For $\omega=e^{2\pi i/3}$ and $\epsilon =e^{2\pi i/5}$, determine $|\mathbb Q[\omega +\epsilon]:\mathbb Q|$.

Question

Let $\omega=e^{2\pi i/3}$ and $\epsilon =e^{2\pi i/5}$, then show the following: (a) $\omega \notin \mathbb Q[\epsilon]$, and (b) determine $|\mathbb Q[\omega +\epsilon]:\mathbb Q|$.

Clearly $\omega,\epsilon$ are primitive $3$rd root of $1$ and primitive $5$th root of $1$, respectively. Since $Gal(\mathbb Q[\epsilon]/\mathbb Q)=\mathbb Z/4 \mathbb Z$, if $\omega \in \mathbb Q[\epsilon]$ by the fundamental theorem of Galois theory and cyclicness of $\mathbb Z/4 \mathbb Z$ we must have $\mathbb Q[\epsilon +\epsilon^{-1}]=\mathbb Q[\omega]$. Note that $(\epsilon +\epsilon^{-1})$ satisfy the irreducible polynomial $x^2+x-1$ over $\mathbb Q$. But $\omega$ is not fixed by the complex conjugation, hence $\omega \notin \mathbb R$. OTOH $(\epsilon + \epsilon^{-1}) \in \mathbb R$. Thus $\mathbb Q[\epsilon +\epsilon^{-1}]\neq \mathbb Q[\omega]$. Therefore $\omega \notin \mathbb Q[\epsilon]$ and part $(a)$ is done. My problem in part $(b)$. We get the tower $\mathbb Q(\omega +\epsilon) \subseteq \mathbb Q(\omega,\epsilon)$. Also $\zeta_{15}\in \mathbb Q(\omega,\epsilon)$. I aimed to show that $\mathbb Q(\omega,\epsilon)=\mathbb Q(\zeta_{15})$. May be there are some other direct way to look at the degree, e.g., looking at the number of conjugates of $\omega +\epsilon$. Give me some hint to complete it. Thanks.

Answer 1

Converting my comment into answer. To reduce typing effort we use $a=\omega, b=\epsilon, c=a+b$.

We have $b^5=1$ and hence $$(c-a) ^5=1$$ The left hand side is of the form $f(c) a+g(c) $ where $f, g$ are polynomials in $c$ with integer coefficients. Why??

Because $a$ is of degree $2$ over $\mathbb {Q} $ and hence all powers of $a$ higher than $1$ can be replaced by a linear combination of the form $pa+q$.

It now follows that $$a=\frac{1-g(c)}{f(c)}\in\mathbb {Q} (c) $$ and then $b=c-a\in\mathbb {Q} (c) $. Hence $\mathbb{Q} (a, b) \subseteq \mathbb {Q} (c) $. And we are done!!

The technique works for any algebraic numbers $a, b$ with $a$ of degree $2$.

Answer 2

Let me apply some rather more advanced notions here. If we can find a prime $p$ such that $[\Bbb Q_p(\omega+\epsilon):\Bbb Q_p]=8$, we will be done.

A natural candidate is $p=5$, since over $\Bbb Q_5$, we know that $\Bbb Q_5(\epsilon)$ is totally ramified, of degree $4$, and $\Bbb Q_5(\omega)$ is unratified quadratic.

To answer the questions, $\omega$ still isn’t in $\Bbb Q_5(\epsilon)$, since the residue field is still $\Bbb F_5$, and (by Hensel) there are no new roots of unity of order prime to $5$. And a quadratic extension of a quartic extension gives an octic extension.

(One could as well have used $p=3$ for this argument.)