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

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 $K[e_1,\cdots,e_n].$

• Do you mean a method for finding a set of generators of $I\cap K[e_1,\ldots,e_n]$ ? How do you give $I$ ? Do you give a set of generators of $I$ ? – Denis Serre Oct 7 '10 at 12:02
• Can you please edit the title to be more informative about the question? – Gerry Myerson Oct 7 '10 at 12:05
• $K[x_1,\cdots,x_n]$ is noetherian ring, any ideal $I$ has a finite set of generators. – tiansong Oct 7 '10 at 13:11
• Yes, I has a finite set of generators. But do you have such a set? An ideal can be described by other means than a set of generators, and it can be non-trivial to find a set of generators from such a description. Of course if you are not interested in a practical solution, just a purely theoretical one, this difference may not matter to you. – Max Horn Oct 7 '10 at 16:08
• The algebraist's cheap answer is, of course, to take a generating set of $I$ (I hope you have one; unless $I$ is given by some very implicit conditions, you can compute one using Gröbner basis methods), you can symmetrize every element of this generating set, and get a generating set for $I\cap K\left[e_1,...,e_n\right]$. Of course, this is cheap for an algebraist, but not computationally cheap since you have to compute a sum over $S_n$. -- Oh, I hope your $K$ has characteristic not dividing $n!$. – darij grinberg Oct 11 '10 at 22:58

## 2 Answers

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.

• No, it does not. – Guntram Oct 8 '10 at 8:21
• Fight!! Fight!! – Gerry Myerson Oct 8 '10 at 11:58
• Again, we're talking here about the subalgebra of symmetric polynomials contained in the ideal $I$, not some sort of ideal. Moreover, you have to suppose I to be a radical ideal to argue geometrically. – Guntram Oct 12 '10 at 6:25
• @ Guntram: I guess I'm confused by the question. I have only ever seen the notation $I\cap k[e_1, \dots, e_n]$ used to refer to an ideal, and so I assumed that tiansong was asking about that. But perhaps $I\cap k[e_1, \dots, e_n]$ is also used as notation for a subalgebra? (For instance, perhaps the question is about how to compute $k[g_1, \dots, g_s]\cap k[e_1,\dots, e_n]$, where $g_1, \dots, g_s$ is a minimal set of generators of $I$?) If so, then my answer is obviously useless. – Daniel Erman Oct 12 '10 at 19:10
• @Guntram: I believe that the algorithm outlined above actually computes the scheme-theoretic closure, and so a radical hypothesis is unnecessary. I updated the answer and the proposed reference to reflect this. Please let me know if you see some error in my understanding of this algorithm. – Daniel Erman Oct 12 '10 at 19:13

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.