Gauss Sums As we have seen, there is a close connection between Legendre symbols of the form 3 and cube roots of unity. Secifically, if is a rimitive cube root of unity, then 2 ± i 3 and hence 2 2 3 In fact, this last equation holds for any element of order 3 in any field F, and hence 3 is a erfect square in any field that has elements of order 3. There are similar considerations for other rimes. For examle, if is a rimitive 5th root of unity, then 2 3 + 4 ± 5. and hence 2 3 + 4 2 5. Again, it is ossible to show that this last equation holds for any element of order 5 in any field F, and therefore 5 is a erfect square in any field that has elements of order 5. Gauss discovered a beautiful generalization of these formulas. Theorem 1 Gauss Sum Formula Let > 2 be rime, and let be a rimitive th root of unity. Then 1 { k ± if 1 mod 4, k ± i if 3 mod 4.
Gauss Sums 2 The sum g 1 k k + 2 2 + 3 3 + + 1 1 is known as a Gauss sum. According to the theorem g 2 { if 1 mod 4, if 3 mod 4, for any rimitive th root of unity. Equivalently, we can write this formula as 1 g 2. EXAMPLE 1 Gauss Sum for 7 It is easy to check that the quadratic residues modulo 7 are {1, 2, 4}, while {3, 5, 6} are quadratic non-residues. Therefore, by the Gauss sum formula + 2 3 + 4 5 6 ± i 7 for any rimitive 7th root of unity. It is not too hard to check that this is correct. Squaring the Gauss sum gives + 2 3 + 4 5 6 2 2 + 2 3 4 + 6 6 7 + 8 10 + 2 11 + 12 and using the identity 7 1 to reduce the owers of simlifies this to + 2 3 + 4 5 6 2 6 + + 2 + 3 + 4 + 5 + 6. But 1 + + 2 + 3 + 4 + 5 + 6 Φ 7 0, and hence + 2 3 + 4 5 6 2 7. EXAMPLE 2 Gauss Sum for 11 It is easy to check that the quadratic residues modulo 11 are {1, 3, 4, 5, 9}, while {2, 6, 7, 8, 10} are quadratic non-residues. Therefore, by the Gauss sum formula 2 + 3 + 4 + 5 6 7 8 + 9 10 ± i 11 for any rimitive 11th root of unity.
Gauss Sums 3 Again, we can use simle algebra to show that this is correct. Squaring the Gauss sum and alying the identity 11 1 gives the formula 2 + 3 + 4 + 5 6 7 8 + 9 10 2 10 + + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 But 1 + + 2 + + 10 Φ 11 0, and hence 2 + 3 + 4 + 5 6 7 8 + 9 10 2 11. EXAMPLE 3 Gauss Sum for 13 It is easy to check that the quadratic residues modulo 13 are {1, 3, 4, 9, 10, 12}, while {2, 5, 6, 7, 8, 11} are quadratic non-residues. Therefore, by the Gauss sum formula 2 + 3 + 4 5 6 7 8 + 9 + 10 11 + 12 ± 13 for any rimitive 13th root of unity. Since 13 1 mod 4, the algebra for checking this goes a little differently. Squaring the Gauss sum and then reducing owers of modulo 13 gives 2 + 3 + 4 5 6 7 8 + 9 + 10 11 + 12 2 12 2 3 4 5 6 7 8 9 10 11 12. But 1 + + 2 + + 12 Φ 13 0, and hence 2 + 3 + 4 5 6 7 8 + 9 + 10 11 + 12 2 13. Of course, the Gauss sum formula gives two ossible values of g in each case, so a natural question to ask is which of these two values g is equal to. For examle, if 1 mod 4, is g equal to or. The answer is that it deends on which rimitive th root of unity we choose. However, in the case where e 2πi/, Gauss was able to rove that { if 1 mod 4, g i if 3 mod 4. For examle, if e 2πi/7 then and if e 2πi/13 then + 2 3 + 4 5 6 i 7, 2 + 3 + 4 5 6 7 8 + 9 + 10 11 + 12 13. This result is actually much more difficult than the Gauss sum formula, and we will not rove it here.
Gauss Sums 4 Proof of the Gauss Sum Formula Throughout this section, let > 2 be a rime, and let be a rimitive th root of unity. Let g x be the Gauss olynomial g x 1 k x k. Our goal is to rove that g 2 1. Extension of the Legendre Symbol For convenience, we will use the convention that a 0 if a. Using this notation, g x 1 k0 k where the sum starts at k 0 instead of k 1. x k, Squaring the Gauss Sum Observe first that 1 1 g 2 j0 k0 j k j+k. Since 1, we can reduce each ower of modulo and then combine like terms. This yields an equation of the form g 2 a 0 + a 1 + a 2 2 + + a 1 1 1 where for each n Z. a n j+k n mod j k 2
Gauss Sums 5 Sum of the Coefficients Note first that g 1 1 k since Z has an equal number of quadratic residues and quadratic non-residues. It follows that g 1 2 0, so the sum of the coefficients in g x 2 is equal to 0. Therefore, 0 a 0 + a 1 + + a 1 0. 3 Value of a 0 It is not hard to determine the value of a 0. By equation 2, we have a 0 j+k 0 mod j k 1 j0 j j. But and thus j j a 0 1 j1 j 2 1 1 0 if j 0, 1 otherwise. 1 1. 4 Equality of the Remaining Coefficients Let n Z. By equation 2, we have that a n j j+k n mod k. If we make the substitution j nj and k nk, then j +k 1 mod, and indeed a n nj nk. j +k 1 mod But nj nk n 2 j k j k
Gauss Sums 6 and hence a n for all n {1,..., 1}. Thus j +k 1 mod j k a 1 a 1 a 2 a 1. 5 End of the Proof Equations 3 and 5 are a 0 + a 1 + + a 1 0 and a 1 a 2 a 1 and combining these together gives a n a 0 1 for each n {1,..., 1}. Substituting in the value of a 1 obtained in 4, we deduce that 1 a n for each n {1,..., 1}. Thus equation 1 becomes 1 g 2 1 2 1. But so and hence 1 + + 2 + + 1 Φ 0 + 2 + + 1 1. g 2 1. This comletes the roof of the Gauss sum formula.
Gauss Sums 7 Symmetry of Gauss Sums The Gauss sum formula tells us that g 2 1 for any rimitive th root of unity. The following formula tells us how the sign of g changes when we use different th roots of unity. Proosition 2 Symmetry of the Gauss Sum Let > 2 be a rime, let be a rimitive th root of unity, and let g x 1 k x k. Then for each n {1,..., 1}, g n n g. PROOF Observe that g n 1 k x nk n 1 nk x nk. But as k runs over the set {1,..., 1}, the roduct m nk also runs over the elements of this set. Hence we can substitute m nk to get g n n 1 m x m m1 n g. EXAMPLE 4 Consider the olynomial If e 2πi/7, then it is easy to check that g 7 x x + x 2 + x 4 x 3 x 5 x 6. g 7 + 2 + 4 3 5 6 i 7.
Gauss Sums 8 According to the formula above, it follows that g 7 n n g 7 for any n Z. For examle, n i 7 g 7 2 2 + 2 2 + 2 4 2 3 2 5 2 6 2 + 4 + 6 3 5 g 7 i 7 since 2 is a quadratic residue modulo 7, and g 7 3 3 + 3 2 + 3 4 3 3 3 5 3 6 3 + 6 + 5 2 4 g 7 i 7 since 3 is a quadratic non-residue modulo 7.