Exercise 13.3.2

Let f ( x ) = x n c 1 x n 1 + . . . + ( 1 ) n c n [ x ] , and let Θ f ( y ) be the resolvent built from φ [ x 1 , . . . , x n ] . Prove that Θ f ( y ) [ y ] .

Proof. Let G be the symmetry group of φ and τ 1 , . . . , τ l be representatives for the left cosets of G in S n . The universal resolvent is Θ ( y ) = i = 1 l ( y ( τ i φ ) ( x 1 , . . . , x n ) ) . Since φ [ x 1 , . . . , x n ] , for each i , 1 i l , ( τ i φ ) ( x 1 , . . . , x n ) [ x 1 , . . . , x n ] , hence the coefficients of Θ ( y ) are in [ x 1 , . . . , x n ] , i.e., Θ ( y ) [ x 1 , . . . , x n ] [ y ] .

Suppose that σ S n , then ( σ Θ ) ( y ) = i = 1 l ( y ( σ τ i ) φ ( x 1 , . . . , x n ) ) . But the set σ τ 1 , . . . , σ τ l is also a set of left coset representatives of G in S n . Thus the application of σ has merely permuted the roots of Θ ( y ) leaving the coefficients fixed. It means that coefficients of Θ ( y ) are symmetric and are polynomials in σ 1 , . . . , σ n with integers coefficients (cf. Ex.9.1.6), i.e., Θ ( y ) [ σ 1 , . . . , σ n ] [ y ] . The application of evaluation map x i α i to Θ ( y ) , so that σ i c i , gives Θ f ( y ) [ c 1 , . . . , c n ] [ y ] = [ y ] . □

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2022-07-19 00:00
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