Primitive root in quadratic field

Let $\displaystyle \omega _{m} = e^{ \frac {2 \pi i} {m}} $, then $\displaystyle \omega _{m} $ is one of the primitive m th roots of unity.

a) Show that $\displaystyle \mathbb {Q} [ \omega _{m} ] \subseteq \mathbb {Q} [ \omega _n ] $ if m|n.

Proof.

Let mt = n for some integer t.

Pick an element $\displaystyle a+b \omega _m \in \mathbb {Q} [ \omega _m ] $, with a and b rational numbers.

Then it equals to $\displaystyle a + b e^ { \frac {2 \pi i } {m} } = a + be^{ \frac {2 \pi i (t) } {n} } = a + be^t + be ^ { \frac {2 \pi i } {n} } $ Now, is this element in $\displaystyle \mathbb {Q} [ \omega _n ] $? I just don't know if $\displaystyle e^t $ is a rational number.

b) Show that $\displaystyle \sqrt {p} \in \mathbb {Q} [ \omega _{4p} ] $.

Proof.

There is a theorem that state: Let $\displaystyle \omega $ be a primitive pth root of unity, p an odd prime. $\displaystyle \sqrt {p} \in \mathbb {Q} [ \omega ] $ if $\displaystyle p \equiv 1 \ (mod \ 4 ) $

Now, is $\displaystyle \omega _{4p}$ a primitive pth roof of unity?

Thank you.