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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!!!!

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    $\begingroup$ What you did is basically a proof by contradiction that the extension you consider is not Galois. In fact, $\mathbb{C}(t,u)=\mathbb{C}(z)$, where $z=u+it$, and this field can't be a Galois extension of its subfield. $\endgroup$ – Alex Gavrilov Sep 10 '16 at 6:32
  • $\begingroup$ please provide a link to the MathSE question $\endgroup$ – YCor Sep 10 '16 at 6:33
  • $\begingroup$ the link is math.stackexchange.com/q/1920483/262757 $\endgroup$ – sunya hu Sep 10 '16 at 6:48
  • $\begingroup$ $\mathbb{C}(t)$ is a proper subfield of $\mathbb{C}(t,u)$ since if $u \in \mathbb{C}(t)$ then $u = f(t)/g(t)$ where $f(x)$, $g(x)$ $ \in \mathbb{C}[x]$ but it yields $(f(t)/g(t))^2 + t^2 = 1$ since $t$ is transcendental hence it is impossible. $\mathbb{C}(t,u)$ is a splitting field over $\mathbb{C}(t)$ of a separable polynomial of $\mathbb{C}(t)[x]$ namely $x^2 + t^2 -1$ hence $\mathbb{C}(t,u)$ is Galois over $\mathbb{C}(t)$ $\endgroup$ – sunya hu Sep 10 '16 at 8:38
  • $\begingroup$ Yes, I was wrong there: there are some exceptions. I will better write an answer instead of a comment. $\endgroup$ – Alex Gavrilov Sep 10 '16 at 10:37
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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?)

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