Constructing expressions

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Suppose you are given all the elementary symmetric functions of $n$ variables $x_1,x_2,...,x_n$ and two rational functions $A(x_1,x_2,...,x_n)$ and $B(x_1,x_2,...,x_n)$ in the same $n$ variables that are both preserved by the same group of permutations of the variables.

I'm looking for constructive proof or an algorithm that allows one to express $B$ with arithmetic operations given the elementary symmetric functions and $A$.

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Of course it is not a complete answer but good point to start.

Let $G$ be a group acting set $\Omega$ where $\Omega$ be set of all elementary symmetric polynomials in n variables i.e. $\Omega=\{e_0,e_1...,e_n\}$.

And set formal sum $$R=\{\sum_{i=0}^nc_ie_i| c_i\in F \}$$ Where $F$ is the field you used.

Then $R$ is a vevtor space over $F$ and you can expand action of $G$ from $\Omega$ to $R$ by defining $$g(\sum_{i=0}^nc_ie_i)=\sum_{i=0}^nc_ig(e_i)$$.

Then, Ever every $g$ correspond to a linear operotor or a $(n+1)(n+1)$ matrices and your functions $A$, $B$ a correspond to vectors in $R$. Moreover, since $gA=A$ and $gB=B$ then $A$ and $B$ are eigenvectors of $g$ for every $g\in G$. Even if I can not see any direct relation between $A$ and $B$ linear algebra is good place to think this problem.

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What I take from your question is that you want a constructive version of the theorem that says that if $K \subset L$ is Galois (here $L = k(x_i)$ and $K = k(\sigma_i)$) with Galois group $G$ (here, $S_n$), every intermediate field (such as $K[A]$) is of the form $L^H$ for some subgroup $H$ of $G$.

If you pick $H$ to be the subgroup fixing $A$, then $K[A]$ will obviously be equal to $L^H$ (it is included in $L^H$, and it can't be $L^{H'}$ for $H \subset H'$ because $A$ is not fixed by such a $H'$). And the surprise falling out of this theorem is that then any expression (such as $B$) that is invariant by $H$ can be expressed as an element of $K[A]$

There is an brute-force algorithm. You look at the sequence of $k$-vector spaces $k[\sigma_i,A]_d$ of polynomials of degree at most $d$ in $k[\sigma_i,A]$. Those are subspaces of $k[x_i]_d$, and you have a linear multiplication-by-$B$ map $k[x_i]_d \to k[x_i]_{d+e}$ where $e$ is the degree of $B$. Eventually, (by checking the respective dimensions of all those vector spaces), $k[\sigma_i,A]_d$ will be big enough inside $k[x_i]_d$ that you can find a nontrivial intersection between $B * k[\sigma_i,A]_d$ and $k[\sigma_i,A]_{d+e}$. Any nonzero element in that intersection correspond to a representation of $B$ as a rational function of the $\sigma_i$ and $A$. Just by computing the dimensions, you can find an upper bound on $d$. It is usually very very large and not the best possible though.

This is more detailed in this answer