Reciprocity Laws by Richard Taylor (Fields Medal Symposium, 2012年10月16日).
Let p be an odd prime dividing n2+1 where n is an integer. Modulo p, the residue class of n is a "square root of -1" which we'll call "i" (as their squares are opposites, i and 1 are distinct, since p ¹ 2).
Modulo a prime number, the residues form a field (a finite ring without divisors of zero is necessarily a field ). The p-1 nonzero residues, may be partitioned into disjoint sets, each containing the opposites and the reciprocals of its own elements. The finest (i.e., "least coarse") such partition has two sets of 2 elements besides sets with 4 elements, viz.
{1,-1} {i,-i} {x,-x,1/x,-1/x} (for any x besides 1, -1, i ,-i )
Thus, p-1 must be a multiple of 4 (conversely, modulo any such prime p, there must be a residue i with a square congruent to -1 or else the lack of the second pair above would mess up the count). This result was first stated in 1632, without proof, by Albert Girard (1595-1632). It's the first line in the following table.
In fact, Girard conjectured that a lot more was true, namely that any prime p of the form 4n+1 could always be written as the sum of two perfect squares, in a unique way). Fermat offered a proof of that in a letter to Mersenne dated 1640年12月25日 (Christmas day). So, the result became known as Fermat's Christmas Theorem.
Why was this visual proof missed for 400 years? (33:58) by Burkar Polster (Mathologer, 2020年01月25日).
Euler conjectured that, for any prime q, any prime factor p of n2-q (besides, possibly, p=2 and p=q) must be of the form 4qk+b2 or 4qk-b2 for some odd integer b. That's the earliest statement of the law of quadratic reciprocity.
Adrien-Marie Legendre (1752-1833)Although special cases had been noted by Euler and Lagrange, the fully general theorem is credited to Legendre, who devised a special notation to express it.
Carl Friedrich Gauss (1777-1855)More than 200 proofs of this deep result have been tallied. Carl Friedrich Gauß found the first of these in 1795 and published it in Section 4 of Disquisitiones Arithmeticae (1796). He came up with no fewer than 8 different proofs over his lifetime. None of these truly satisfied him, though... Apparently, the multitude of proofs serves only to increase the mystery shrouding the law of quadratic reciprocity, which Gauß dubbed Aureum Theorema (the golden theorem). The following articles present this golden topic, which once stirred the enthusiasm of the great Gauss.
A quadratic residue is simply the congruence class of a square. The term is usually reserved to nonzero residues. By contrast, a residue which is not congruent to any square is sometimes called a quadratic nonresidue [sic].
Modulo an odd prime p, there are just (p-1)/2 quadratic residues, namely:
12, 22, ... [(p-1)/2]2 (modulo p)
Those are all distinct because the residue ring Z / pZ is known to be a field (a finite ring can be proven to be a field if there are no divisors of zero in it). In a field, x and y have the same square if and only if
(x-y) (x+y) = 0
This implies that y is either x or -x (these two are distinct, since the field at hand is not of characteristic 2). The opposite of a residue x being of the form p-x, the above does enumerate all the quadratic residues modulo p.
Leonhard Euler (1707-1783) devised the first effective test to tell quickly whether a given residue is a quadratic residue or not. This is known as Euler's criterion : Modulo an odd prime p, the nonzero residue a is a quadratic residue if and only if the following equation is true, modulo p :
a (p-1)/2 = 1
This is easily established by considering a primitive root g (i.e., a residue generating multiplicatively the group of the p-1 nonzero residues modulo p). The even powers of g are quadratic residues and the odd powers are not.
In particular, g itself can't be a quadratic residue (the order of g must be p-1, and it would be at most (p-1)/2 if g was congruent to the square of some x, since the order of x divides p-1, by Fermat's little theorem).
If a is an even power of g , the above relation holds (as the left-hand side is something raised to the power of p-1). Conversely, if a is an odd power of g, then the left-hand side is congruent to g raised to (p-1)/2 which must be congruent to -1 (as it can't be congruent to 1 while its square is). QED
Part of the secret of analysis is the art of using notations well.
Gottfried Wilhelm von Leibniz
(1646-1716)
At first, the Legendre symbol (a|p)
was defined as follows,
for any integer
a and [only] for an odd prime p.
For typographical reasons, we'll use the "horizontal" version (a|p) of the symbol, although the more traditional "vertical" version is preferred in print.
