I've been doing a little bit of field theory for number fields but not much with function fields. The question originally asked says "For some field F, show that the field $F(u_1,\ldots, u_n)$ is a Galois extension of $F(s_1,\ldots,s_n)$ where the $s_i$ are the $i^{th}$ elementary symmetric polynomials in $n$ variables and show the Galois group is $S_n$."

Immediately I replaced $F(s_1,\ldots,s_n)$ with rational symmetric functions because of the fundamental theorem of symmetric polynomials. Then, it seems definitional almost that the Galois group is $S_n$ because I think the way symmetric polynomials are defined is by their invariance under the action of $S_n$ on the order of the variables. However I am confused about how to show this rigorously.

For now I feel like my biggest roadblock is understanding how to work with function fields. For number fields, I would make a formal polynomial like $X^2-2$ over the rationals and say this polynomial is irreducible and then adjoin $\sqrt{2}$ and this would be a Galois extension. It seems like if I were working with function fields of a single variable, $F(t)$ Galois extensions may look the same, $X^2-t$ and this would irreducible over the rational functions and I can adjoin $\sqrt{t}$ and this extension is Galois. But for function fields over several variables, does my formal polynomial have like several formal variables? Please help!

  • 4
    $\begingroup$ The question should not be "For some field $F$...", but rather "For every field $F$...". To answer your question at the end, your polynomial will have coefficients in the base field, whatever it happens to be (in your case, $F(s_1,\dots,s_n)$). Have you seen the theorem that if a finite group $G$ acts by automorphisms on a field $E$, then $E/E^G$ is Galois with Galois group $G$, where $E^G$ is the subfield of $E$ fixed pointwise by $G$? $\endgroup$ – KCd Mar 31 '12 at 14:19
  • $\begingroup$ So it's trivial? By your lemma, given a finite group $G$ of automorphisms on a field K, then K is a Galois extension of $K^G$ and $G$ is a the Galois group of $K/K^G$. Since the by the symmetric functions theorem that says every symmetric polynomial with coefficients in a ring can be written as a polynomial in the elementary symmetric functions, the field generated by the elementary symmetric polynomials in $n$ variables is precisely the symmetric rational functions in $n$ variables. Now let $S_n$ act on the field of rational functions of $n$ variables by permuting the variables. $\endgroup$ – Steven-Owen Mar 31 '12 at 15:27
  • $\begingroup$ The intermediate field fixed by this is trivially the symmetric rational functions. So the extension is clearly Galois with Galois group $S_n$. $\endgroup$ – Steven-Owen Mar 31 '12 at 15:27
  • $\begingroup$ If you grant the theorem on symmetric functions, then the result you ask is a simple consequence of that general result I mentioned from Galois theory. However, there is still something more going on here: the field $F(s_1,\dots,s_n)$ has transcendence degree $n$ over $F$, i.e., the polynomials $s_1,\dots,s_n$ are algebraically independent over $F$. So you could write $F(s_1,\dots,s_n)$ as $F(y_1,\dots,y_n)$ and the result you're asking about is showing that the "generic" polynomial $t^n - y_1t^{n-1} + ... \pm y_n$ has Galois group $S_n$ over $F(y_1,\dots,y_n)$. $\endgroup$ – KCd Mar 31 '12 at 23:22

Note: Your polynomials are still polynomials in one variable. It's the coefficients of the polynomial that are rational functions in several variables, but these coefficients are constants of your field.

To show that $F(u_1,\ldots,u_n)$ is Galois over $F(s_1,\ldots,s_n)$, it suffices to show that it is the splitting field of some separable polynomial with coefficients in $F(s_1,\ldots,s_n)$. Let $f(t)\in F(s_1,\ldots,s_n)[t]$ be the polynomial $$f(t) = t^n - s_1t^{n-1} + s_2t^{n-2} + \cdots + (-1)^ns_n.$$ Since $$f(t) = (t-u_1)(t-u_2)\cdots(t-u_n),$$ it follows that $F(u_1,\ldots,u_n)$ is indeed the splitting field of $f(t)$ over $F(s_1,\ldots,s_n)$.

Note also that $f(t)$ has no repeated roots, since the $u_i$ are pairwise distinct indeterminates, so $f(t)$ is separable. Thus, $F(u_1,\ldots,u_n)$ is the splitting field of a separable polynomial over $F(s_1,\ldots,s_n)$, and so it is a Galois extension of $F(s_1,\ldots, s_n)$.

To finish off, we need to show the Galois group is precisely $S_n$. Note that $S_n$ acts on $F(u_1,\ldots,u_n)$ by fixing $F$ and letting $\sigma$ map $u_i$ to $u_{\sigma(i)}$; these automorphisms leave $F(s_1,\ldots,s_n)$ fixed, and they are pairwise distinct, so this proves that $G=\mathrm{Gal}(F(u_1,\ldots,u_n)/F(s_1,\ldots,s_n))$ contains a subgroup isomorphic to $S_n$. On the other hand, an element of $G$ is completely determined by what it does to $u_1,\ldots,u_n$, and it must map $u_i$ to a root of $f(t)$ above, hence elements of $G$ act as permutations of $u_1,\ldots,u_n$, showing that $G$ must be isomorphic to a subgroup of $S_n$. These two conclusions tell us that $G$ must in fact be isomorphic to $S_n$, as desired.

Another method is to note, as KCd remarks, that $S_n$ acts on $F(u_1,\ldots,u_n)$, hence, $F(u_1,\ldots,u_n)$ is Galois over the fixed field of $S_n$, with Galois group $S_n$. The fixed field of $S_n$ under this action is precisely the field of symmetric rational functions, which by the Fundamental Therorem of symmetric functions is generated over $F$ by $s_1,\ldots,s_n$; that is, the fixed field is $F(s_1,\ldots,s_n)$, which shows that $F(u_1,\ldots,u_n)/F(s_1,\ldots,s_n)$ is Galois with Galois group $S_n$, as desired.

  • $\begingroup$ You could skip the first of this using KCd's theorem, right? Not that it's bad to have other reasons. $\endgroup$ – Dylan Moreland Mar 31 '12 at 21:00
  • $\begingroup$ @Dylan: Yes, you are right. So I've added an alternative argument that does not require us to invoke that theorem, to go with the direct proof that the extension is Galois. $\endgroup$ – Arturo Magidin Mar 31 '12 at 21:23
  • $\begingroup$ @ArturoMagidin, Thanks! I saw the alternative proof, but I couldn't see where to get a more thorough understanding of how the polynomials would actually work--for that, I am extra grateful!! $\endgroup$ – Steven-Owen Apr 1 '12 at 2:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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