Zolotarev’s lemma∗
mathcam†
2013-03-21 15:41:49
We will identify the ring Zn of integers modulo n, with the set {0, 1, . . . n−1}.
Lemma 1 (Zolotarev). For any prime number p and any m ∈ Z∗p , the Legendre
symbol m
is equal to the signature of the permutation τm : x 7→ mx of Z∗p .
p
Proof. We write (σ) for the signature of any permutation σ. If σ is a circular
permutation on a set of k elements, then (σ) = (−1)k−1 . Let i be the order of
m in Z∗p . Then the permutation τm consists of (p − 1)/i orbits, each of size i,
whence
(τm ) = (−1)(i−1)(p−1)/i
If i is even, then
i p−1
i
m(p−1)/2 = m 2
= (−1)
p−1
i
= (τm )
And if i is odd, then 2i divides p − 1, so
m(p−1)/2 = mi
p−1
2i
= 1 = (τm ).
In both cases, the lemma follows from Euler’s criterion.
Lemma 1 extends easily from the Legendre symbol to the Jacobi symbol
for odd n. The following is Zolotarev’s penetrating proof of the quadratic
reciprocity law, using Lemma 1.
m
n
Lemma 2. Let λ be the permutation of the set
Amn = {0, 1, . . . , m − 1} × {0, 1, . . . , n − 1}
which maps the kth element of the sequence
(0, 0)(0, 1) . . . (0, n − 1)(1, 0) . . . (1, n − 1)(2, 0) . . . (m − 1, n − 1),
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to the kth element of the sequence
(0, 0)(1, 0) . . . (m − 1, 0)(0, 1) . . . (m − 1, 1)(0, 2) . . . (m − 1, n − 1),
for every k from 1 to mn. Then
(λ) = (−1)m(m−1)n(n−1)/4
and if m and n are both odd,
(λ) = (−1)(m−1)(n−1)/4 .
Proof. We will use the fact that the signature of a permutation of a finite totally
ordered set is determined by the number of inversions of that permutation. The
sequence (0, 0), (0, 1) . . . defines on Amn a total order ≤ in which the relation
(i, j) < (i0 , j 0 ) means
i < i0 or (i = i0 and j < j 0 ).
But λ(i0 , j 0 ) < λ(i, j) means
j 0 < j or (j 0 = j and i0 < i).
The only pairs ((i, j), (i0 , j 0 )) thatget inverted are, therefore, the ones with i < i0
n
and j > j 0 . There are indeed m
2
2 such pairs, proving the first formula, and
the second follows easily.
And finally, we proceed to prove quadratic reciprocity. Let p and q be distinct
odd primes. Denote by π the canonical ring isomorphism Zpq → Zp × Zq .
Define two permutations α and β of Zp × Zq by α(x, y) = (qx + y, y) and
β(x, y) = (x, x + py). Finally, define a map λ : Zpq → Zpq by λ(x + qy) = px + y
for x ∈ {0, 1, . . . q − 1} and y ∈ {0, 1, . . . p − 1}. Evidently λ is a permutation.
Note that we have π(qx + y) = (qx + y, y) and π(x + py) = (x, x + py), so
therefore
π ◦ λ ◦ π −1 ◦ α = β.
Let us compare the signatures of the two sides. The permutation m 7→ qx+y
is the composition of m 7→ qx and m 7→ m + y. The latter has signature 1,
whence by Lemma 1,
q q
q
(α) =
=
p
p
and similarly
p p
p
=
.
(β) =
q
q
By Lemma 2,
(π ◦ λ ◦ π −1 ) = (−1)(p−1)(q−1)/4 .
Thus
(−1)
(p−1)(q−1)/4
2
q
p
=
p
q
which is the quadratic reciprocity law.
Reference
G. Zolotarev, Nouvelle démonstration de la loi de réciprocité de Legendre,
Nouv. Ann. Math (2), 11 (1872), 354-362
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