Edited: the proof below assumes $k$ is algebraically closed. The proof for the multiplicity inequality has been added.
Given $x \in X := V(f_1, \ldots, f_m)$, let $k$ be the local dimension of $X$ at $x$ (i.e. $k$ is the maximum of the dimension of all irreducible components of $X$ containing $x$).
Claim: $mult_x(X) \leq $ the product of the largest $n-k$ elements of the sequence $\deg(f_1), \dots, \deg(f_n)$.
If $k = 0$, then the claim holds due to Bézout's theorem. This estimate is also sharp: given degrees $d_1, \ldots, d_n$, take generic homogeneous polynomials of the specified degrees in $n$-variables - they intersect only at the origin, and the multiplicity at the origin is precisely $\prod_i d_i$ due to Bézout's theorem.
In the case that $n > k \geq 1$, we will prove the following
Sub-claim 1: For each $j = 1, \ldots, n-k$, there are $g_1, \ldots, g_j$ such that each $g_j$ is a linear combination of the $f_i$, and $V(g_1, \ldots, g_j)$ is a complete intersection on a neighborhood of $x$.
Proof: Proceed by induction on $j$. For $j = 1$, set $g_1 := f_1$. If $k = n - 1$, then stop. Otherwise, for $2 \leq j \leq n - k$, we can assume by induction we found $g_1, \ldots, g_{j-1}$ such that $V(g_1, \ldots, g_{j-1})$ is a complete intersection on a neighborhood of $x$. We claim that there is a linear combination of the $f_1, \ldots, f_n$ which does not identically vanish on any irreducible component of $V(g_1, \ldots, g_{j-1})$. Indeed, otherwise since $k$ is infinite, we can find $m$ linearly independent linear combinations of $f_1, \ldots, f_m$ which vanish identically on one of the irreducible components of $V(g_1, \ldots, g_{j-1})$ containing $x$, which would mean that the local dimension of $X$ at $x$ is $n - j + 1 > k$, a contradiction. Now let $g_j$ be that linear combination of the $f_i$ which does not identically vanish on any irreducible component of $V(g_1, \ldots, g_{j-1})$, and repeat. $\square$
Once you find $g_1, \ldots, g_{n-k}$ as above, let $Y := V(g_1, \ldots, g_{n-k})$. The second observation is that $mult_x(X) \leq mult_x(Y)$, which follows from the following general fact:
Sub-claim 2: Let $X \subseteq Y \subseteq k^n$ be a chain of closed subschemes and $x$ be a closed point of $X$ such that $X$ and $Y$ have the same local dimension at $x$. Then $mult_x(X) \leq mult_x(Y)$.
Proof: For each $q \geq 0$, there is a surjection
$$O_{x,Y}/m_x^qO_{x,Y} \to O_{x,X}/m_x^qO_{x,X}$$
where $m_x$ is the ideal of $x$. Treating this surjection as an $O_{x,Y}$-module homomorphism, it follows that
$$l(O_{x,Y}/m_x^qO_{x,Y}) \geq l(O_{x,X}/m_x^qO_{x,X})$$
where $l$ is the "length" (see e.g. Atiyah-Macdonald, Proposition 6.9). By the assumption on dimension, for $q \gg 1$, both of these lengths are given by polynomials in $q$ of degree $d$, where $d := \dim_x(X) = \dim_x(Y)$. It follows that the coefficient of $q^d$ in the polynomial providing the values of $l(O_{x,Y}/m_x^qO_{x,Y})$ can not be smaller than the corresponding coefficient of the polynomial for $l(O_{x,X}/m_x^qO_{x,X})$. $\square$
The third observation is that if one of the $f_i$ appears in linear combinations defining more than one $g_j$, say $g_{j_1}, g_{j_2}$, then replacing $g_{j_2}$ by an element of the form $g_{j_2} + ag_{j_1}$ for an appropriate $a \in k$ we may ensure that $f_i$ does not appear in the linear combination defining $g_{j_2}$. In this way we can find appropriate "invertible" linear combinations $g'_1, \ldots, g'_{n-k}$ of $g_1, \ldots, g_{n-k}$ such that
- $V(g_1, \ldots, g_{n-k}) = V(g'_1, \ldots, g'_{n-k})$, and
- there is a reordering of $f_1, \ldots, f_n$ such that $\deg(g'_j) \leq \deg(f_j)$, $j = 1, \ldots, n-k$.
The final observation is that $mult_x(Y) = mult_x(Y \cap H)$ for a generic affine subspace of dimension $k$ containing $x$. On the other hand, if $h_1, \ldots, h_k$ are linear polynomials such that $x$ is an isolated point of $V(g'_1, \ldots, g'_{n-k}, h_1, \ldots, h_k)$, then we are done by the $k = 0$ case.
Best Answer
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
Zariskischeme-theoretic closure of $\text{Spec}(k[x_1, \dots, x_n]/I)$ under the map $\phi$.So the
Zariski closureKernel 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.