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.


1 Answer 1


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 $f\in K[\sigma_1,\dots,\sigma_n]$.

  • $\begingroup$ @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 $\endgroup$
    – user114539
    Jan 30, 2016 at 18:45
  • $\begingroup$ Do you do this using transcendence basis, showing that $\sigma_i$ are algebraically independent over $K$? $\endgroup$
    – user114539
    Jan 30, 2016 at 19:13
  • $\begingroup$ Yes, transcendence bases make the proof quick. Dummit and Foote has an explanation of the algebraic independence of elementary symmetric polynomials in the chapter on transcendental field extensions. (I realize this post is very old—just a comment in case anyone stumbles upon it, like I did.) $\endgroup$ Feb 21, 2023 at 3:28

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