I'm teaching myself some mathematics, so post question here sometimes is my last resort to get an answer, i have already posted this question on Mathematics Stack Exchange But no one answers, and I really want to know the answer.


Let $E$ be an extension of $\mathbb{C}$ such that $E$ = $\mathbb{C}(t,u)$ where $t$ is transcendental over $\mathbb{C}$ and $u$ satisfies the equation $u^2+t^2=1$ over $\mathbb{C}(t)$. 


let $n = 2m + 1$ and let $\mathbb{C}(t^n,u^n)$ be an extension field over $\mathbb{C}$, then $\mathbb{C}(t,t^n,u^n)$ is a splitting field over $\mathbb{C}(t^n,u^n)$ of $x^n-t^n$ ($\mathbb{C}$ contains nth roots of unity) in fact $\mathbb{C}(t,t^n,u^n)$ = $\mathbb{C}(t,u)$ since $u^{2m} = (1-t^2)^{m}$ then $u = u^n/(1-t^2)^{m}$ hence $\mathbb{C}(t,u)$ is a splitting field over $\mathbb{C}(t^n,u^n)$ of $x^n-t^n$ we can apply same discussion to $u$ to get $\mathbb{C}(u,t^n,u^n)$ = $\mathbb{C}(t,u)$ and $\mathbb{C}(t,u)$ is a splitting field over $\mathbb{C}(t^n,u^n)$ of $x^n-u^n$


let $\eta \in Gal\mathbb{C}(t,u)/\mathbb{C}(t^n,u^n)$ then $\eta(t) = zt$ and $\eta(u) = qu$ where $z$ and $q$ are nth roots of unity


BUT $\eta(u^2) + \eta(t^2) = 1$ implies $q^2u^2 + z^2t^2 = 1$ implies $q^2(1-t^2) + z^2t^2 = 1$ since $t$ is transcendental over $\mathbb{C}$ hence $q^2 = z^2$ and $q^2 = 1$ which is absurd when $m \gt 0$, therefore $Gal\mathbb{C}(t,u)/\mathbb{C}(t^n,u^n) = \{1\}$ when $m \gt 0$ this means$\mathbb{C}(t,u) = \mathbb{C}(t^n,u^n)$ 


so I must have made some grotesque error above (is $sinx$ expressible rationally with complex coefficients in terms of $cos^nx$ and $sin^nx?$ $n = 2m + 1 $, $ m \gt 0 $), who can help me figure it out? thanks!!!!



Extension field $\mathbb{C}(t,u)$ over $\mathbb{C}(t^n,u^n)$
What the above argument proves is that the extension is either trivial, or not Galois. My first guess was that it isn't Galois, but it was a stupid mistake! In fact, it is trivial.


Denote $z=u+it$. Then $2u=z+z^{-1}$, hence $\mathbb{C}(t,u)=\mathbb{C}(z)$. Consider the simplest case m=1. 
Let $A=u^3-it^3,\,B=u^3+it^3$. Then  

$$z=\frac{3}{4}\cdot\frac{27A^3-7AB^2-16A^2B^3-4B}{27A^4-16A^3B^3-6A^2B^2-3AB-2}.$$
Thus, $\mathbb{C}(t^3,u^3)=\mathbb{C}(z)$. 


The same is true for $m>1$, but the computations grow tedious, and a
proof I have in mind is a bit complicated. (Does anyone know a simple one?) 


