# Question on Galois Theory

• April 22nd 2009, 02:41 PM
Jose27
Question on Galois Theory
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)

• April 22nd 2009, 06:32 PM
NonCommAlg
Quote:

Originally Posted by Jose27
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}

i got $k$ as the fixed field!! (Wondering) before i give my solution, do you have any example of a non-constant $r(x) \in K$ which is in the fixed field of {F,G}?
• April 22nd 2009, 07:17 PM
Jose27
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.
• April 22nd 2009, 08:17 PM
NonCommAlg
Quote:

Originally Posted by NonCommAlg

i got $k$ as the fixed field!! (Wondering)

never mind! i just found a mistake in my solution! i knew what i got was strange ... anyway, i might think about it a bit more later.
• April 23rd 2009, 10:47 PM
ThePerfectHacker
Quote:

Originally Posted by Jose27
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)

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 $k$ be a field with $K=k(t)$ (the rational function field). Let $\alpha \in K - k$ and write $\alpha = \tfrac{g(t)}{h(t)}$ in reduced terms. Let $F = k(\alpha)$. Then $[K:F]$ is the maximum of $\deg(f),\deg (g)$ (Surprised).

Let $F,G$ be the automorphisms you defined above. Notice that if $L$ is the fixed field of $\{ F,G\}$ then $L$ is also the fixed field of $\left< F,G \right>$ (the subgroup generated by $F$ and $G$ of $\text{Aut}(K)$). Therefore $L = K^{\left< F,G\right>}$, now we use Artin's result that since $L$ is a fixed field of a finite group of automorphisms it means that $K/L$ is Galois with Galois group, $\text{Gal}(K/L) = \left< F,G\right>$.

Now we will determine the group structure of $\left$. Note that any automorphism of $K$ is determined on $x$ for $K = k(x)$ (i.e. the field generated by $x$). Notice that $F^2(x) = F(\tfrac{1}{x}) = x$ so $F^2 = \text{id}$. Notice that $G^2(x) = G(1-x) = 1-(1-x) = x$ so $G^2 = \text{id}$.
Now, $FGF(x) = FG(\tfrac{1}{x}) = F(\tfrac{1}{1-x}) = \tfrac{1}{1-\tfrac{1}{x}} = \tfrac{x}{x-1} = 1 - \tfrac{1}{1-x}$
And, $GFG(x) = GF(1-x) = G(1-\tfrac{1}{x}) = 1 - \tfrac{1}{1-x}$.
Thus, $FGF = GFG$, by group presentations it means that $\left \simeq S_3$.
Now we know that $[K:F] = |S_3| = 6$ (thus NonCommAlg's initial idea is wrong).

By Luroth's theorem we know there exists an $h\in K$ so that $L = k(h)$, this theorem is not really here because the problem is kinda hinting that $L$ is a simple extension over $k$, but anyway we will find such an $h$. The idea is to use the trace $\text{Tr}_{K/L}$. We will pick $r\in K$ and find $h=\text{Tr}_{K/L}(r)$, notice that $h\in L$. Write $h=\tfrac{p}{q}$ in reduced form, IF the max of $\deg (p) \text{ and }\deg (q) = 6$ then by first paragraph $k\subseteq k(r)\subseteq L$ where $[k(r):k] = 6$ but $[L:k]=6$ and so $[L:k(r)]=1\implies k(r) = L$. This approach requires some guesswork because if you pick such an $r$ so that the trace in reduced form has both polynomials of degrees less than 6 then it means $k(h)$ would just be a proper subfield of $L$.

Notice $\text{id},F,G,FG,GF,FGF$ are the six automorphisms. Pick $r=x$ then its trace is:
$x + \tfrac{1}{x} + (1-x) + (\tfrac{1}{1-x}) + (1 - \tfrac{1}{x}) + (1 - \tfrac{1}{1-x} ) = 3$ ---> 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.
• April 24th 2009, 03:18 AM
NonCommAlg
it's easy to prove that $L \subset M,$ where $M$ is the set of all $r(x)=(x(1-x))^{2\ell} \frac{\sum_{i=0}^n a_i(x(1-x))^i}{\sum_{i=0}^m b_i(x(1-x))^i},$ where $\ell \in \mathbb{Z}, \ a_i \in k, \ b_i \in k, \ a_0b_0a_nb_m \neq 0,$ and $m - n=3 \ell.$

clearly for every $r(x) \in M$ we have $r(x)=r(1-x).$ the question is: exactly which $r(x) \in M$ also satisfy $r(x)=r(1/x)$? (Thinking)
• April 24th 2009, 08:09 AM
ThePerfectHacker
Quote:

Originally Posted by NonCommAlg
it's easy to prove that $L \subset M,$ where $M$ is the set of all $r(x)=(x(1-x))^{2\ell} \frac{\sum_{i=0}^n a_i(x(1-x))^i}{\sum_{i=0}^m b_i(x(1-x))^i},$ where $\ell \in \mathbb{Z}, \ a_i \in k, \ b_i \in k, \ a_0b_0a_nb_m \neq 0,$ and $m - n=3 \ell.$

clearly for every $r(x) \in M$ we have $r(x)=r(1-x).$ the question is: exactly which $r(x) \in M$ also satisfy $r(x)=r(1/x)$? (Thinking)

You should have the approach above that uses the trace, it works out.

Again, $\text{Gal}(K/L) = \{ \text{id},F,G,GF,FG,GFG\}$.
The six automorphisms are: $x\mapsto x, \tfrac{1}{x},\tfrac{1}{1-x},1-\tfrac{1}{x}, 1-\tfrac{1}{1-x}$.

If $r$ is a rational function in $K$ then $\text{Tr}_{K/L}(r) \in L$, thus the trace allows us to produce elements in $L$.

Just for illustration, let us use $r=x$ again, then:
$\text{Tr}_{K/L}(x) = x + \tfrac{1}{x}+ (1-x) + \tfrac{1}{1-x}+(1-\tfrac{1}{x}) + (1-\tfrac{1}{1-x})=3$
Certainly $3$ is fixed by $F,G$ but that is constant and uninteresting.

Let us make it more interesting by taking $r=x^2$ then:
$\text{Tr}_{K/L}(x^2) = x^2 + (\tfrac{1}{x})^2+ (1-x)^2 + (\tfrac{1}{1-x})^2+(1-\tfrac{1}{x})^2 + (1-\tfrac{1}{1-x})^2 = \tfrac{2x^6 - 6x^5 + 9x^4 - 8x^3+9x^2-6x+2}{x^4 - 2x^3 + x^2}$

By what was explained above it follows that: $L = k\left( \tfrac{2x^6 - 6x^5 + 9x^4 - 8x^3+9x^2-6x+2}{x^4 - 2x^3 + x^2} \right)$.

So in general an element is fixed by $F,G$ if and only if it has the form $\frac{\sum_i a_i h^i}{\sum_j b_j h^j}$ where $h= \tfrac{2x^6 - 6x^5 + 9x^4 - 8x^3+9x^2-6x+2}{x^4 - 2x^3 + x^2}$.
(Of course the coefficients $b_j$ are such that the lower function is non-zero).