# Cubic reciprocity

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Cubic reciprocity

Cubic reciprocity is a collection of theorems in elementary and algebraic number theory that state conditions under which the congruence x3 ≡ p (mod q) is solvable; the word "reciprocity" comes from the form of the main theorem, which states that if p and q are primary numbers in the ring of Eisenstein integers, both coprime to 3,

The congruence x3p (mod q) is solvable if and only if x3q (mod p) is.

## History

Sometime before 1748 Euler made the first conjectures about the cubic residuacity of small integers, but they were not published until 1849, after his death.

Gauss's published works mention cubic residues and reciprocity three times: there is one result pertaining to cubic residues in the Disquisitiones Arithmeticae (1801). In the introduction to the fifth and sixth proofs of quadratic reciprocity (1818) he said that he was publishing these proofs because their techniques (Gauss's lemma and Gaussian sums, respectively) can be applied to cubic and biquadratic reciprocity. Finally, a footnote in the second (of two) monographs on biquadratic reciprocity (1832) states that cubic reciprocity is most easily described in the ring of Eisenstein integers.

From his diary and other unpublished sources, it appears that Gauss knew the rules for the cubic and quartic residuacity of integers by 1805, and discovered the full-blown theorems and proofs of cubic and biquadratic reciprocity around 1814. Proofs of these were found in his posthumous papers, but it is not clear if they are his or Eisenstein's.

Jacobi published several theorems about cubic residuacity in 1827, but no proofs. In his Königsberg lectures of 1836–37 Jacobi presented proofs. The first published proofs were by Eisenstein (1844).

## Integers

A cubic residue (mod p) is any number congruent to the third power of an integer (mod p). If x3a (mod p) does not have an integer solution, a is a cubic nonresidue (mod p).

As is often the case in number theory, it is easiest to work modulo prime numbers, so in this section all moduli p, q, etc., are assumed to be positive, odd primes.

The first thing to notice when working within the ring Z of integers is that if the prime number q is ≡ 2 (mod 3) every number is a cubic residue (mod q). Let q = 3n + 2; since 0 = 03 is obviously a cubic residue, assume x is not divisible by q. Then by Fermat's little theorem, $x^q = x^{3 n + 2} \equiv x \pmod{q}\;\mbox{ and }\;x^{q - 1} = x^{3 n + 1}\equiv 1 \pmod{ q},\mbox{ so }$ $x = 1 \cdot x \equiv x^q x^{q - 1} = x^{3n + 2} x^{3n+1} = x^{6n + 3} = (x^{2n+1})^3 \pmod{ q}$

is a cubic residue (mod q).

Therefore, the only interesting case is when the modulus p ≡ 1 (mod 3).

In this case, p ≡ 1 (mod 3), the nonzero residue classes (mod p) can be divided into three sets, each containing (p−1)/3 numbers. Let e be a cubic nonresidue. The first set is the cubic residues; the second one is e times the numbers in the first set, and the third is e2 times the numbers in the first set. Another way to describe this division is to let e be a primitive root (mod p); then the first (respectively second, third) set is the numbers whose indices with respect to this root are ≡ 0 (resp. 1, 2) (mod 3). In the vocabulary of group theory, the first set is a subgroup of index 3 (of the multiplicative group   Z/pZ ×), and the other two are its cosets.

### Primes ≡ 1 (mod 3)

A theorem of Fermat states that every prime p ≡ 1 (mod 3) is the sum of a square and three times a square: p = a2 + 3b2 and (except for the signs of a and b) this representation is unique.

Letting m = a + b and n = ab, we see that this is equivalent to p = m2mn + n2 (which equals (nm)2 − (nm)n + n2 = m2 + m(nm) + (nm)2, so m and n are not determined uniquely). Thus, \begin{align}4p &= (2m-n)^2 + 3n^2 \\ &= (2n-m)^2 + 3m^2 \\ &= (m+n)^2 + 3(m-n)^2, \end{align}

and it is a straightforward exercise to show that exactly one of m, n, or mn is a multiple of 3, so $p = \tfrac14 \left(L^2+ 27M^2\right),$   and this representation is unique up to the signs of L and M.

