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Gauss' lemma (Theorem)

Gauss' lemma on quadratic residues is:

Proposition 1: Let $p$ be an odd prime and let $n$ be an integer which is not a multiple of $p$ . Let $u$ be the number of elements of the set $$\left\{n,2n,3n,\ldots,\frac{p-1}{2}n\right\}$$ whose least positive residues, modulo $p$ , are greater than $p/2$ . Then $$\legsymp{n}=(-1)^u$$ where $\legsymp{n}$ is the Legendre symbol.

That is, $n$ is a quadratic residue modulo $p$ when $u$ is even and it is a quadratic nonresidue when $u$ is odd.

Gauss' Lemma is the special case $$ S=\left\{1,2,\ldots,\frac{p-1}{2}\right\ $$ of the slightly more general statement below. Write $ \mathbbmss{F}_p$ for the field of $p$ elements, and identify $ \mathbbmss{F}_p$ with the set $\{0,1,\ldots,p-1\}$ , with its addition and multiplication mod $p$ .

Proposition 2: Let $S$ be a subset of $ \mathbbmss{F}_p^{*}$ such that $x\in S$ or $-x\in S$ , but not both, for any $ x\in \mathbbmss{F}_p^{*}$ . For $ n\in \mathbbmss{F}_p^{*}$ let $u(n)$ be the number of elements $k\in S$ such that $kn\notin S$ . Then $$\legsymp{n} = (-1)^{u(n)}.$$ Proof: If $a$ and $b$ are distinct elements of $S$ , we cannot have $an=\pm bn$ , in view of the hypothesis on $S$ . Therefore $$\prod_{a\in S}an=(-1)^{u(n)}\prod_{a\in S}a.$$ On the left we have $$n^{(p-1)/2}\prod_{a\in S}a=\legsymp{n}\prod_{a\in S}a$$ by Euler's criterion. So $$\legsymp{n}\prod_{a\in S}a=(-1)^{u(n)}\prod_{a\in S}a$$ The product is nonzero, hence can be cancelled, yielding the proposition.

Remarks: Using Gauss' Lemma, it is straightforward to prove that for any odd prime $p$ :

\begin{displaymath} \left(\frac{-1}{p}\right)= \begin{cases} 1 & \textrm{ if $p\... ...pmod 4$} \ -1 & \textrm{ if $p\equiv -1\pmod 4$} \end{cases} \end{displaymath}

\begin{displaymath} \left(\frac{2}{p}\right)= \begin{cases} 1 & \textrm{ if $p\e... ...d 8$} \ -1 & \textrm{ if $p\equiv \pm 3\pmod 8$} \end{cases} \end{displaymath}

The condition on $S$ can also be stated like this: for any square $ x^2\in \mathbbmss{F}_p^{*}$ , there is a unique $y\in S$ such that $x^2=y^2$ . Apart from the usual choice $$S=\{1,2,\ldots,(p-1)/2\}\;,$$ the set $$\{2,4,\ldots,p-1\}$$ has also been used, notably by Eisenstein. I think it was also Eisenstein who gave us this trigonometric identity, which is closely related to Gauss' Lemma: $$\legsymp{n}=\prod_{a\in S}\frac{\sin{(2\pi an/p)}}{\sin{(2\pi a/p)}}$$

It is possible to prove Gauss' Lemma or Proposition 2 ``from scratch'', without leaning on Euler's criterion, the existence of a primitive root, or the fact that a polynomial over $ \mathbbmss{F}_p$ has no more zeros than its degree.




"Gauss' lemma" is owned by drini. [ full author list (3) | owner history (2) ]
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See Also: Euler's criterion, Zolotarev's lemma

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Cross-references: degree, polynomial, primitive root, trigonometric identity, square, product, Euler's criterion, hypothesis, proof, subset, multiplication, addition, field, quadratic nonresidue, even, Legendre symbol, residues, positive, elements, number, multiple, integer, prime, odd, quadratic residues
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This is version 11 of Gauss' lemma, born on 2002-02-14, modified 2006-12-22.
Object id is 1955, canonical name is GaussLemma.
Accessed 8825 times total.

Classification:
AMS MSC11A15 (Number theory :: Elementary number theory :: Power residues, reciprocity)

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