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Abstract Algebra

Exercise 2.2.13 (Action of $S_n$ on $\mathbb{Z}[x_1,x_2,\ldots,x_n]$)

Let n be a positive integer and let R be the set of all polynomials with integer coefficients in the independent variables x 1 , x 2 , … , x n , i.e., the members of R are finite sums of elements of the form a x 1 r 1 x 2 r 2 ⋯ x n r n , where a is any integer and r 1 , … , r n are nonnegative integers. For each σ ∈ S n define a map

σ : R → R by σ ⋅ p ( x 1 , x 2 , … , x n ) = p ( x σ ( 1 ) , x σ ( 2 ) , … , x σ ( n ) ) .

Prove that this defines a (left) group action of S n on R .

Answers

This is a generalization of Exercise 12 part (b).

Proof. First 1 ⋅ p = p . For all σ , τ in S n , if j = τ ( i ) then σ ⋅ x τ ( i ) = σ ⋅ x j = x σ ( j ) = x σ ( τ ( i ) ) , therefore

σ ⋅ ( τ ⋅ x i ) = σ ⋅ x τ ( i ) = x σ ( τ ( i ) ) = x ( στ ) ( i ) = ( στ ) ⋅ ( x i ) ( i = 1 , 2 , … , n ) . (1)

By definition

σ ⋅ p ( x 1 , x 2 , … , x n ) = p ( x σ ( 1 ) , x σ ( 1 ) , … , x σ ( n ) ) ,

then

σ ⋅ p ( x 1 , x 2 , … , x n ) = p ( σ ⋅ x 1 , σ ⋅ x 2 , … , σ ⋅ x n ) . (2)

Therefore

σ ⋅ ( τ ⋅ p ) = σ ⋅ ( τ ⋅ p ( x 1 , x 2 , … , x n ) ) = σ ⋅ p ( τ ⋅ x 1 , τ ⋅ x 2 , … , τ ⋅ x n ) (by (2)) = p ( σ ⋅ ( τ ⋅ x 1 ) , σ ⋅ ( τ ⋅ x 2 ) , … , σ ⋅ ( τ ⋅ x n ) ) (by (2)) = p ( ( στ ) ⋅ x 1 , ( στ ) ⋅ x 2 , … , ( στ ) ⋅ x n ) (by (1)) = ( στ ) ⋅ p ( x 1 , x 2 , … , x n ) (by (2)) = ( στ ) ⋅ p .

So φ is a left action of S n on R □

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2025-10-15 10:57
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