The proof of the Lagrange's Rational Function Theorem Lagrange's rational function theorem states that if one has two rational functions in multiple variables $f(x_1,x_2,...x_n)$ and $g(x_1,x_2,...,x_n)$ then one can can express $f$ as a rational function in $g$ if and only if the set of permutations that keep $g$ unchanged is a subset of the set of permutations that preserve $f$. 
A slightly more precise statement of the theorem can be found here in the first paragraph of this paper here.
Is anyone familiar with the proof of this theorem? While it is fairly clear that if $f$ can be expressed in terms of $g$ the set of permutations that keep $g$ unchanged has to be the subset of those that keep $f$ unchanged, the converse is far from obvious. 
 A: Consider the field $K = \mathbb{Q}(x_1,x_2,\ldots, x_n)$, and consider $K_g \subset K$ to be the subfield generated by $g$. Define
$$
H_g = Aut(K/K_g) = \{\sigma \in Aut(K) : \sigma(\alpha) = \alpha \quad\forall \alpha\in K_g\}
$$
Similarly, define $K_f$ and $H_f$. Then you want to show that
$$
K_f\subset K_g \Leftrightarrow H_g \subset H_f
$$
If all the hypotheses are satisfied, which I think they are, this is merely the Fundamental Theorem of Galois Theory
A: In Harold M. Edwards' Galois Theory pp. 33-34, he includes a translation of Section 104 of Lagrange's Réflexions (1771) in which Lagrange first presents the theorem and includes a proof. The theorem, as originally stated by Lagrange, is translated as:

"If $t$ and $y$ are any two [polynomials] in the roots $x', x'', x''', \dots$ of $x^\mu + mx^{\mu-1} + nx^{\mu-2}+[px^{\mu-3}+]\cdots = 0$ and if these [polynomials] are such that every permutation of the roots $x', x'', x''', \dots$ which changes $y$ also changes $t$, one can, generally speaking, express $y$ rationally in terms of $t$ and $m, n, p, \dots$, so that when one knows a value of $t$ one will also know immediately the corresponding value of $y$"

Edwards goes on to prove this by using Cramer's rule (c. 1750) though he notes Lagrange takes a more general approach (related to his use of generally speaking).
Let $t$ and $y$ be the polynomials in the $n$ roots and $t_1, t_2, \dots, t_k$ denote the distinct polynomials from $t$ under the permutation of the roots. Since we assume a permutation that fixes $t$ fixes $y$, there are at most $k$ distinct polynomials $y$ can take under permutation of the roots. Call them $y_1, y_2, \dots, y_k$ such that the permutations that carry $t$ to $t_i$ carry $y$ to $y_i$. Now consider the $k$ polynomials
$$
y_1 + y_2 + \cdots + y_k, \\
t_1y_1 + t_2y_2 + \cdots + t_ky_k, \\
t^2_1y_1 + t^2_2y_2 + \cdots + t^2_ky_k, \\
\vdots \\
t^{k-1}_1y_1 + t^{k-1}_2y_2 + \cdots + t^{k-1}_ky_k.
$$
Observe that these polynomials are symmetric in the roots $x', x'', x''', \dots$ since any permutation of the roots results in a rearrangement of the summands. Now because they are symmetric in the roots, they are expressible in the elementary symmetric polynomials (i.e. the coefficients $m, n, p, \dots$ of our original equation) by the Fundamental Theorem of Symmetric Polynomials. Call them $s_1, s_2, \dots, s_k$.
What is left to show is that $y_1 = y$ can be rationally expressed in terms of $t_1 = t$ and the coefficients $m, n, p, \dots$. Here he uses Cramer's rule with our $k$ equations $\sum_i t_i^j y_i = s_j$ where $s_j$ is expressible in $m, n, p, \dots$. Then $y_i = D_i / \Delta$ where $\Delta$ is the Vandermonde determinant and $D_i$ is the determinant from replacing the $i$th column of $\Delta$ by the column of $s_1, \dots, s_k$. We want to how that $y_i = D_i/\Delta = D_i\Delta / \Delta^2$ is rationally expressible in $t_1$.
Edwards makes the stipulation that $\Delta^2 = \prod_{k<i}(t_i - t_k)^2$, the discriminant of $F(X) = \prod (X-t_i)$, is nonzero, i.e. that $F(X)$ has no multiple roots. He indicates that Lagrange handles this case in a more general argument but Edwards does not present his proof.
Observe that any permutation of the $t_i$ other than $t_1$ changes the sign of the both $D_1$ and $\Delta$ but leaves their product $D_1\Delta$ unchanged. So $D_1\Delta$ can be viewed as a polynomial in $t_1$ that is symmetric in $t_2, \dots, t_k$, and is expressible in the elementary symmetric polynomials
$$\begin{align}
\tau_1 &= t_2 + t_3 + \cdots + t_k \\
\tau_2 &= t_2t_3 + t_2t_4 + \cdots + t_{k-1}t_k \\
&\vdots\\
\tau_{k-1} &= t_2t_3\cdots t_k\end{align}.
$$
But then we can express these in terms of the elementary symmetric polynomials in all $t_1, \dots, t_k$ called $\sigma_1, \dots, \sigma_k$
$$
\begin{align}
\tau_1 &= \sigma_1 - t_1 \\
\tau_2 &= \sigma_2 - t_1\sigma_1 + t^2_1 \\
&\vdots\end{align}.
$$
Thus $D_1\Delta$ and further $y = y_1 = D_1\Delta/\Delta^2$ is rationally expressible in $t_1$ and $\sigma_i$'s, which are in turn expressible in $m, n, p, \dots$.
