The following is stated in the book Abstract Algebra by Robert Ash, the book is free and the relevant chapter could be seen here. In 6.1.4. it is shown that every symmetric polynomial could be expressed as a polynomial in the elementary symmetric polynomials, and from this in 6.1.5. the following corollary is derived ($F$ denotes a field):

If $g$ is a polynomial in $F[X]$ and $f(\alpha_1, \ldots, \alpha_n)$ is any symmetric polynomial in the roots $\alpha_1, \ldots, \alpha_n$, then $f \in F[X]$.

Proof. We may assume without loss of generality that $g$ is monic. Then in a splitting field of $g$ we have $$ g(X) = (X-\alpha_1)\ldots (X-\alpha_n) = X^n - e_1 X^{n-1} + \ldots + (-1)^n e_n. $$ By (6.1.4), $f$ is a polynomial in the $e_i$, and since the $e_i$ are simply $\pm$ the coefficients of $g$, the coefficients of $f$ are in $F$. $\square$

I cannot make sense of it, why should $f$ be even a polynomial in $X$? Where does the $X$ comes from? I see that it is a polynomial in the $e_i$ (which themselve depend on the $\alpha_i$), so that it could be written as a polynomial in terms of $\pm$ the coefficients of $g$ (viewing them as indeterminates), so if we view them as values we get elements from $F$, i.e. that would give $f \in F$ after "plugin in" the values of the coefficients. But this does not makes any sense either?

  • $\begingroup$ Over any field $F$, if $g(X) = \prod_{i=1}^n (X-\alpha_i) \in F[X]$ is a separable polynomial (so the $\alpha_i \in \overline{F}^{sep}$) and $f(\alpha_1,\ldots,\alpha_n,X) \in F[\alpha_1,\ldots,\alpha_n][X]$ is invariant under permutation of the $\alpha_i$, then $f \in F[X]$. $\endgroup$ – reuns Sep 27 '17 at 19:44
  • $\begingroup$ @reuns Do you think that the book missed to state that $X$ should be also an indeterminante in $f$? $\endgroup$ – StefanH Sep 27 '17 at 19:46
  • $\begingroup$ I don't know but what I wrote is the answer to your question :) If you don't like symmetric polynomials then let $K = F[\alpha_1,\ldots,\alpha_n]$ so that $K/F$ is a Galois extension and with $G = \text{Gal}(K/ F)$ then $F = K^G = \{ a \in K, \forall \sigma \in G, \sigma(a) = a\}$ the fixed field of $G$. $\endgroup$ – reuns Sep 27 '17 at 19:47
  • $\begingroup$ @reuns Okay, but I cannot make sense of your proof, could you please give a few more details why this proves the claim... $\endgroup$ – StefanH Sep 27 '17 at 19:48
  • $\begingroup$ Galois theory : $f \in K[X]$ and its coefficients are invariant under $G$ thus $f \in K^G[X] = F[X]$. Otherwise use the decomposition in elementary symmetric polynomials. Also when $g$ is non-separable it gets more complicated ! $\endgroup$ – reuns Sep 27 '17 at 19:50

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