Euler's criterion, as discussed above, may thus be simply expressed:
If p is an odd prime, then (a|p) = a (p-1)/2 (modulo p).
The Legendre symbol was introduced specifically to stress the symmetrical relationship between (m|n) and (n|m) when m and n are both odd primes. This relation (the law of quadratic reciprocity) is expressed by the last two of the following fundamental properties of the Legendre symbol :
[ Skip the rest of this article on first reading. ]
It was later realized that it would be nice to consistently define the meaning of the symbol for less restricted arguments, while maintaining the above.
Although it's traditionally done, there's no compelling reason to call a generalized version of the Legendre symbol by any other name... The Jacobi symbol and Kronecker symbol are just to the restricted Legendre symbol what real and complex exponents are to integral exponents [ for a positive base, of course ].
Jacobi has defined (a|k) for any odd integer k by adding the simple rule:
Kronecker defined the symbol for all integers by adding the rules:
Some authors argue that the above needlessly defines the symbol (n|2) when n is congruent to 2, 3, 6 or 7 modulo 8. They'd rather leave such cases undefined.
(2 | q) = (-1)(q2-1)/8
For two distinct odd primes p and q, the following statements are either both true or both false, unless p and q are both congruent to 3 modulo 4, in which case one statement is true and the other is not:
The late John Conway has observed (video) that the exact same statement also holds if either p or q is equal to -1. For this and other reasons, Conway liked to consider -1 to be an odd prime which has only two powers (+1 and -1) instead of infinitely many. (This viewpoint allows a natural extension of the fundamental theorem of arithmetic to negative integers.)
Indeed, q = -1 is clearly a quadratic residue modulo an odd prime p when p isn't congruent to 3 modulo 4 (Girard, 1632). Conversely, when p is congruent to 3 modulo 4 (so that both p and q are) then -1 isn't a quadratic residue modulo p. Thus, the law stated above holds if and only if the following statement is properly interpreted so that it holds true:
p is a quadratic residue modulo -1.
That's trivially the case because there's only one residue class "modulo -1" and it's therefore equal to its own square!
Theorem
of the Day #29 by Robin Whitty
Quadratic Reciprocity (56:36)
by Burkar Polster (Mathologer, 2020年03月14日).
For a given odd prime p, let's denote < a > the integer congruent to a which is between -p/2 and +p/2. Gauss' lemma states that
Proof : Since the result is trivial when p divides a (the sign of 0 being 0) we may assume that a is invertible modulo p.
The absolute values | < j a > | are all distinct, since a is invertible modulo p [otherwise, distinct values of j would be either equal or opposite modulo p, which is ruled out in the range from 1 to (p-1)/2]. These absolute values thus assume all the values from 1 to (p-1)/2, once and only once (the pigeonhole principle ) and their product F is the factorial of (p-1)/2. Therefore, (using Euler's criterion) we have the following equalities, modulo p:
The result follows by cancelling F (which is coprime to p) and observing that equality modulo p certainly implies equality in the range {-1,+1}. QED
The idea and the result are similar to Gauss's lemma, except that the nonzero residues (modulo an odd prime p) are split into even and odd ones, instead of being split into low and high ones, as in Gauss' original lemma.
If r is the (least positive) remainder of ja modulo p, then (-1)r r has an even least positive remainder modulo p
To use Gauss' lemma when a is between 1 and p-1, we may use the following pseudocode to count the number of times k for which x = < j a > is negative:
x := 0 ; k := 0 ;
for j := 1 to (p-1)/2
x := x + a
if x > p then x := x - p
if x > p/2 then k := k+1
next j
Despite its simple appearance, this procedure is much lesss efficient than a fast implementation of Euler's criterion. Nevertheless, it makes things easier to tally from a theoretical standpoint.
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Cubic reciprocity
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Eisenstein integers
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Unique factorization domain (UFD)
"...entiers complexes cubiques" (1908)
Raymond Le Vavasseur (1862-1930; ENS 1888">ENS 1883).
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Quartic reciprocity | Gaussian integers | Carl F. Gauss (1777-1855)
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Eisenstein reciprocity | Gotthold Eisenstein (1823-1852)
This is a partial solution to Hilbert's ninth problem.
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Artin's Reciprocity and Mersenne Primes (pdf) by H.W. Lenstra Jr. and P. Stevenhagen (March 2000).
