Gauss's lemma (number theory)

This article is about Gauss's lemma in number theory. See also Gauss's lemma (polynomial).

Gauss's lemma in number theory, named after Carl Friedrich Gauss, gives a condition for an integer to be a quadratic residue. Although it is not useful computationally, it has theoretical significance, being involved in some proofs of quadratic reciprocity.

Statement of the lemma

For any odd prime p let a be an integer that is coprime to p.

Consider the integers

$a,2a,3a,\dots ,{\frac {p-1}{2}}a$

and their least positive residues modulo p. (These residues are all distinct, so there are (p−1)/2 of them.)

Let n be the number of these residues that are greater than p/2. Then

$\left({\frac {a}{p}}\right)=(-1)^{n}$

where (a/p) is the Legendre symbol.

Example

Taking p = 11 and a = 7, the relevant sequence of integers is

7, 14, 21, 28, 35.

After reduction modulo 11, this sequence becomes

7, 3, 10, 6, 2.

Three of these integers are larger than 11/2 (namely 6, 7 and 10), so n = 3. Correspondingly Gauss's lemma predicts that

$\left({\frac {7}{11}}\right)=(-1)^{3}=-1.$

This is indeed correct, because 7 is not a quadratic residue modulo 11.

Proof

A fairly simple proof of the lemma, reminiscent of one of the simplest proofs of Fermat's little theorem, can be obtained by evaluating the product

$Z=a\cdot 2a\cdot 3a\cdot \cdots \cdot {\frac {p-1}{2}}a$

modulo p in two different ways. On one hand it is equal to

$Z=a^{(p-1)/2}\left(1\cdot 2\cdot 3\cdot \cdots \cdot {\frac {p-1}{2}}\right)$

The second evaluation takes more work. If x is a nonzero residue modulo p, let us define the "absolute value" of x to be

$|x|={\begin{cases}x&{\mbox{if }}1\leq x\leq {\frac {p-1}{2}},\\-x&{\mbox{if }}{\frac {p+1}{2}}\leq x\leq p-1.\end{cases}}$

Since n counts those multiples ka which are in the latter range, and since for those multiples, −ka is in the first range, we have

$Z=(-1)^{n}\left(|a|\cdot |2a|\cdot |3a|\cdot \cdots \cdots \left|{\frac {p-1}{2}}a\right|\right).$

Now observe that the values |ra| are distinct for r = 1, 2, ..., (p−1)/2. Indeed, if |ra| = |sa|, then ra = ±sa, and therefore r = ±s (because a is invertible modulo p), so r = s because they are both in the range 1 ≤ r ≤ (p−1)/2. But there are exactly (p−1)/2 of them, so they must just be some rearrangement of the integers 1, 2, ..., (p−1)/2. Therefore

$Z=(-1)^{n}\left(1\cdot 2\cdot 3\cdot \cdots \cdot {\frac {p-1}{2}}\right).$

Comparing with our first evaluation, we may cancel out the nonzero factor

$1\cdot 2\cdot 3\cdot \cdots \cdot {\frac {p-1}{2}}$

and we are left with

$a^{(p-1)/2}=(-1)^{n}.$

This is the desired result, because the left hand side is just an alternative expression for the Legendre symbol (a/p).

Applications

Gauss's lemma finds its main application in proving quadratic reciprocity. The "supplementary law"

$\left({\frac {-1}{p}}\right)=(-1)^{(p-1)/2},$

which is part of the statement of quadratic reciprocity, can be deduced from Gauss's lemma by taking a = −1. With more care, the case a = 2 gives the other supplementary law.

Relation to the transfer in group theory

Let G be the multiplicative group of nonzero residue classes in Z/pZ, and let H be the subgroup {+1, −1}. Consider the following coset representatives of H in G,

$1,2,3,\dots ,{\frac {p-1}{2}}.$

Applying the machinery of the transfer to this collection of coset representatives, we obtain the transfer homomorphism

$\phi :G\to H,$

which turns out to be the map that sends a to (-1)ⁿ, where a and n are as in the statement of the lemma. Gauss's lemma may then be viewed as a computation that explicitly identifies this homomorphism as being the quadratic residue character.