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For the solution of congruences $x^2\equiv p \pmod q$ and $x^2\equiv q \pmod p$, we have the law of quadratic reciprocity. Is there any solution for the above congruences?

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I'm guessing you are not really demanding the same $x$ work in both equations... –  Will Jagy Jul 22 '10 at 19:29
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I think it would be reasonable to interpret this question as asking for generalizations of quadratic reciprocity. The most down-to-earth reference I know for this is Ireland and Rosen's Classical Introduction to Modern Number Theory. –  Qiaochu Yuan Jul 22 '10 at 19:39
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@Will By the Chinese remainder theorem, it doesn't matter whether we take x to be the same in both cases or not. –  David Speyer Jul 22 '10 at 20:08
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There is no article on the higher reciprocity law on wikipedia, but at least the cubic case, en.wikipedia.org/wiki/Cubic_reciprocity, is elaborated in quite details. Then you can guess that in general one has to go to cyclotomic extensions and generalised Gauss sums. –  Wadim Zudilin Jul 22 '10 at 23:45
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I think what you're looking for is known as the Eisenstein Reciprocity Law. A good source for this and related material is Reciprocity Laws: From Euler to Eisenstein by Franz Lemmermeyer. –  Kevin Ventullo Jul 23 '10 at 3:15
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1 Answer

Let me interpret the question as follows: is there a simple relationship between the solvability of $x^n \equiv p \bmod q$ and that of $x^n \equiv q \bmod p$ for primes $p \equiv q \equiv 1 \bmod n$?

Here the answer is yes if $n = 2$, and no otherwise. The reason is the following: the congruence $x^n \equiv p \bmod q$ for primes $q \equiv 1 \bmod n$ is solvable if and only if the primes above $q$ in the field $K$ of n-th roots of unity split completely in the Kummer extension $L = K(\sqrt[n]{p})$. This splitting behaviour depends only on the residue classes modulo primes above $p$ by class field theory, the reason being that $L/K$ is abelian. This gives rise to a reciprocity law, which basically states a very close connection between $(p/{\mathfrak q})_n$ and $({\mathfrak q}/p)_n$ (the second symbol has to be suitably interpreted, which is possible if $K$ has class number $h$ coprime to $n$).

What this means is that any reasonable connection between $(p/q)_n$ and $(q/p)_n$ will have to use information on the principal ideals ${\mathfrak p}^h$ and ${\mathfrak q}^h$ . In the simplest nontrivial case $n = 4$, the connection is provided by "Burde's" reciprocity law $(p/q)_4 (q/p)_4 = (ac-bd/p)$, where $p = a^2 + b^2$ and $q = c^2 + d^2$, with everything suitably normalized.

Moreover, you can't expect something similar to hold over the rationals, because the extension ${\mathbb Q}(\sqrt[n]{p})$ is not abelian over ${\mathbb Q}$ except when $n = 2$.

There is a similar (but already more complicated) law for $n = 3$; the case $n = 4$ is special because $i^2 = -1$ is rational.

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I oppose the word "simple" in the interpretation. –  Dror Speiser Jul 24 '10 at 9:11
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