We claim that $K$ is the quartic subfield of ${\mathbb Q}(\zeta_m)$, the field of $m$th roots of unity. This will follow from the theorem of Kronecker and Weber once we have seen that no prime $\ne m$ is ramified in $K/{\mathbb Q}$. But the identity $$\label{E2}
 2(A+B\sqrt{m})(A+C\sqrt{m}) = (A+B\sqrt{m}+C\sqrt{m})^2$$ shows that $K = k\left(\sqrt{2(A+C\sqrt{m})}\right)$, and so the only odd primes that are possibly ramified in $K/k$ are common divisors of $A^2-mB^2 = mC^2$ and $A^2-mC^2 = mB^2$. Since $B$ and $C$ are assumed to be prime to each other, only $2$ and $m$ can ramify. Now $\sqrt{m} \equiv 1 \bmod 2$ (since $m \equiv 1 \bmod 4$) implies $B\sqrt{m} \equiv B \bmod 4$, and we see $A+B\sqrt{m} \equiv A+B 
\equiv 1 \bmod 4$, which shows that $2$ is unramified in $K/k$ (and therefore also in $K/{\mathbb Q}$).

The reciprocity formula will follow by comparing the decomposition laws in $K/{\mathbb Q}$ and ${\mathbb Q}(\zeta_m)/{\mathbb Q}$: if $(m/p) = +1$, then $p$ splits in $k/{\mathbb Q}$; if $f > 0$ is the smallest natural number such that $p^f \equiv 1 \bmod m$ (here we have to assume that $p>0$), then $p$ splits into exactly $g = (m-1)/f$ prime ideals in ${\mathbb Q}(\zeta_m)$, and we see $$\begin{array}{rl} \left(\frac{p}{m} \right)_4 = 1 
    & \iff  p^{(m-1)/4} \equiv 1 \bmod m \\
    & \iff  f \text{ divides } \frac14(m-1) = \frac14fg \\
    & \iff  g \equiv 0 \bmod 4 \\
    & \iff  \text{the degree of the decomposition field } Z \text{ of } p 
            \text{ is divisible by 4} \\
    & \iff Z \text{ contains } K \text { (because }
             Gal({\mathbb Q}(\zeta_m)/{\mathbb Q}) \text { is cyclic)} \\
    & \iff  p \text{ splits completely in } K/{\mathbb Q}\\
    & \iff  p \text{ splits completely in } K/k \text{ (since } p
              \text{ splits completely in } k/{\mathbb Q}\text{)} \\
    & \iff \left(\frac{A+B\sqrt m}{p}\right) = 1. 
\end{array}$$

This completes the proof of the theorem.

Letting $m=2$ and replacing the quartic subfield of ${\mathbb Q}(\zeta_m)$ used above by the cyclic extension ${\mathbb Q}\left(\sqrt{2+\sqrt{2}}\,\right)$ contained in ${\mathbb Q}(\zeta_{16})$ yields the equivalence $$\label{E3}
 \left(\frac{A+B\sqrt 2}{p}\right) = 1 \iff p \text{ splits in }
   {\mathbb Q}\left(\sqrt{2+\sqrt{2}}\,\right) \iff p \equiv \pm 1 \bmod 16,$$ stated in a slightly different way in [@11].

Formula (\[E1\]) differs from the one given in [@11], which reads $$\label{E4} 
\left(\frac{A+B\sqrt m}{p}\right) = 
   (-1)^{(p-1)(m-1)/8}\left(\frac{2}{p}\right)
                  \left(\frac{p}{m}\right)_4,$$ where $A, B, C > 0,\, B$ is odd, and $C$ is even. Formula (2) shows that $$\left(\frac{A+B\sqrt m}{p}\right) = \left(\frac{2}{p}\right)
   \left(\frac{A+C\sqrt m}{p}\right),$$ and so, for $B$ even and $C$ odd, (4) is equivalent to $$\label{E5}
\left(\frac{A+B\sqrt m}{p}\right) = (-1)^{(p-1)(m-1)/8}
        \left(\frac{p}{m}\right)_4.$$

Now $A \equiv 1 \bmod 4$ since $A^2 = m(B^2+C^2)$ is the product of $m \equiv 1 \bmod 4$ and of a sum of two relatively prime squares, and we have $A+B \equiv 1 \bmod 4 \iff {4 \mid B} \iff m \equiv 1 \bmod 8.$ The sign of $B$ is irrelevant, therefore $$\left(\frac{-1}{p}\right)^{B/2} = (-1)^{(p-1)(m-1)/8}.$$ This finally shows that (\[E1\]) is in fact equivalent to (\[E4\]).

Another version of (\[E1\]) which follows directly from (\[E5\]) is $$\label{E6}
\left(\frac{A+B\sqrt m}{p}\right) = 
  \left(\frac{p^*}{m}\right)_4,$$ where $A, B > 0$ and $p^* = (-1)^{p-1)/2}p$.

