5

For number $n\ge2$, let $\xi$ be a primitive $n$-th root of unity.

The determinant of circulant matrix is a symmetric polynomial in $x_0,\dots,x_{n-1}$ $$f_n=\prod_{j=0}^{n-1}\sum_{i=0}^{n-1}ΞΎ^{ij}x_i$$ so after expansion, all coefficients are integer.

Is the following true?

For prime number $n$, all coefficients of$$f_n-\sum_{i=0}^{n-1}x_i^n$$are divisible by $n$.

Using SageMath I verified it for $n=2,3,5,7$.

For $n=5$:

from sympy import symbols
x = symbols('x_:5')

f5 = prod(sum(exp(I2pi/5ji) * x[i] for i in (0..4)) for j in (0..4))
poly = expand(f5- sum(x[i]**5 for i in (0..4))).maxima_methods().rootscontract()

output: $-5x_{0} x_{1}^{3} x_{2} + 5x_{0}^{2} x_{1} x_{2}^{2} + 5x_{0}^{2} x_{1}^{2} x_{3} - 5x_{0}^{3} x_{2} x_{3} - 5x_{1} x_{2}^{3} x_{3} + 5x_{1}^{2} x_{2} x_{3}^{2} + 5x_{0} x_{2}^{2} x_{3}^{2} - 5x_{0} x_{1} x_{3}^{3} - 5x_{0}^{3} x_{1} x_{4} + 5x_{1}^{2} x_{2}^{2} x_{4} - 5x_{0} x_{2}^{3} x_{4} - 5x_{1}^{3} x_{3} x_{4} - 5x_{0} x_{1} x_{2} x_{3} x_{4} + 5x_{0}^{2} x_{3}^{2} x_{4} - 5x_{2} x_{3}^{3} x_{4} + 5x_{0} x_{1}^{2} x_{4}^{2} + 5x_{0}^{2} x_{2} x_{4}^{2} + 5x_{2}^{2} x_{3} x_{4}^{2} + 5x_{1} x_{3}^{2} x_{4}^{2} - 5x_{1} x_{2} x_{4}^{3} - 5x_{0} x_{3} x_{4}^{3}$

poly/5 in ZZ[x]

output: True

hbghlyj
  • 5,361

1 Answers1

2

We work in $K=\mathbf{Z}/n\mathbf{Z}$. The circulant matrix $J$ verifies $J^n=I_n.$ Since we are in characteristic $n$ this matrix $J$ only $1$ as eigenvalues. Now as it stands $\phi_n=\sum_{k=0}^{n-1} X_kJ^k\in K[X_1,\dots,X_n] ,$ and $\sum_{k=0}^{n-1} X_k^n=(\sum_{k=0}^{n-1}X_k)^n.$ Consider the matrix : $$N:=\sum_{k=0}^{n-1} X_k (J^k-I_n)$$ It is nilpotent, since $J^k$ are unipotent. Therefore $\phi_n = (\sum_{k=0}^{n-1}X_k)I_n +N$. Hence by taking determinants $$f_n\mod n =\det \phi_n=\det \left((\sum_{k=0}^{n-1}X_k)I_n+ N\right)= (\sum_{k=0}^{n-1} X_k)^n= \sum_{k=0}^{n-1}X_k^n.$$ This translates to $f_n-\sum_{k=0}^{n-1}X^n_k \equiv 0 $ in $\mathbf{Z}/n\mathbf{Z}$. Hope that helps.

HomerSimpson
  • 56