For relatively-prime integers m and n define the rational cubic residue symbol as $\left[\frac{m}{n}\right]_3 = \begin{cases} &+1 \mbox{ if }m\mbox{ is a cubic residue }\pmod{n}\\ &-1\mbox{ if }m\mbox{ is a cubic nonresidue }\pmod{n} \end{cases}$

### Euler

Euler's conjectures are based on the representation p = 3a2 + b2. The symbol m|n is read "m divides n" and means there is an a such that n = ma. \begin{align} \left[\frac{2}{p}\right]_3 =1 &\mbox{ if and only if } 3|a\\ \left[\frac{3}{p}\right]_3 =1 &\mbox{ if and only if } 9|a; \mbox{ or }9|(a\pm b)\\ \left[\frac{5}{p}\right]_3 =1 &\mbox{ if and only if } 15|a; \mbox{ or }3|a \mbox{ and }5|b; \mbox{ or } 15|(a\pm b); \mbox{ or } 15|(a\pm 2b)\\ \left[\frac{6}{p}\right]_3 =1 &\mbox{ if and only if } 9|a; \mbox{ or }9|(2a\pm b)\\ \left[\frac{7}{p}\right]_3 =1 &\mbox{ if and only if } 21|a; \mbox{ or }3|a\mbox{ and }7|b;\mbox{ or }21|(a\pm b);\mbox{ or }7|(4a\pm b);\mbox{ or }7|(a\pm 2b) \end{align}

The first two can be restated as

Let p ≡ 1 (mod 3) be a positive prime. Then 2 is a cubic residue of p if and only if    p = a2 +  27b2.
Let p ≡ 1 (mod 3) be a positive prime. Then 3 is a cubic residue of p if and only if  4p = a2 + 243b2.

### Gauss

Gauss proves that if $p = 3n + 1= \tfrac14 \left(L^2+ 27M^2\right),$  then $L(n!)^3\equiv 1 \pmod{p},$    from which $\left[\frac{L}{p}\right]_3 = \left[\frac{M}{p}\right]_3 =1$ is an easy deduction.

### Jacobi

Jacobi stated (without proof)

Let qp ≡ 1 (mod 6) be positive primes, $p = \tfrac14 \left(L^2+ 27M^2\right),$    and let x be a solution of x2 ≡ −3 (mod q). Then $\left[\frac{q}{p}\right]_3 =1 \mbox{ if and only if } \left[\frac{\frac{L+3Mx}{2}p}{q}\right]_3 =1 \mbox{ if and only if } \left[\frac{(\frac{L+3Mx}{L-3Mx})}{q}\right]_3 =1.$

(The "numerator" in the last expression is an integer (mod q), not a Legendre symbol).

If $q = \tfrac14 \left(L'^2+ 27M'^2\right),$    then $x\equiv\pm \frac{L'}{3M'}\pmod{q}$, and we have $\left[\frac{q}{p}\right]_3 =1 \mbox{ if and only if } \left[\frac{(\frac{LM'+L'M}{LM'-L'M})}{p}\right]_3 =1.$

Along the same lines, von Lienen proved $\left[\frac{p}{q}\right]_3 \left[\frac{q}{p}\right]_3 = \left[\frac{(\frac{LM'+L'M}{2M})}{q}\right]_3^2.$

### Other theorems

Emma Lehmer proved

Let $q\mbox{ and }p = \tfrac14 \left(L^2+ 27M^2\right)$   be primes. $\left[\frac{q}{p}\right]_3 = 1 \mbox{ if and only if } \begin{cases} q|LM\mbox{ or }\\ L\equiv\pm \frac{9r}{2u+1} M\pmod{q}, \;\;\;\mbox{ where }\\ \;\;\;\;\; u\not\equiv 0,1,-\frac12, -\frac13 \pmod{q} \;\;\;\mbox{ and } \\ \;\;\;\;\;3u+1 \equiv r^2 (3u-3)\pmod{q} \end{cases}$

Note that the first condition implies:

Any number that divides L or M is a cubic residue (mod p).