Formula (\[E1\]) can be extended to composite values of $m$ (where the prime factors of $m$ satisfy certain conditions given in [@11]) in very much the same way as Jacobi extended the quadratic reciprocity law of Gauss; this extension, however, is not needed for deriving the known rational reciprocity laws of K. Burde [@1], E. Lehmer [@6; @7] and A. Scholz [@9]. These follow from (\[E1\]) by assigning special values to $A$ and $B$, in other words: they all stem from the observation that the quartic subfield $K$ of ${\mathbb Q}(\zeta_m)$ can be generated by different square roots over $k={\mathbb Q}(\sqrt{m}\,)$.

The fact that (\[E1\]) is valid for primes $p \mid ABC$ (which has not been proved in [@11]) shows that we no longer have to exclude the primes $q \mid ab$ in Lehmer’s criterion (as was necessary in [@11]), and it allows us to derive Burde’s reciprocity law in a more direct way: let $p$ and $q$ be primes $\equiv 1 \bmod 4$ such that $p = a^2+b^2$, $q=c^2+d^2$, $2\mid b$, $2\mid d$, $(p/q) = +1$, and define $$A = pq, \quad B = b(c^2-d^2)+2acd, \quad C = a(c^2-d^2)-2bcd, \quad
   m = q.$$ Then $2 \mid B,\, B \equiv 2d(ac+bd) \bmod q$ (since $c^2 \equiv 
-d^2 \bmod q$), the sign of $A$ does not matter (since $q \equiv 
1 \bmod 4$), and so formula (\[E1\]) yields $$\left(\frac{q}{p}\right)_4 = \left(\frac{A+B\sqrt p}{q}\right) 
   = \left(\frac{B}{q}\right)\left(\frac{p}{q}\right)_4.$$ Now the well known $(\frac{2d}{q}) = +1$ implies Burde’s law $$\label{E7} 
\left(\frac{p}{q}\right)_4 \left(\frac{q}{p}\right)_4 
  = \left(\frac{ac-bd}{q}\right).$$

A rational reciprocity law equivalent to Burde’s has already been found by T. Gosset [@5], who showed that, for primes $p$ and $q$ as above, $$\label{E8} 
\left(\frac{q}{p}\right)_4 \equiv 
  \left(\frac{a/b-c/d}{a/b+c/d}\right)^{(q-1)/4} \bmod q.$$ Multiplying the numerator and denominator of the term on the right side of (8) by $a/b+c/d$ and observing that $c^2/d^2 \equiv 
-1 \bmod q$ yields $$\begin{aligned}
 \Big(\frac{q}{p}\Big)_4 & \equiv 
 \Big(\frac{a^2/b^2+1}{q}\Big)_4\Big(\frac{a/b+c/d}{q}\Big)
 = \Big(\frac{p}{q}\Big)_4 \Big(\frac{b}{q}\Big)
    \Big(\frac{a/b+c/d}{q}\Big) \\
 & = \Big(\frac{p}{q}\Big)_4 \Big(\frac{a+bc/d}{q}\Big)
   = \Big(\frac{p}{q}\Big)_4 \Big(\frac{d}{q}\Big)
   \Big(\frac{ad+bc}{q}\Big) \bmod q, \end{aligned}$$ which is Burde’s reciprocity law since $(\frac{2d}{q}) = +1$.

A more explicit form of Burde’s reciprocity law for composite values of $p$ and $q$ has been given by L. Rédei [@8]; letting $n = pq = A^2+B^2$ in [@8 (17)], we find $A = ac-bd, \, B = ad+bc,$ and his reciprocity formula [@8 (23)] gives (7).

Yet another version of Burde’s law is due to A. Fröhlich [@4]; he showed $$\label{E9}
 \left(\frac{p}{q}\right)_4 \left(\frac{q}{p}\right)_4 
  =\left(\frac{a+bj}{q}\right) = \left(\frac{c+di}{p}\right),$$ where $i$ and $j$ denote rational numbers such that $i^2 \equiv -1
\bmod p$ and $j^2 \equiv -1 \bmod q$. Letting $i=a/b$ and $j=c/d$ and observing that $(\frac{a}{p}) = (\frac{c}{q}) = +1$, we find that (9) is equivalent to (7).

The reciprocity theorem of Lehmer [@6; @7] is even older; it can be found in Dirichlet’s paper [@2] as Théorème I and II; Dirichlet’s ideas are reproduced in the charming book of Venkow [@10] and may be used to give proofs for other rational reciprocity laws using nothing beyond quadratic reciprocity.

[**Remark.**]{} The author has recently generalized Scholz’s reciprocity law to all number fields with odd class number in the strict sense.

[99]{}

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