Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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

share|cite|improve this answer
@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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.