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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.

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    $\begingroup$ I should've checked what I wrote in my now-deleted answer; I had a vague memory of an argument from integral closure but now I can't actually think of it and am likely misremembering, mea culpa. $\endgroup$ Commented Sep 17, 2022 at 5:18
  • $\begingroup$ Qiaochu, thank you so much for taking the time to think about it. And I still learned something because I was unfamiliar with the crucial integral closure concept. In any case, we know that the matter is far subtler than Artin implies. $\endgroup$ Commented Sep 17, 2022 at 5:20

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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.

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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.

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  • $\begingroup$ Thanks very much, Acrobatic. The main issue with your approach is that it uses the result that every field has an algebraic closure. Since the symmetry discussion in Artin's book comes right after he proves that every polynomial has a splitting field, he couldn't possibly have had in mind a proof that uses the much more powerful result of algebraic closure. Maybe the issue is just much more involved than Artin realised? Interesting to learn about the Nullstellensatz though, and your post inspired me to reread a proof that all fields have algebraic closures. $\endgroup$ Commented Sep 17, 2022 at 12:47

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