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arxiv: 1310.7608 · v1 · pith:YW6O4YCVnew · submitted 2013-10-28 · 🧮 math.RA

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

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keywords mathbbcharinvariantoverlinepolynomialssigmadotsgenerated
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Let $K$ be a field and let $\mathbb N = \{1,2, \dots \}$. Let $R_n=K[x_{ij} \mid 1\le i\le n, j\in \mathbb N]$ be the ring of polynomials in $x_{ij}$ $(1 \le i \le n, j \in \mathbb N)$ over $K$. Let $S_n = Sym (\{1,2, \ldots, n \})$ and $Sym (\mathbb N)$ be the groups of the permutations of the sets $\{1,2,\dots, n \}$ and $\mathbb N$, respectively. Then $S_n$ and $Sym (\mathbb N)$ act on $R_n$ in a natural way: $\tau (x_{ij})=x_{\tau(i)j}$ and $\sigma (x_{ij})=x_{i\sigma (j)}$ for all $\tau \in S_n$ and $\sigma \in Sym(\mathbb N)$. Let $\overline{R}_n$ be the subalgebra of the symmetric polynomials in $R_n$, \[ \overline{R}_n = \{f \in R_n \mid \tau (f) = f \mbox{for each} \tau \in S_n \} . \] In 1992 the second author proved that if $char (K)= 0$ or $char(K)=p > n$ then every $Sym (\mathbb N)$-invariant ideal in $\overline{R}_n$ is finitely generated (as such). In this note we prove that this is not the case if $char (K)=p\le n$. We also survey some results about $Sym (\mathbb N)$-invariant ideals in polynomial algebras and some related results.

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