Filling in the gap of a proof of the Fundamental Theorem of Symmetric Polynomials.

Question

Suppose we can assume that the set of symmetric rational functions in $k(X_1,...,X_n)$, where $k$ is a field, is the same as the field generated by $k(a_1,...,a_n)$ where the $a_i$ are the elementary symmetric polynomials, how do we move from there to equating the symmetric polynomials in $n$ variables with the polynomials in the $a_i$? I'm asking because I'm learning this from Emil Artin's monograph on Galois theory where he seems to move from the first assertion (which he does prove clearly) to the second with only a few words of justification which I don't follow. Basically, I don't see an easy move from the field case to the ring case. Please note that this question is specifically about filling in a gap. There are a huge number of published proofs of both these assertions; my question is merely about moving from the first to the second. Thank You.

Answer 1

Someone else had exactly the same question and the answer is covered in another thread: A proof of the fundamental theorem of symmetric polynomials That totally resolves my issue. It does use integral closure, and it might be what Qiaochu was trying to remember. Thanks a lot to everyone who has contributed to this thread.

Answer 2

Fix an symmetric polynomial $f$, and write $f=\frac{g(a_1,\dots ,a_n)}{h(a_1,\dots ,a_n)}$ in elementary symmetric polynomials. Then $fh=g$. Substituting $(X_1,\dots ,X_n)$ by the roots of the equation $X^n+b_1 X^{n-1}+\dots +b_n$ where $(b_i)_i \in \bar{k}^n$ is a zero of $h$ in an algebraic closure of $k$, we see that $(b_i)_i $ is also a zero of $g$. By the Nullstellensatz, $g \in \sqrt{h}$, and if we choose $g,h$ coprime, $h$ cannot have a zero. But since this method uses Nullstellensatz, it might not be what you would like.