How to show $\sqrt[3]{X-i}\notin \mathbb{C}(X,\sqrt[3]{X+i})$

Question

I'm trying to show $\sqrt[3]{X-i}\notin \mathbb{C}(X,\sqrt[3]{X+i})$. But this is harder than I expected.

Is there any easy way to show this?

Answer 1

Its easier for me to give my solution over a general field. Let $k$ be a field with a cube root of unity $\zeta$. In our case $k=\mathbb{C}(x)$ Let $a,b \in k$ and assume that $\sqrt[3]{b} \in k(\sqrt[3]{a})$. Now the congugates of $\sqrt[3]{a}$ are $\sqrt[3]{a},\zeta\sqrt[3]{a},\zeta^2\sqrt[3]{a}$ so in fact $\text{Tr} (\sqrt[3]{a})=0$.

If we have $$\sqrt[3]{b}=\alpha+\beta \sqrt[3]{a}+\gamma(\sqrt[3]{a})^2$$ then taking the trace of both sides, gives $3\alpha=0$ so we really have $$\sqrt[3]{b}=\beta \sqrt[3]{a}+\gamma(\sqrt[3]{a})^2$$ let $\sigma $ be the automorphism such that $\sigma(\sqrt[3]{a})=\zeta\sqrt[3]{a}$ so we have either $$\sigma(\sqrt[3]{b})=\sqrt[3]{b} \ \ \text{or} \ \ \zeta\sqrt[3]{b} \ \ \text{or} \ \ \zeta^2\sqrt[3]{b}$$ Say $\sigma(\sqrt[3]{b})=\zeta\sqrt[3]{b}$ then by applying $\sigma$ to the above equation we have $$\zeta\sqrt[3]{b}=\zeta\beta \sqrt[3]{a}+\zeta^2\gamma(\sqrt[3]{a})^2$$, however if the multiply the same equation by $\zeta$ and subtract we get $\gamma=0$. (the first case is seen to be impossible and the third gives a similar result) So we have now $$\sqrt[3]{b}=\beta \sqrt[3]{a}$$ and cubing we have $$\frac{b}{a}=\beta^3 $$

So to prove our result we need only remark that $\frac{x-i}{x+i}$ is not the third power of a rational function in $\mathbb{C}(x)$