Is $i$ an element of $\mathbb{Q}(\sqrt[4]{5},w)$ where $w=e^{2πi/3}$?

Question

I am trying to check if $i$ is an element of $\mathbb{Q}(\sqrt[4]{5},w)$ where $w=e^{2πi/3}$. How can I check if this is the case? Would it be correct to express my field as $a\mathbb(\sqrt[4]{5}) + b (-1/2 + i (\sqrt{3}/2))$ for rationals $a$ and $b$ and show that no combination of these gives me the element $i$?

Answer 1

The elements of the field are of the form:

$$\alpha=a_{00}+a_{01}w+a_{10}\sqrt[4]5+a_{11}\sqrt[4]5w+a_{20}\sqrt[4]{5^2}+a_{21}\sqrt[4]{5^2}w+a_{30}\sqrt[4]{5^3}+a_{31}\sqrt[4]{5^3}w$$ where the $a_{ij}$ are rational numbers.

The imaginary component of this is:

$$\operatorname{Im}\alpha =\frac{\sqrt3}2\left(a_{01}+a_{11}\sqrt[4]5+a_{21}\sqrt[4]{5^2}+a_{31}\sqrt[4]{5^3}\right)$$

If $\operatorname{Im}\alpha =1,$ then you get, by multiplying both sides by $\sqrt{3}:$

$$\sqrt{3}=\frac32\left(a_{01}+a_{11}\sqrt[4]5+a_{12}\sqrt[4]{5^2}+a_{13}\sqrt[4]{5^3}\right)\in\mathbb Q(\sqrt[4]5).$$

So if you can show $\sqrt 3\notin \mathbb Q(\sqrt[4]5),$ you will have shown $i$ can't be in your field.

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You can prove the other way - if $z$ is a real number and $w$ is the value in the original problem, then $\mathbb Q(z,w)$ contains $i$ if and only if $\sqrt3\in\mathbb Q(z).$

Essentially, if $K$ is a complex field, $K(w)$ can be seen to be equivalent to $K(i\sqrt 3).$ If $K$ is a real field, then the imaginary component of $K(w)$ are all of the form $a\sqrt 3$ for $a\in K.$

In this question $K=\mathbb Q(\sqrt[4]5).$