When is $i$ contained in $Q(\zeta)$ As part of a problem I have to determine for which values of $n$ is $i$ contained in $\Bbb Q(\zeta)$, were $\zeta$ is a primitive $n^{th}$ root of the unity. Clearly, if $n$ is multiple of $4$, it is contained since it is in fact a primitive root of the unity.
But I don't know how to prove that it is not contained in other cases  (I don't even know if it is contained).
Particularly I am interested the case when $n=p$ a prime number. 
 A: No number field can contain an infinite number of roots of unity. And any finite subgroup of the multiplicative subgroup of $\mathbf{C}^*$ is cyclic. As $i$ an element of order $4$ in the multiplicative group then $4$ must divide $n$, where $n$ is the order of $\zeta$.
A: If $\zeta$ is a primitice root of $n$th order with $n$ odd, then $i\zeta$ is a primitive root of $4n$th order. We have $[\Bbb Q[\zeta]:\Bbb Q]=\varphi(n)$ and $[\Bbb Q[i\zeta]:\Bbb Q]=\varphi(4n)=2\varphi(n)\ne \varphi(n)$ 
A: In the case $n=p$ prime we can have a good understanding of the situation. Let $p>2$ be a prime and notice that $\mathrm{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})\simeq (\mathbb{Z}/p\mathbb{Z})^\times$ and thus it contains only one subextension of degree $2$: that is, the fixed field of the only subgroup of index $2$ in $\mathrm{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$. But what is that subgroup? This subgroup can be identified with the square elements modulo $p$, since $x\mapsto x^2$ is a homomorphism in $\mathbb{Z}/p\mathbb{Z}$ with kernel $\{-1,1\}$ and thus the image (ie the squares) is the only subgroup of index $2$. We want to find an element that is fixed by this subgroup.
Recall that $\sigma\in\mathrm{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$ is determined by the integer $b$ such that $\sigma(\zeta)=\zeta^b$ (that is, we can explicitly write the isomorphism of $\mathrm{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$ and $\mathbb{Z}/p\mathbb{Z}$). Consider the quadratic Gauss sum: $$\gamma = \displaystyle\sum_{a=0}^{p-1} \left(\dfrac{a}{p}\right) \zeta^a$$ where $\left(\dfrac{a}{p}\right)$ is the Legendre symbol (ie $1$ if $a$ is a square modulo $p$, $-1$ if it is not and $0$ in case $p\mid a$). Clearly $\gamma$ is fixed by the only subgroup of index $2$, and thus $\gamma$ is in the only subextension of degree $2$. If we prove that $\gamma\notin\mathbb{Q}$, we will have that $\mathbb{Q}(\gamma)$ is the only subextension of degree $2$.
A computation (try it, it is not that difficult) can show that $\gamma^2 = p\left(\dfrac{-1}{p}\right)$ which is not a square in $\mathbb{Q}$ and thus shows that $\gamma\notin\mathbb{Q}$. But it also shows that $\mathbb{Q}(\gamma)=\mathbb{Q}\left(\sqrt{p\left(\dfrac{-1}{p}\right)}\right)$ which tells us that if $-1$ is a square modulo $p$ then the extension is totally real and if $-1$ is not a square then $i\in\mathbb{Q}(\gamma)$. That is, we have shown that the only quadratic subextension of the $p$-th cyclotomic extension is $$\begin{cases}\mathbb{Q}(\sqrt{p}) \text{ if } p\equiv 1\pmod{p}\\ \mathbb{Q}(i\sqrt{p})\text{ if } p\equiv 3\pmod{p}\end{cases}$$ since it is known that $-1$ is a square modulo an odd prime $p$ iff $p\equiv 1\pmod{4}$.
Thus if $i\in\mathbb{Q}(\zeta)$ we would have that $\mathbb{Q}(i)$ is a quadratic subextension of $\mathbb{Q}(\zeta)$. Since $\mathbb{Q}(\sqrt{p})$ is totally real, $i$ can't be inside there, and if $i\in\mathbb{Q}(i\sqrt{p})$ then $\mathbb{Q}(\sqrt{p})\subseteq\mathbb{Q}(i\sqrt{p})$ and since both have degree $2$, they must be equal and that can't happen since one is totally real and the other one is not.
Hence, $i$ can't be inside $\mathbb{Q}(\zeta)$ for $\zeta$ any $p$-th primitive root.
