The Legendre symbol is a tool for measuring whether or not $$ x^2 \equiv a \text{ } (p) $$ has a solution in $\mathbb{F}_p$ for some fixed integer $a$. Does the Legendre symbol generalize to higher degrees? For example, can I define a law $$ \left( \frac{\cdot}{p} \right)_k:\mathbb{F}_p^* \to ??? $$ telling me whether or not $$ \frac{\mathbb{F}_p[x]}{(x^k - a)} $$ is a field, and if it is not, how far is it from being a field? Also, if there is such a rule, are there reciprocity laws which can be found?

  • 2
    $\begingroup$ Yes, it does, but the progress of our understanding is slower the higher the degree. Suzanne Rousseau's thesis for Eastern Washington on quadratic and cubic reciprocity might at least shed some light on the cubic equivalent of the Legendre symbol: dc.ewu.edu/cgi/viewcontent.cgi?article=1026&context=theses $\endgroup$
    – Lisa
    Jun 2, 2017 at 19:04
  • $\begingroup$ Darn, so there is no overarching generalization known yet? $\endgroup$
    – 54321user
    Jun 2, 2017 at 19:12
  • $\begingroup$ The discrete Fourier transform of $f_m(n) = 1_{n \equiv x^m \bmod p}$ is $F_m(k) = C_m \sum_{n=1}^{p-1} e^{-2i \pi n^m k/p}$ where $C_m$ is the number of solutions of $x^m \equiv 1 \bmod p$. See Gauss sum for the case $m=2$. $\endgroup$
    – reuns
    Jun 2, 2017 at 19:13
  • $\begingroup$ @user1952009 Sorry, I do not understand the significance of your statement. $\endgroup$
    – 54321user
    Jun 2, 2017 at 19:14
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    $\begingroup$ I'm not sure I'm answering the exact question you are asking, but there has been a lot of work on reciprocity laws for such symbols. I suggest you could start looking at cubic reciprocity, Eisenstein reciprocity, and generally a lot of algebraic number theory from there. Things are not as clean as in the quadratic case, though. $\endgroup$ Jun 2, 2017 at 19:36

1 Answer 1


Your question about a generalization of the Legendre symbol/the quadratic reciprocity law was known as Hillbert’s 9th problem and is solved completely by class field theory , CFT for short. Here is a quick – but necessarily not short ! - glimpse at that vast magnificent landscape (for a history of CFT, see e.g. Cassels-Fröhlich, chapter XI):

Given a number field $K$, CFT aims to « describe » the abelian extensions $L/K$ in terms of the « arithmetic » of $K$ alone. This formulation is rather vague, but let us start from the quadratic reciprocity law. The initial problem was the study of the congruence $x^2 - a \equiv 0 $ mod $p$, where $p$ is a prime not dividing $a$. Gauss’ reciprocity means that the existence of a solution depends only on the arithmetic progression mod $4a$ to which $p$ belongs. But a well known result says that, for a quadratic field $L=\mathbf Q (\sqrt d)$, an odd prime $p$ splits completely in $L$ iff $p$ does not divide $d$ and $d$ is a square mod $p$ (this is the second part of your question).The splitting of the prime $p$ is the link with the aforementioned goal of CFT, because of the following classical theorem : let $L_i, i = 1, 2$ be two Galois extensions (not necessarily abelian) extensions of $K$, and let $ Spl(L_{i}/K)$ be the set of primes of $K$ which split completely in $L_i$ ; then $L_1 = L_2$ iff $ Spl(L_{1}/K) = Spl(L_{2}/K) $. Thus CFT for quadratic number fields, in view of Gauss’ reciprocity, can be expressed in terms of the arithmetic of $\mathbf Q$, more precisely in terms of congruences.

If we stick to the base field $\mathbf Q$, CFT in the above sense can actually be derived entirely from the Kronecker-Weber theorem, which asserts that any abelian extension $L$ of $\mathbf Q$ is contained in a cyclotomic field $\mathbf Q(\zeta_m)$. The integer $m$ is called a defining modulus for $L$, and the smallest such modulus, denoted $f_L$, is the conductor of $L$. For $a \in \mathbf Z$, coprime to $m$, the Artin symbol ($L/a$) is defined as the restriction to $L$ of the automorphism of $C_m := (\mathbf Z/m\mathbf Z)^* := Gal(\mathbf Q(\zeta_m)/ \mathbf Q)$ which sends $\zeta_m$ to $\zeta_{m}^{a}$. The symbol $(L/.)$ is a generalization of the Legendre and Jacobi symbols, and it gives an exact sequence $1 \to I_{L,m} \to C_m \to Gal(L/\mathbf Q) \to 1$, where the kernel $ I_{L,m}$ is defined tautologically. This sequence, usually called the Artin reciprocity law, generalizes Gauss’ reciprocity. A prime $p$ splits completely in $L$ iff $p$ represents a class of $ I_{L,m}$ mod $f_L$ .

To get genuine « higher » symbols and reciprocity, you have to start from a number field $K \neq \mathbf Q$, but then comes the very hard part of CFT. Let us only say that, given an abelian $L/K$, the notion of a defining modulus $\mathcal M$ for $L/K$ can be adequately generalized (here, if one « neglects » the infinite primes, $\mathcal M$ is an ideal of the ring of integers $A_K$ of $K$), $C_m$ is replaced by $C_{\mathcal M}$ := $A_\mathcal M / R_{\mathcal M}$, where $A_\mathcal M$ is the group of fractional ideals coprime to $\mathcal M$ and $R_{\mathcal M}$ is the ray subgroup mod $\mathcal M$, the Artin symbol $(L/K,.)$ is defined by $(L/K, \mathcal P) (\alpha) \equiv \alpha^{N\mathcal P}$ mod $\mathcal P . A_L$ (here $N\mathcal P$ is the norm of the ideal $\mathcal P$ of $A_K$), and the Artin reciprocity law is the exact sequence $1 \to I_{L/K, \mathcal M} := N_{L/K} (C_{L,\mathcal M}) \to C_{K,\mathcal M} \to Gal(L/K) \to 1$

Concerning specifically your question, let me refer only to the exercises 1 and 2 of Cassels-Fröhlich. For a number field $K$ containing the group of $m$-th roots of 1, an $m$-th power residue symbol is defined in ex. 1.1, and is shown to give a direct extension of the Legendre symbol/quadratic residue (which correspond to $K=\mathbf Q$ and $m=2$). Similarly, the Hilbert symbols are defined in ex. 2, and the product formula proved in ex. 2.9 also generalizes directly the quadratic reciprocity law ./.

  • $\begingroup$ What is the arithmetic progression mod 4a you mention in the first paragraph? $\endgroup$
    – 54321user
    Jun 4, 2017 at 5:00
  • $\begingroup$ It is just a "down to earth" way to express the quadratic reciprocity law: if $p$ and $q$ are two odd primes not dividing $a$, and $p \equiv q$ mod $4a$, then $x^2 \equiv a$ mod $p$ has a solution iff $x^2 \equiv a$ mod $q$ has a solution. $\endgroup$ Jun 4, 2017 at 6:59

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