Algebra, IV, p. 98
Bourbaki marks this one as harder than the rest.
Let K be a commutative field and f∈K[X1,…,Xn,Y1,…,Yn]; for every permutation σ∈Sn denote by σf the rational fraction
σf(X1,…,Xn,Y1,…,Yn)=f(Xσ(1),…,Xσ(n),Yσ(1),…,Yσ(n)). Then
f is said to be symmetric with respect to the pairs
(Xi,Yi) if
uf=f for all
σ∈Sn. a) For every element
ν=(A,…,A,μ1,…,μn) of
N2n and every permutation
σ∈Sn write
σ−1(ν)=(λσ(1),…,λσ(n),μσ(1),…,μσ(n)),
so that Sn in this way becomes a group of operators on N2n. Let γ be any orbit of Sn in N2n, and denote by uγ the symmetric polynomial
∑X1λ1X2λ2…XnλnY1μ1…Ynμn
where (λ1,…,λn,μ1,…,μn) ranges over γ. Show that the polynomials uγ form a basis over K of the vector space of all polynomials that are symmetric in the (Xi,Yi).
\* b) Suppose that K has characteristic 0. Show that every polynomial uγ is equal to a polynomial with rational coefficients in the symmetric functions
vλμ=i=1∑nXiλiYiμi
(argue by induction on the number of pairs (Ai,μi) which are not (0,0) in ν=(λ1,…,λn,μ1,…,μn) in considering the products u,vλμ).
c) By elementary symmetric functions of the (Xi,Yi) we shall understand the n(n+3)/2 polynomials whk=u, corresponding to the orbits γ of elements (Ai,…,Ai,μ1,…,μn) other than (0,…,0) and such that h of the pairs (λi,μi) are equal to (1,0), k others are (0,1) and the remaining n−h−k are (0,0). Show that if K is of characteristic 0, then every symmetric polynomial in the (Xi,Yi) is equal to a polynomial in the elementary symmetric functions with coefficients in K (consider the sums ∑i=1n(UXi+VYi)k, where U and V are two indeterminates, and use b)).
d) If K is a field of characteristic 3, and n≥4, show that ∑i=1nXi2Yi2 is not equal to any polynomial in the symmetric functions whk with coefficients in K. \*