Finding generators of subalgebra of polynomial algebra $K[x_1,\cdots,x_n]$ that are invariant under the action of symmetric group

Question

Let $I =\langle f_1,\cdots,f_m\rangle \subset K[x_1,\cdots,x_n]$be an ideal, where $f_k\in K[x_1,\cdots,x_n].$

$K[e_1,\cdots,e_n]$ the polynomial algebra generated by the elementary symmetric polynomials $e_1,\cdots,e_n\in K[x_1,\cdots,x_n].$

Is there any method(algorithm) to compute the K-algebra $I \cap K[e_1,\cdots,e_n]$ of the intersection of $I$ and $K[e_1,\cdots,e_n]? $

Since $K[e_1,\cdots,e_n]$ is not an ideal of $ K[x_1,\cdots,x_n],$ it fails to compute elimination ideal. Anyway, $I \cap K[e_1,\cdots,e_n]$ is an ideal of the ring $K[e_1,\cdots,e_n].$ What I want to do is to give the generating sets of this ideal by polynomials in $K[e_1,\cdots,e_n].$

Answer 1

UPDATE: Sorry! As pointed out in a comment, my previous answer was incorrect. So I've edited my answer. The following simpler algorithm seems to me that it should work (at least assuming I'm understanding the question\dots).

From the geometric perspective, the inclusion map $k[e_1, \dots, e_n]\subseteq k[x_1, \dots, x_n]$ corresponds to the quotient $\phi: \mathbb A^n\to \mathbb A^n/S_n$. The ideal $J:= I\cap k[e_1, \dots, e_n]$ is the contraction of the ideal $I$. Hence, geometrically, it seems to me that computing $J$ is equivalent to computing the Zariski scheme-theoretic closure of $\text{Spec}(k[x_1, \dots, x_n]/I)$ under the map $\phi$.

So the Zariski closure Kernel of a Ring Map algorithm on page 84 of Greuel and Pfister's "A Singular Introduction to Commutative Algebra" would seem to be applicable. To apply the algorithm, you define a ring $R:=k[x_1,\dots,x_n, t_1, \dots, t_n]$ and then define an ideal $N\subseteq R$ by

$ N:=I+\langle t_1-e_1(\mathbf{x}), \dots, t_n-e_n(\mathbf{x}) \rangle $

where $e_i(\mathbf{x})$ is the $i$'th symmetric polynomial. Compute the elimination ideal $N\cap k[t_1, \dots, t_n]$, and say that it equals $\langle p_1(\mathbf{t}), \dots, p_r(\mathbf{t})\rangle$. Then $I\cap k[e_1, \dots, e_n]$ will equal the ideal $\langle p_1(\mathbf{e}), \dots, p_r(\mathbf{e})\rangle$.

I hope this helps.

Answer 2

I have a positive reference to your question under special case:

Let $I=(x_1^{d+1},x_2^{d+1},\dots,x_n^{d+1})$ be a monomial complete intersection ideal in the polynomial ring $R=\mathbb{C}[x_1,\dots,x_n]$. Easy to see $I$ is fixed by group $G=S_n$.

So $S_n$ acts on the quotients ring $A=R/I$. we see $A^G=R^G/I^G$, where $R^G=\mathbb{C}[e_1,\dots,e_n]$, is the ring of symmetric functions and $I^G=I \cap R^G$.

You are looking for $I \cap R^G$.

See the paper http://arxiv.org/abs/0801.2662 by Aldo Conca, Christian Krattenthaler, Junzo Watanabe. In this paper they show that $I \cap R^G$ is generated by power sum symmetric polynomials $p_{d+1},p_{d+2},\dots,p_{d+n}$.

Treatment here is only for $I=(x_1^{d+1},x_2^{d+1},\dots,x_n^{d+1})$. But I hope, this would help.