# A proof of the fundamental theorem of symmetric polynomials

I'm reading Exploratory Galois Theory by John Swallow. On page 123 he gives the following remark / alternate proof of the fundamental theorem of symmetric polynomials:

Let $K$ be a field and $L$ be the field of rational functions $K(X_1,\dots,X_n)$. Now consider the subfield $K(\sigma_1,\dots,\sigma_n)$ generated over $K$ by the elementary symmetric polynomials. Then $L$ is the splitting field of $X^n − \sigma_1X^{n−1} +\cdots +(−1)^n\sigma_n$, since this polynomial is equal to the product $(X − X_1)(X − X_2) \cdots (X − X_n)$. The Galois group must be a subgroup of $S_n$; on the other hand, every permutation in $S_n$ gives a different automorphism of $L$ over $K(\sigma_1,\dots,\sigma_n)$. Hence $K(X_1,\dots,X_n)/K(\sigma_1,\dots,\sigma_n)$ is Galois with group $S_n$, and $K(\sigma_1,\dots,\sigma_n)$ is the fixed field of $S_n$. To perform the final step – to say that every symmetric polynomial is a polynomial in the elementary symmetric functions, that is, that each symmetric polynomial lies not only in $K(\sigma_1,\dots,\sigma_n)$ but $K[\sigma_1,\dots,\sigma_n]$ – requires a notion of integrality beyond the scope of this text.

Could anyone explain how to finish this proof? I am familiar with integral ring extensions but I'm not sure what to do with it.

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Let $f\in K[X_1,\dots,X_n]$ be a symmetric polynomial. Then $$f\in K(X_1,\dots,X_n)^{S_n}=K(\sigma_1,\dots,\sigma_n).$$ We want to prove that $f\in K[\sigma_1,\dots,\sigma_n]$. The ring extension $K[\sigma_1,\dots,\sigma_n]\subset K[X_1,\dots,X_n]$ is integral since $X_i$ is integral over $K[\sigma_1,\dots,\sigma_n]$ for all $i=1,\dots,n$. (Note that $X_i$ is a root of the monic polynomial $X^n − \sigma_1X^{n−1} +\cdots +(−1)^n\sigma_n\in K[\sigma_1,\dots,\sigma_n]$.) In particular, $f$ is integral over $K[\sigma_1,\dots,\sigma_n]$. Since $f\in K(\sigma_1,\dots,\sigma_n)$ and $K[\sigma_1,\dots,\sigma_n]$ is integrally closed (why?) we get $f\in K[\sigma_1,\dots,\sigma_n]$.
@user26857 Is it easy to see that $K[\sigma_1,\ldots, \sigma_n]\cong K[x_1,\ldots, x_n]$? Or is there another way to answer the why you ask above? Thanks – user114539 Jan 30 at 18:45
Do you do this using transcendence basis, showing that $\sigma_i$ are algebraically independent over $K$? – user114539 Jan 30 at 19:13