TITLE: Extension field $\mathbb{C}(t,u)$ over $\mathbb{C}(t^n,u^n)$
QUESTION [5 upvotes]: 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!!!!

REPLY [3 votes]: You did nothing wrong.
$K = \Bbb C(t^n,u^n) \cap \Bbb C(t) $ contains $t^n$ and $u^{2n} + t^{2n} = (1-t^2)^n + t^{2n} = f(t)$ where $f$ is a polynomial of degree $2n-2$, which is coprime with $n$.
Then $[K : \Bbb C(t)]$ has to divide both $n$ and $2n-2$, so it must be $1$.
For example let $n=3$ and write $T = t^3, U = u^3$.
we have $(T^2 + U^2 - 1) - 3tT + 3t^2 = 0$ and $T - t^3 = 0$.
Then this is just a matter of elimination to get smaller annihilating polynomials (if you compute the resolvant of the two polynomials you get the equation satisfied bewteen $U$ and $T$ be we want the step just before that) :
$t(T^2+U^2-1) - 3t^2T + 3t^3$
$3T + t(T^2+U^2-1) - 3t^2T $
$T(T^2+U^2-1) - 3tT^2 + 3t^2T$
$T(T^2+U^2+2) + t(-2T^2+U^2-1)$ 
hence $t = \frac {T(T^2+U^2+2)}{2T^2-U^2+1}$
I am pretty sure the numerator and denominator can be obtained systematically with a well chosen determinant just like the resolvant.