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

$\omega$ will denote $\frac{-1+i\sqrt{3}}{2}$ , which is one of the complex 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 $N:K\to \Z$ by $$N(a+b\omega)=a^2-ab+b^2$$ or equivalently $$N(z)=z\overline{z}\;.$$

Whereas $\Z$ has only two units (meaning invertible elements), namely $\pm 1$ , $K$ has six, namely all the sixth roots of 1: $$\pm 1\qquad\pm\omega\qquad\pm\omega^2$$ and we know $\omega^2=-1-\omega$ . Two nonzero elements $\alpha$ and $\beta$ of $K$ are said to be associates if $\alpha=\beta\mu$ for some unit $\mu$ . This is 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 $\rho=\alpha\beta$ implies that $\alpha$ or $\beta$ , but not both, is a unit. It turns out that the irreducible elements of $K$ are (up to multiplication by units):

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

- positive real integers $q\equiv 2\pmod{3}$ which are prime in $\Z$ . Such integers are called rational primes in $K$ .

- complex numbers $q=a+b\omega$ where $N(q)$ is a prime in $Z$ and $N(q)\equiv 1\pmod{3}$ .

For example, $3+2\omega$ is a prime in $K$ because its norm, 7, is prime in $\Z$ 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 associates is 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 $N(q)\equiv 1\pmod{3}$ is prime in $\Z$ and \begin{eqnarray*} a&\equiv& 2\pmod{3} \\ b&\equiv& 0\pmod{3}\;. \end{eqnarray*}One can verify that this selection exists and is unambigous.

Next, we seek a three-valued function analogous to the two-valued quadratic residue character $x\mapsto\legsymp{x}$ . Let $\rho$ be a prime in $K$ , with $\rho\ne\pi$ . If $\alpha$ is any element of $K$ such that $\rho\nmid\alpha$ , then $$\alpha^{N(\rho)-1}\equiv 1\pmod{\rho}\;.$$ Since $N(\rho)-1$ is a multiple of 3, we can define a function $$\chi_\rho:K\to\{1,\omega,\omega^2\}$$ by \begin{eqnarray*} \chi_\rho(\alpha)&\equiv&\alpha^{(N(\rho)-1)/3}\text{ if }\rho\nmid\alpha \\ \chi_\rho(\alpha)&=&0\text{ if }\rho\mid\alpha\;. \end{eqnarray*}$\chi_\rho$ is a character, called the cubic residue character mod $\rho$ . We have $\chi_\rho(\alpha)=1$ if and only if $\alpha$ is a nonzero cube mod $\rho$ . (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 distinct primes in $K$ , neither of them $\pi$ , then $$\chi_\rho(\sigma)=\chi_\sigma(\rho)\;.$$

The quadratic reciprocity law has two ``supplements'' which describe $\legsymp{-1}$ and $\legsymp{2}$ . 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 \begin{eqnarray*} m&=&(\rho+1)/3\qquad\text{ if $\rho$ is a rational prime} \\ m&=&(a+1)/3\qquad\text{ if $\rho=a+b\omega$ is a complex prime.} \end{eqnarray*} Remarks:Some writers refer to our ``irreducible'' elements as ``primes'' in $K$ ; what we have called primes, they call ``primary primes''.

The quadratic reciprocity law would take a simpler form if we were to make a different convention on what is a prime in $\Z$ , a convention similar to the one in $K$ : a prime in $\Z$ is either 2 or an irreducible element $x$ of $\Z$ such that $x\equiv 1\pmod 4$ . The primes would then be 2, -3, 5, -7, -11, 13, ...and the QRL would say simply $$\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=1$$ for any two distinct odd primes $p$ and $q$ .