aa r X i v : . [ m a t h . A C ] N ov A nice involution for multivariable polynomial rings
Wiland Schmale ∗
25. November 2020
Abstract
The principal minors of the Toeplitz matrix ( x i − j +1 ) ≤ i,j, ≤ n , where x = 1 , x k = 0if k ≤ −
1, directly determine an involution of the polynomial ring R [ x , ..., x n ] overany commutative ring R . keywords: involution, multivariable polynomial ring, principal Toeplitz minorsIn [1] principal minors have been used to transform a conjecture on certain Toeplitzpencils into an equivalent algebraic conjecture. A closer look on this transition to minorsinstead of the original variables shows that it is actually an involution.Let R be any commutative ring, R [ x , ..., x n ] the polynomial ring in the independentvariables x , ..., x n , n ≥
1, over R and T ( k ) = x . . . x x . . . x k − x k − . . . x k x k − . . . . . . x , m k = m k ( x , ..., x k ) = det( T ( k )) for ≤ k ≤ n, Note that T ( k ) is a submatrix and m k a principal minor of T ( n ). One has m = x , m = x − x and so on.Let now the substitution ring homomorphism ϕ of R [ x , ..., x n ] over R be defined as ϕ : R [ x , ..., x n ] → R [ x , ..., x n ] , x k m k , for 1 ≤ k ≤ n. The following can then be observed:
Involution property: ϕ ◦ ϕ = id R [ x ,...,x n ] = identity map on R [ x , ..., x n ]The proof will be based on the following recursive properties of principal minors of T ( n ). Recursions for m k m k m k and x k x k x k : For 1 ≤ k ≤ nm k = k X i =1 ( − i − x i m k − i , m := 1 (1)and x k = k X i =1 ( − i − m i x k − i , x := 1 (2) ∗ University of Oldenburg (retired), Germany, email: [email protected] roof: In order to obtain (1) we will prove the following slightly more general relationfor any first column entries a , ..., a k :det a . . . a x . . . a k − x k − . . . a k x k − . . . . . . x = k X i =1 ( − i − a i m k − i , where m := 1 . This implies relation (1) as a special case. The instance “ k = 1” beeing trivial we proceedinductively and obtain for 1 ≤ k ≤ n − a . . . a x . . . a k x k − . . . a k +1 x k . . . . . . x = a · det T ( k ) − det a . . . a x . . . a k x k − . . . a k +1 x k − . . . . . . x = a m k − k X i =1 ( − i − a i +1 m k − i (by induction)= a m ( k +1) − + k +1 X i =2 ( − i − a i m k +1 − i = k +1 X i =1 ( − i − a i m ( k +1) − i In order to obtain (2) one only has to solve (1) for x k which actually is the last term inthe sum up to the factor ( − k − :We obtain( − k − x k = m k x − k − X i =1 ( − i − x i m k − i = k − X i =1 ( − i x i m k − i + m k x x k = k − X i =0 ( − i + k − x i m k − i = k X j =1 ( − k − j + k − m j x k − j , where k − j + k − ≡ j − ✷ The recursions (1) and (2) lead us directly to the involution property . It will besufficient to show that ϕ ( ϕ ( x k )) = x k for 1 ≤ k ≤ n . Since ϕ ( ϕ ( x )) = ϕ ( m )) = ϕ ( x ) = m = x