# Are fixed fields of finite subgroups generated by symmetric polynomials on Galois conjugates?

Let $F/K$ be an extension of fields, let $G=\{\sigma_1,\dots,\sigma_n\}$ be a finite subgroup of $\operatorname{Gal}(F/K)$.

If $\alpha\in F$, then any symmetric polynomial $\varphi\in K[\sigma_1(\alpha),\dots,\sigma_n(\alpha)]$ satisfies $\varphi\in F^G$, the fixed field of $G$.

Is it true that $F^G=K(\{\varphi_\alpha\in K[\sigma_1(\alpha),\dots,\sigma_n(\alpha)]: \varphi_\alpha \mbox{ is elementary symmetric}, \alpha\in F\})$? The above assertion says that the easy inclusion $\supset$ holds, but do these suffice? Note that we can take the polynomials to be elementary symmetric by the fundamental theorem of symmetric polynomials.

This question arose naturally to me when computing some fixed fields, since a useful strategy for determining them is to find some nice $\alpha$ to try (e.g. a set of generators for $F/K$), and start adjoining symmetric polynomials on the $\sigma_i(\alpha)$ to $K$ until you reach the necessary degree. So the question in other words is, does this procedure always work?

In a finite Galois extension, already trace is provably surjective (this proves non-degeneracy of the trace pairing $\langle \alpha,\beta\rangle=\hbox{trace}_{F/K}(\alpha\cdot \beta)$). This is standard, and found in many places, e.g., http://www.math.umn.edu/~garrett/m/number_theory/overheads/noth-10-05-2011.pdf

• Dear Paul: thank you for your answer. The link is broken, though. Also, what about the case where $F/K$ is not finite Galois? Nov 26, 2012 at 16:04
• @Bruno: there aren't any nonconstant symmetric polynomials in infinitely many variables.
– user14972
Nov 26, 2012 at 17:37
• @Hurkyl: I don't understand your comment. The group $G$ is finite. Nov 26, 2012 at 18:33
• For $F/K$ genuinely infinite, there may not be so many finite subgroups of the profinite group that is the proper $\hbox{Gal}(F/K)$. Maybe that part of the question needs refinement? Nov 26, 2012 at 18:47
• @paulgarrett: I'm sorry, I don't know anything about topological groups. For me, $\operatorname{Gal}(F/K)$ is the group of $K$-automorphisms of $F$. In any case, the question of $F/K$ finite but not Galois remains. Nov 26, 2012 at 21:39

Your statement is true. Take any element $x\in F^{G}$, then $\sum_{i=1}^{n}\sigma_{i}(x/n)$ is a $G$-elementary symmetric function of $\alpha=x/n\in F^{G}$. This equals to $x$, since $x$ is fixed by the action of $G$. (This proof is only valid provided your field is not of characteristic $n$.)

But for your problem of finding fixed fields, you can only take $\alpha$ to be some known numbers you have in hand, e.g. from some generating set $S$ of the extension $F/K$: $F=K(S)$.

More concrete example is the Galois group of $f(X)=X^4-2$. Let $S$ be the set of all its roots: $S=\{ \, \sqrt{2},\sqrt{2}\sqrt{-1},-\sqrt{2}, -\sqrt{2}\sqrt{-1}\, \}$. The Galois group $\mathrm{Gal}_{\mathbb{Q}}(\mathbb{Q}(S))$ is the dihedral group of order $8$, and it has a unique cyclic subgroup $G$ of order four. The corresponding fixed field of $G$ is $\mathbb{Q}(\sqrt{-1})$. Note that $\sqrt{-1}=(\sqrt{2}\sqrt{-1})/\sqrt{2}$ is not a symmetric polynomial of $\sqrt{2}$. For $\prod_{g\in G}(X-g\sqrt{2})=X^4-2$ only has rational coefficients.

So the above proved proposition is sometimes misleading, but indeed elementary symmetric functions are good candidates. If symmetric functions don't suffice, carefully write down the action of $G$ to your well understood $\alpha$'s and construct numbers that is fixed by the action of $G$ using your experiences of previously worked out examples, untill you reach your degree of the extension.