The first few examples of this are equivalent to Euler's conjectures: \begin{align} \left[\frac{2}{p}\right]_3 =1 &\mbox{ if and only if } &L \equiv M &\equiv 0 \pmod{2} \\ \left[\frac{3}{p}\right]_3 =1 &\mbox{ if and only if } &M &\equiv 0 \pmod{3}\\ \left[\frac{5}{p}\right]_3 =1 &\mbox{ if and only if } &LM &\equiv 0 \pmod{5}\\ \left[\frac{7}{p}\right]_3 =1 &\mbox{ if and only if } &LM &\equiv 0 \pmod{7}\\ \left[\frac{11}{p}\right]_3 =1 &\mbox{ if and only if } &LM(L-3M)(L+3M) &\equiv 0 \pmod{11}\\ \left[\frac{13}{p}\right]_3 =1 &\mbox{ if and only if } &LM(L-2M)(L+2M) &\equiv 0 \pmod{13} \end{align}

Martinet proved

Let pq ≡ 1 (mod 3) be primes, $pq = \tfrac14 \left(L^2+ 27M^2\right).$   Then $\left[\frac{L}{p}\right]_3 \left[\frac{L}{q}\right]_3 =1 \;\;\mbox{ if and only if } \;\;\left[\frac{q}{p}\right]_3 \left[\frac{p}{q}\right]_3 =1$

Sharifi proved

Let p = 1 + 3x + 9x2 be prime. Then

Any divisor of x is a cubic residue (mod p).

## Eisenstein integers

### Background

In his second monograph on biquadratic reciprocity, Gauss says:

The theorems on biquadratic residues gleam with the greatest simplicity and genuine beauty only when the field of arithmetic is extended to imaginary numbers, so that without restriction, the numbers of the form a + bi constitute the object of study ... we call such numbers integral complex numbers. [bold in the original]

These numbers are now called the ring of Gaussian integers, denoted by Z[i]. Note that i is a fourth root of 1.

The theory of cubic residues must be based in a similar way on a consideration of numbers of the form a + bh where h is an imaginary root of the equation h3 = 1 ... and similarly the theory of residues of higher powers leads to the introduction of other imaginary quantities.

In his first monograph on cubic reciprocity Eisenstein developed the theory of the numbers built up from a cube root of unity; they are now called the ring of Eisenstein integers. Eisenstein said (paraphrasing) "to investigate the properties of this ring one need only consult Gauss's work on Z[i] and modify the proofs". This is not surprising since both rings are unique factorization domains.

The "other imaginary quantities" needed for the "theory of residues of higher powers" are the rings of integers of the cyclotomic number fields; the Gaussian and Eisenstein integers are the simplest examples of these.

### Facts and terminology

Let $\omega = \frac{-1 + i\sqrt 3}{2} = e^\frac{2\pi i}{3}$ be a complex cube root of unity. The Eisenstein integers Z[ω] are all numbers of the form a + bω where a and b are ordinary integers.

Since ω3 − 1 = (ω − 1)(ω2 + ω + 1) = 0 and ω ≠ 1, we have ω2 = − ω − 1 and ω = − ω2 − 1. Since $\omega^3 = \omega \omega^2 = \omega \overline{\omega} =1, \;\;\ \overline{\omega} = \omega^2$   and $\overline{\omega^2} = \omega$   where the bar denotes complex conjugation. Also, $\omega -\overline{\omega} = i\sqrt{3}.$

If λ = a + bω and μ = c + dω,

λ + μ = (a + c) + (b + d)ω and
λ μ = ac + (ad + bc)ω + bdω2 = (acbd) + (ad + bcbd)ω.

