cubic reciprocity law

In a ring $\mathbb{Z}/n\mathbb{Z}$, a cubic residue is just a value of the function$x^3$ for some invertible element $x$ of the ring. Cubic residues displaya reciprocity phenomenon similar to that seen with quadraticresidues. But we need some preparation in order to state the cubicreciprocity law.

$\omega$ will denote $\frac{-1+i\sqrt{3}}{2}$, which is one of thecomplex cube roots of $1$.$K$ will denote the ring $K=\mathbb{Z}[\omega ]$. The elements of$K$ are the complex numbers $a+b\omega$where $a$ and $b$ are integers. We define the norm $N:K\to \mathbb{Z}$ by

$$N(a+b\omega )=a^2-ab+b^2$$

or equivalently

$$N(z)=z\overline{z}.$$

Whereas $\mathbb{Z}$ has only two units (meaning invertible elements), namely$\pm 1$, $K$ has six, namely all the sixth roots of 1:

$$\pm 1\mathit{\hspace{1em}\hspace{1em}}\pm \omega \mathit{\hspace{1em}\hspace{1em}}\pm {\omega }^2$$

and we know ${\omega }^2=-1-\omega$. Two nonzero elements $\alpha$and $\beta$ of $K$ are saidto be associates if $\alpha =\beta \mu$ for some unit $\mu$. Thisis an equivalence relation, and any nonzero element has six associates.

$K$ is a principal ring, hence has unique factorization. Let us call$\rho \in K$ “irreducible” if the condition $\rho =\alpha \beta$ impliesthat $\alpha$ or $\beta$, but not both, is a unit.It turns out that the irreducible elements of $K$ are (up to multiplicationby units):

– the number $1-\omega$, which has norm 3. We will denote it by $\pi$.

– positive real integers $q\equiv 2(mod3)$ which are prime in $\mathbb{Z}$.Such integers are called rational primes in $K$.

– complex numbers $q=a+b\omega$ where $N(q)$ is a prime in $Z$ and$N(q)\equiv 1(mod3)$.

For example, $3+2\omega$ is a prime in $K$ because its norm, 7, isprime in $\mathbb{Z}$ and is 1 mod 3; but 7 is not a prime in $K$.

Now we need some convention whereby at most one of any six associatesis called a prime. By convention, the following numbers are nominated:

– the number $\pi$.

– rational primes (rather than their negative or complex associates).

– complex numbers $q=a+b\omega$ where $N(q)\equiv 1(mod3)$ isprime in $\mathbb{Z}$ and

$a$$\equiv$$2(mod3)$

$b$$\equiv$$0(mod3).$

One can verify that this selection exists and is unambigous.

Next, we seek a three-valued function analogous to thetwo-valued quadratic residue character $x\mapsto \left(\frac{x}{p}\right)$.Let $\rho$ be a prime in $K$, with $\rho \ne \pi$. If $\alpha$ is anyelement of $K$ such that $\rho \nmid \alpha$, then

$${\alpha }^{N(\rho )-1}\equiv 1(mod\rho ).$$

Since $N(\rho )-1$ is a multiple of 3, we can define a function

$${\chi }_{\rho }:K\to \{1,\omega ,{\omega }^2\}$$

by

${\chi }_{\rho }(\alpha )$$\equiv$${\alpha }^{(N(\rho )-1)/3}\text{ if }\rho \nmid \alpha$

${\chi }_{\rho }(\alpha )$$=$$0\text{ if }\rho \mid \alpha .$

${\chi }_{\rho }$ is a character, called the cubic residue character mod $\rho$.We have ${\chi }_{\rho }(\alpha )=1$ if and only if $\alpha$ is a nonzero cubemod $\rho$. (Compare Euler’s criterion.)

At last we can state this famous result of Eisenstein and Jacobi:

Theorem (Cubic Reciprocity Law): If $\rho$ and $\sigma$ are any two distinctprimes in $K$, neither of them $\pi$, then

$${\chi }_{\rho }(\sigma )={\chi }_{\sigma }(\rho ).$$

The quadratic reciprocity law has two “supplements” which describe$\left(\frac{-1}{p}\right)$ and $\left(\frac{2}{p}\right)$. Likewise the cubic law has this supplement,due to Eisenstein:

Theorem: For any prime $\rho$ in $K$, other than $\pi$,

$${\chi }_{\rho }(\pi )={\omega }^{2m}$$

where

$m$$=$$(\rho +1)/3\mathit{\hspace{1em}\hspace{1em}}\text{ if }\rho \text{ is a rational prime}$

$m$$=$$(a+1)/3\mathit{\hspace{1em}\hspace{1em}}\text{ if }\rho =a+b\omega \text{ is a complex prime.}$

Remarks: Some writers refer to our “irreducible” elementsas “primes” in $K$; what we have called primes, they call “primary primes”.

The quadratic reciprocity law would take a simpler form if we were tomake a different convention on what is a prime in $\mathbb{Z}$, a conventionsimilar to the one in $K$: a prime in $\mathbb{Z}$ is either 2 or anirreducible element $x$ of $\mathbb{Z}$ such that $x\equiv 1(mod4)$.The primes would then be 2, -3, 5, -7, -11, 13, …and the QRLwould say simply

$$\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=1$$

for any two distinct odd primes $p$ and $q$.