Exercise 2.2.15

Given a polynomial f F [ x 1 , , x n ] , let deg i ( f ) be the maximal exponent of x i which appears in f . Thus f = x 1 3 x 2 + x 1 x 2 4 has degree deg 1 ( f ) = 3 and deg 2 ( f ) = 4 .

(a)
If f is symmetric, explain why the deg i ( f ) are the same for i = 1 , , n .
(b)
Show that deg i ( σ 1 a 1 σ n a n ) = a 1 + a 2 + + a n for i = 1 , , n .

Answers

Proof.

(a)
If x 1 appears in a term c x 1 a 1 x 2 a 2 x n a n of f , then the transposition τ = ( 1 , 2 ) applied to f show that c x 2 a 1 x 1 a 2 x n a n is a term of f , so x 2 appears in a term of f with the same exponent. Thus the maximal exponent is the same for the two variables : deg 1 ( f ) = deg 2 ( f ) ,

and the same is true for any pair of variables.

(b)
As σ k = 1 i 1 < < i k n x i 1 x i 2 x i k , deg i ( σ k ) = 1 . For polynomial of one variable x , deg ( pq ) = deg ( p ) + deg ( q ) , and deg 1 ( f ) is the degree in x 1 of f as an element of k [ x 2 , , x n ] [ x 1 ] , so deg i ( fg ) = deg i ( f ) + deg i ( g ) .

Therefore deg i ( σ 1 a 1 σ n a n ) = a 1 deg i ( σ 1 ) + + a n deg i ( σ n ) = a 1 + + a n .

User profile picture
2022-07-19 00:00
Comments