This shows that Z[ω] is closed under addition and multiplication, making it a ring.

The units are the numbers that divide 1. They are ±1, ±ω, and ±ω2. They are similar to 1 and −1 in the ordinary integers, in that they divide every number. The units are the powers of −ω, a sixth (not just a third) root of unity.

Given a number λ = a + bω, its conjugate means its complex conjugate a + 2 = (ab) − bω  (not abω), and its associates are its six unit multiples: \begin{align} \lambda &= a + b\omega \\ \omega\lambda &= -b + (a -b)\omega \\ \omega^2\lambda &= (b-a)-a\omega \\ -\lambda &= -a-b\omega \\ -\omega\lambda &= b + (b -a)\omega \\ -\omega^2\lambda &= (a -b) + a\omega \end{align}

The norm of λ = a + bω is the product of λ and its conjugate $\mathrm{N} \lambda = \lambda\overline{\lambda}=a^2-ab+b^2.$ From the definition, if λ and μ are two Eisenstein integers, Nλμ = Nλ Nμ; in other words, the norm is a completely multiplicative function. The norm of zero is zero, the norm of any other number is a positive integer. ε is a unit if and only if Nε = 1. Note that the norm is always ≡ 0 or ≡ 1 (mod 3).

Z[ω] is a unique factorization domain. The primes fall into three classes:

• 3 is a special case: 3 = −ω2(1 − ω)2. It is the only prime in Z divisible by the square of a prime in Z[ω]. In algebraic number theory, 3 is said to ramify in Z[ω].
• Positive primes in Z ≡ 2 (mod 3) are also primes in Z[ω]. In algebraic number theory, these primes are said to remain inert in Z[ω].
• Positive primes in Z ≡ 1 (mod 3) are the product of two conjugate primes in Z[ω]. $p=\mathrm{N} \pi = \mathrm{N} \overline{\pi}= \pi \overline{\pi}$   In algebraic number theory, these primes are said to split in Z[ω].

Thus, inert primes are 2, 5, 11, 17, ... and a factorization of the split primes is

7 = (3 + ω) × (2 − ω),
13 = (4 + ω) × (3 − ω),
19 = (3 − 2ω) × (5 + 2ω),
31 = (1 + 6ω) × (−5 − 6ω), ...

The associates and conjugate of a prime are also primes.

Note that the norm of an inert prime q is Nq = q2 ≡ 1 (mod 3).

In order to state the unique factorization theorem, it is necessary to have a way of distinguishing one of the associates of a number. Eisenstein defines a number to be primary if it is ≡ 2 (mod 3). It is straightforward to show that if gcd(Nλ, 3) = 1 then exactly one associate of λ is primary. A disadvantage of this definition is that the product of two primary numbers is the negative of a primary.

Most modern authors say that a number is primary if it is coprime to 3 and congruent to an ordinary integer (mod (1 − ω)2), which is the same as saying it is ≡ ±2 (mod 3). There are two reasons to do this: first, the product of two primaries is a primary, and second, it generalizes to all cyclotomic number fields. Under this definition, if gcd(Nλ, 3) = 1 one of λ, ωλ, or ω2λ is primary. A primary under Eisenstein's definition is primary under the modern one, and if λ is primary under the modern one, either λ or −λ is primary under Eisenstein's. Since −1 is a cube, this does not affect the statement of cubic reciprocity, but it does affect the unique factorization theorem. This article uses the modern definition, so

The product of two primary numbers is primary and the conjugate of a primary number is also primary.

The unique factorization theorem for Z[ω] is: if λ ≠ 0, then $\lambda = \omega^\mu(1-\omega)^\nu\pi_1^{\alpha_1}\pi_2^{\alpha_2}\pi_3^{\alpha_3} \dots$

where   0 ≤ μ ≤ 2,   ν ≥ 0,   each πi is a primary prime, and each αi ≥ 1, and this representation is unique, up to the order of the factors.

