Symmetric polynomials and non-finitely generated Sym(N)Sym (\mathbb N)-invariant ideals

preprint2013arXivOpen access

Abstract

Let KK be a field and let N={1,2,… }\mathbb N = \{1,2, \dots \}. Let Rn=K[xij∣1≤i≤n,j∈N]R_n=K[x_{ij} \mid 1\le i\le n, j\in \mathbb N] be the ring of polynomials in xijx_{ij} (1≤i≤n,j∈N)(1 \le i \le n, j \in \mathbb N) over KK. Let Sn=Sym({1,2,…,n})S_n = Sym (\{1,2, \ldots, n \}) and Sym(N)Sym (\mathbb N) be the groups of the permutations of the sets {1,2,…,n}\{1,2,\dots, n \} and N\mathbb N, respectively. Then SnS_n and Sym(N)Sym (\mathbb N) act on RnR_n in a natural way: τ(xij)=xτ(i)jτ(x_{ij})=x_{τ(i)j} and σ(xij)=xiσ(j)σ(x_{ij})=x_{iσ(j)} for all τ∈Snτ\in S_n and σ∈Sym(N)σ\in Sym(\mathbb N). Let R‾n\overline{R}_n be the subalgebra of the symmetric polynomials in RnR_n, \[ \overline{R}_n = \{f \in R_n \mid τ(f) = f \mbox{for each} τ\in S_n \} . \] In 1992 the second author proved that if char(K)=0char (K)= 0 or char(K)=p>nchar(K)=p > n then every Sym(N)Sym (\mathbb N)-invariant ideal in R‾n\overline{R}_n is finitely generated (as such). In this note we prove that this is not the case if char(K)=p≤nchar (K)=p\le n. We also survey some results about Sym(N)Sym (\mathbb N)-invariant ideals in polynomial algebras and some related results.

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