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.

havesuch 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. $\endgroup$ – Max Horn Oct 7 '10 at 16:081more comment