The notions of congruence and greatest common divisor are defined the same way in Z[ω] as they are for the ordinary integers Z. Because the units divide all numbers, a congruence (mod λ) is also true modulo any associate of λ, and any associate of a GCD is also a GCD.

### Cubic residue character

An analogue of Fermat's little theorem is true in Z[ω]: if α is not divisible by a prime π, $\alpha^{\mathrm{N} \pi - 1} \equiv 1 \pmod{\pi}$

Now assume that Nπ ≠ 3, so that Nπ ≡ 1 (mod 3).

Then $\alpha^{\frac{\mathrm{N} \pi - 1}{3}}$     makes sense, and $\alpha^{\frac{\mathrm{N} \pi - 1}{3}}\equiv \omega^k \pmod{\pi}$     for a unique unit ωk.

This unit is called the cubic residue character of α (mod π) and is denoted by $\left(\frac{\alpha}{\pi}\right)_3 = \omega^k \equiv \alpha^{\frac{\mathrm{N} \pi - 1}{3}} \pmod{\pi}.$

It has formal properties similar to those of the Legendre symbol.

The congruence $x^3 \equiv \alpha \pmod{\pi}$    is solvable in Z[ω] if and only if $\left(\frac{\alpha}{\pi}\right)_3 = 1.$ $\Bigg(\frac{\alpha\beta}{\pi}\Bigg)_3=\Bigg(\frac{\alpha}{\pi}\Bigg)_3\Bigg(\frac{\beta}{\pi}\Bigg)_3$ $\overline{\Bigg(\frac{\alpha}{\pi}\Bigg)_3}=\Bigg(\frac{\overline{\alpha}}{\overline{\pi}}\Bigg)_3$     where the bar denotes complex conjugation.
if π and θ are associates, $\Bigg(\frac{\alpha}{\pi}\Bigg)_3=\Bigg(\frac{\alpha}{\theta}\Bigg)_3$
if α ≡ β (mod π), $\Bigg(\frac{\alpha}{\pi}\Bigg)_3=\Bigg(\frac{\beta}{\pi}\Bigg)_3$

The cubic character can be extended multiplicatively to composite numbers (coprime to 3) in the "denominator" in the same way the Legendre symbol is generalized into the Jacobi symbol. Like the Jacobi symbol, if the "denominator" of the cubic character is composite, then if the "numerator" is a cubic residue mod the "denominator" the symbol will equal 1, if the symbol does not equal 1 then the "numerator" is a cubic nonresidue, but the symbol can equal 1 when the "numerator" is a nonresidue: $\left(\frac{\alpha}{\lambda}\right)_3 = \left(\frac{\alpha}{\pi_1}\right)_3^{\alpha_1} \left(\frac{\alpha}{\pi_2}\right)_3^{\alpha_2} \dots$   where $\lambda = \pi_1^{\alpha_1}\pi_2^{\alpha_2}\pi_3^{\alpha_3} \dots$
If a and b are ordinary integers, gcd(a, b) = gcd(b, 3) = 1, then $\left(\frac{a}{b}\right)_3 = 1.$

### Statement of the theorem

Let α and β be primary. Then $\Bigg(\frac{\alpha}{\beta}\Bigg)_3 = \Bigg(\frac{\beta}{\alpha}\Bigg)_3.$

There are supplementary theorems for the units and the prime 1 − ω:

Let α = a + bω be primary, a = 3m + 1 and b = 3n. (If a ≡ 2 (mod 3) replace α with its associate −α; this will not change the value of the cubic characters.) Then $\Bigg(\frac{\omega}{\alpha}\Bigg)_3 = \omega^\frac{1-a-b}{3}= \omega^{-m-n},\;\;\; \Bigg(\frac{1-\omega}{\alpha}\Bigg)_3 = \omega^\frac{a-1}{3}= \omega^m,\;\;\; \Bigg(\frac{3}{\alpha}\Bigg)_3 = \omega^\frac{b}{3}= \omega^n.$

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