Hey, I have been working on this problem for a while now, but I just can't solve it, so I'm posting it to see if anyone has any idea how to work it out:
Let k be an arbitrary field and K=k(x) be the fraction field of the polynomial ring in one variable. Define F,G:K--->K by F(f(x)/g(x))=f(1/x)/g(1/x) and G(f(x)/g(x))=f(1-x)/g(1-x).
1) Find the fixed field L of {F,G}
2)Determine Gal(K/L) (the Galois group of the extension)
3)Find an h in L such that L=k(h)
Thanks in advance.
No, I have problems describing the fixed functions of G, and so I am unable to even talk about it's fixed field. I got that no Möbius-type function can be fixed by G, but following a similar treatment led to really ugly equations even for second degree functions.
It is helpful to know this result.
Let me rephrase the result here so that you do not have to jump back and forth between pages to see it, go there to see the proof. Let be a field with (the rational function field). Let and write in reduced terms. Let . Then is the maximum of .
Let be the automorphisms you defined above. Notice that if is the fixed field of then is also the fixed field of (the subgroup generated by and of ). Therefore , now we use Artin's result that since is a fixed field of a finite group of automorphisms it means that is Galois with Galois group, .
Now we will determine the group structure of . Note that any automorphism of is determined on for (i.e. the field generated by ). Notice that so . Notice that so .
Now,
And, .
Thus, , by group presentations it means that .
Now we know that (thus NonCommAlg's initial idea is wrong).
By Luroth's theorem we know there exists an so that , this theorem is not really here because the problem is kinda hinting that is a simple extension over , but anyway we will find such an . The idea is to use the trace . We will pick and find , notice that . Write in reduced form, IF the max of then by first paragraph where but and so . This approach requires some guesswork because if you pick such an so that the trace in reduced form has both polynomials of degrees less than 6 then it means would just be a proper subfield of .
Notice are the six automorphisms. Pick then its trace is:
---> this does not help.
I am too tired to find a polynomial that works but just go through a few of them (by computer for example) and see if you can find one that gives the desired form.
You should have the approach above that uses the trace, it works out.
Again, .
The six automorphisms are: .
If is a rational function in then , thus the trace allows us to produce elements in .
Just for illustration, let us use again, then:
Certainly is fixed by but that is constant and uninteresting.
Let us make it more interesting by taking then:
By what was explained above it follows that: .
So in general an element is fixed by if and only if it has the form where .
(Of course the coefficients are such that the lower function is non-zero).