1. Funadamental theorem of Galois Theory

$k(x)$ is an automorphism defined by $\sigma_1(x)=x, \sigma_2(x)=1-x$, and $\sigma_3(x)=1/x$ Let $I(x)=\frac{(x^2-x+1)^3}{x^2(x-1)^2}$

Assuming the fundamental theorem of Galois Theory, exhibit all intermediate fields between E=k(x) and F=k(I).

2. Originally Posted by dabien
$k(x)$ is an automorphism defined by $\sigma_1(x)=x, \sigma_2(x)=1-x$, and $\sigma_3(x)=1/x$ Let $I(x)=\frac{(x^2-x+1)^3}{x^2(x-1)^2}$

Assuming the fundamental theorem of Galois Theory, exhibit all intermediate fields between E=k(x) and F=k(I).
I'll assume that k is a field and k(x) is a field of rational functions in the variable x. Unfortunately, $\sigma_1(x)=x, \sigma_2(x)=1-x$, and $\sigma_3(x)=1/x$ do not form a group of automorphisms (it is not even closed), so you can't find the corresponding field. However, it seems like $\sigma_2(x)=1-x$, and $\sigma_3(x)=1/x$ generate a group A of automorphisms (isomorphic to a symmetric group $S_3$), $A=\{\sigma_1, \sigma_2, \sigma_3, \sigma_4, \sigma_5, \sigma_6\}$, where $\sigma_4(x)=1-1/x$, $\sigma_5(x)=1/(1-x)$, and $\sigma_6(x)=x/(x-1)$.

So your base field is F=k(I) where $I=I(x)=\frac{(x^2-x+1)^3}{x^2(x-1)^2}$. Now consider the splitting field $E=k(x)$ for an irreducible polynomial f in k(I) such that $f(X) ={(X^2 - X + 1 )}^3 -IX^2{(X-1)}^2$, where $f(X) \in k(I)[X]$ (verify this). We see that $Gal(E/F)=S_3$.

Now we shall find an intermediate field corresponding a subgroup of $Gal(E/F)$. Let $H=\{\sigma_1, \sigma_3\}$. It is easy to check that H is a subgroup of $Gal(E/F)$. The field corresponding H is $S=k(x+\frac{1}{x})$, which is fixed by elements of H.

I'll leave it to you to find all other intermediate fields, because it seems like a lengthy computation is required to find and verify it.