Gauss dubbed it his ‘golden theorem’. In this post we will replace ${\mathbb{Z}}$
by the structurally similar ring ${\mathbb{F}_{p}[X]}$ and prove higher versions of quadratic reciprocity, thereby turning gold into platinum.

The golden theorem

Remember that the classic quadratic reciprocity states that for ${p}$ and ${q}$ distinct and odd prime numbers, we have:
$$
\bigg( \frac{p}{q} \bigg) \bigg( \frac{q}{p} \bigg)=(-1)^{(\frac{p-1}{2})(\frac{q-1}{2})}= \left\{ \begin{array}{ll} 1 & \textrm{if $p \equiv 1$ mod $4$ or $q \equiv 1$ mod $4$}\\ -1 & \textrm{if $p \equiv q \equiv 3$ mod $4$} \end{array} \right.
$$
where $\big( \frac{p}{q} \big)$ is the Legendre symbol, with value equal to $1$ if the equation $x^{2} \equiv p$ has a solution mod $q$, and $-1$ if it doesn’t. To see why this is useful, consider the following question: find all odd primes $p$ for which $5$ is a quadratic residue mod $p$. Using quadratic reciprocity, the solution is easy:
$$
\bigg( \frac{p}{5} \bigg) \bigg( \frac{5}{p} \bigg)=(-1)^{(\frac{p-1}{2})(\frac{5-1}{2})}=1,
$$
which holds iff $(5/p)=(p/5)=1$ iff $p=\pm 1$ mod $5$. There are a number of other interesting applications, such as finding integer solutions to degree $2$ polynomial equations.

Legendre revisited

Suppose you have an irreducible polynomial ${P}$ that does not divide a polynomial ${a}$, and ${d}$ a divisor of ${p-1}$, then by some elementary computations, the equation ${x^{d} \equiv a}$ has a solution mod ${P}$ iff ${a^{\frac{\vert P \vert -1}{d}} \equiv 1}$ mod ${P}$, with ${\vert P \vert = p^{deg(P)}}$. In general we will denote by ${(a/P)_{d}}$ the unique element in ${\mathbb{F}_{p}^{*} \subset (\mathbb{F}_{p}[X]/(P))^{*}}$, such that
$$ a^{\frac{\vert P \vert -1}{d}} \equiv \bigg( \frac{a}{P} \bigg)_{d} \ \bmod \ P \ \ \ \ \ $$
If ${P\,\vert\, a}$ we define ${(a/P)_{d}=0}$. This symbol is called the ${d}$-th power residue symbol.

Carlitz’ version

Using this newly introduced symbol, we can state and prove the ${d}$-th power reciprocity law:
Theorem  Let ${P}$ and ${Q}$ be monic, irreducible polynomials in ${\mathbb{F}_{p}[X]}$ of degrees ${\delta}$ and ${\nu}$ respectively. Then,
$$ \bigg( \frac{Q}{P} \bigg)_{d} = (-1)^{\frac{p-1}{d}\delta \nu} \bigg( \frac{P}{Q} \bigg)_{d}. \ \ \ \ \ $$

Proof: The first step will consist of a simple reformulation. Define ${(a/P)=(a/P)_{q-1}}$. If we prove that
$$ \bigg( \frac{Q}{P} \bigg) = (-1)^{\delta \nu} \bigg( \frac{P}{Q} \bigg), \ \ \ \ \ $$
then the result will follow by raising to the ${(p-1)/d}$ power and noting that ${(a/P)_{d}=(a/P)^{\frac{p-1}{d}}}$.

Take ${\alpha}$ and ${\beta}$ roots of ${P}$ and ${Q}$ respectively. Over the field ${\mathbb{F}(\alpha,\beta)}$ we have
$$
P(X)=(X-\alpha)(X – \alpha^{p}) \cdots (X-\alpha^{p^{\delta-1}})
$$
and
$$
Q(X)=(X-\beta)(X-\beta^{p}) \cdots (X-\beta^{p^{\nu-1}}), $$using Fermat’s little theorem and the Frobenius map.

Noting that for a polynomial ${f}$ over ${\mathbb{F}(\alpha,\beta)}$, the equality ${f(X) \equiv f(\alpha)}$ holds mod ${(X-\alpha)}$, and for a polynomial ${g \in \mathbb{F}_{p}[X]}$ we have that ${g(X^{p})=g(X)^{p}}$, we can compute ${(Q/P)}$.
\begin{align*} \bigg( \frac{Q}{P} \bigg) & \equiv Q(X)^{1+p+\cdots+p^{\delta-1}} \\ & \equiv Q(X)Q(X^{p}) \cdots Q(X^{p^{\delta -1}}) \\ & \equiv Q(\alpha)Q(\alpha^{p}) \cdots Q(\alpha^{p^{\delta-1}}) \bmod (X-\alpha) \end{align*}
By invoking symmetry, this condition will also hold mod ${(X-\alpha^{p^{i}})}$ and thus also mod ${P}$. Plugging in the equation for $Q(X)$, and noting that both sides of the resulting congruence,
$$ \bigg( \frac{Q}{P} \bigg) \equiv \prod_{i=0}^{\delta -1} \prod_{j=0}^{\nu -1} (\alpha^{p^{i}} – \beta^{p^{j}}) \ \bmod \ P, \ \ \ \ \ $$
are in ${\mathbb{F}(\alpha,\beta)}$ (and thus equal), we get
$$ \bigg( \frac{Q}{P} \bigg) = (-1)^{\delta \nu} \prod_{i=0}^{\delta -1} \prod_{j=0}^{\nu -1} (\beta^{p^{j}} – \alpha^{p^{i}}) = (-1)^{\delta \nu} \bigg( \frac{P}{Q} \bigg), \ \ \ \ \ $$
completing the proof. $\Box$

About Theo Raedschelders

dabbling in geometry, algebra and physics in and around Brussels and Antwerp

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