PUMaC 2011 · 代数（A 组） · 第 7 题
PUMaC 2011 — Algebra (Division A) — Problem 7
题目详情
- [ 7 ] Let α , α , . . . , α be a fixed labeling of the complex roots of x − 1. Find the num- 1 2 6 ber of permutations { α , α , . . . , α } of these roots such that if P ( α , . . . , α ) = 0, then i i i 1 6 1 2 6 P ( α , . . . , α ) = 0, where P is any polynomial with rational coefficients. i i 1 6 11
解析
- Let ζ = e . Without loss of generality, let α = ζ for each i from 1 to 6. Then we have i α = − 1 and α = 1. Therefore, the equations f ( α , . . . , α ) = α + 1 = 0 and g ( α , . . . , α ) = 3 6 1 6 3 1 6 α − 1 = 0 show that α and α must be fixed by any such permutation. 6 3 6 5 2 4 We also have that ζ + ζ = 1 and ζ + ζ = − 1. Therefore we can see that f ( α , ..., α ) = 1 6 α + α − 1 = 0 and g ( α , ..., α ) = α + α + 1 = 0 are also polynomials of the desired form, 1 5 1 6 2 4 so these polynomials must also be zero upon permutation, and therefore ( α , α ) → ( α , α ) 2 4 2 4 or ( α , α ) → ( α , α ). Similarly, ( α , α ) → ( α , α ) or ( α , α ) → ( α , α ). 2 4 4 2 1 5 1 5 1 5 5 1 Suppose α and α are fixed by a permutation that also swaps α and α , and consider 2 4 5 1 2 the polynomial f ( α , ..., α ) = α − α = 0. This polynomial permutes to f ( α , ..., α ) = 1 6 2 i i 1 1 6 2 10 2 α − α = ζ − ζ 6 = 0. Similarly, the permutation that fixes α and α but reverses α and 2 5 1 2 5 α does not work due to the same polynomial as above. Finally, we need to show that the 4 final two permutations do work. Clearly the identity permutation works. It remains to show 5 2 4 that the permutation that fixes the roots ± 1 and swaps the pairs of roots ( ζ, ζ ) and ( ζ , ζ ) satisfies the conditions of the problem. This permutation is simply complex conjugation. Since we know that P ( α , α , α , α , α , α ) = 0, we have 1 2 3 4 5 6 P ( α , α , α , α , α , α ) = P ( α , α , α , α , α , α ) = 0 . 1 2 3 4 5 6 1 2 3 4 5 6 and thus both of these permutations work, and the answer is 2 .