Radicals in a Cyclotomic Field Extension I was looking for the theorems which describe about all the kinds of radicals contained in a cyclotomic extension. With radical I mean the number, say $x$, is not in $\mathbb{Q}$ but some power, say $x^n$, is in $\mathbb{Q}$. For example, $\sqrt{2}$ is contained in $\mathbb{Q}(\zeta_8)$. Definitely there could be other radicals existing.
Now, is there a theorem which describes or classifies a general scenario saying that some/any $n$th root of a rational (we can rather talk about real $n$th root of an integer (because we can always rationalize the denominator)) is contained in a specific cyclotomic extension? In other words, what are all the radicals present in a cyclotomic extension?
Please correct me if I'm stating anything wrong.

For $a \in \mathbf Z$, $\sqrt{a}$ is contained in some cyclotomic field (specifically, $\mathbf Q(\zeta_{4|a|})$. For nonzero $a \in \mathbf Z$, if $x^n - a$ is irreducible over $\mathbf Q$ and $n > 1$ is not a power of $2$ then $\sqrt[n]{a}$ (that means an $n$th root of $a$, i.e., a root of $x^n - a$) is not contained in a cyclotomic field.
Each $n > 2$ has an odd prime factor or is divisible by $4$.
If $d \mid n$ and $d > 0$ then $\sqrt[n]{a}^{n/d}$ is a $d$th root of $a$ (a root of $x^d - a$), so if an $n$th root of $a$
is contained in a cyclotomic field then so is a $d$th root of $a$ for each positive factor $d$ of $n$. And since $x^n - a$ is irreducible over $\mathbf Q$, so is $x^d - a$ for each positive divisor $d$ of $n$: see Theorem 4.1 here.
Therefore in order to prove no $n$th root of $a$ is contained in a cyclotomic field when $n$ is not a power of $2$ and $x^n - a$ is irreducible over $\mathbf Q$, it suffices to show this when $n = p$ is an odd prime.
We want to show $\sqrt[p]{a}$ is not in a cyclotomic field when $x^p - a$ is irreducible over $\mathbf Q$.
Cyclotomic fields are abelian Galois extensions of $\mathbf Q$, so by Galois theory subfields of cyclotomic fields are also (abelian) Galois extensions of $\mathbf Q$. Therefore if $\sqrt[p]{a}$ is in a cyclotomic field, $\mathbf Q(\sqrt[p]{a})$ would be Galois over $\mathbf Q$, and that implies there are $p$ roots of $x^p - a$ in $\mathbf Q(\sqrt[p]{a})$ by irreducibility of $x^p - a$ over $\mathbf Q$. A ratio of different $p$th roots of $a$ is a nontrivial $p$th root of unity, so $\mathbf Q(\zeta_p)$ is a subfield of $\mathbf Q(\sqrt[p]{a})$. Computing field degrees over $\mathbf Q$, $(p-1) \mid p$ and that is a contradiction.  (Another reason $\mathbf Q(\sqrt[p]{a})$ is not Galois over $\mathbf Q$ is that it has a real embedding since $p$ is odd and there is only one real $p$th root of $a$, which contradicts the existence of $p$ roots of $x^p-a$ in $\mathbf Q(\sqrt[p]{a})$.)
What happens if $n$ is a power of $2$ and $n > 2$?  Things get tricky here, as they often do when working with $x^n - a$ and $n$ is a power of $2$ (consider the difference between $p$-power cyclotomic extensions of $\mathbf Q$ for prime $p$ when $p = 2$ and $p \not= 2$). I leave it to someone else to sort things out completely, and will just point out one restriction by reducing to the case $n = 4$.
When $n = 2^k$ for $k \geq 2$, irreducibility of $x^n - a$ over $\mathbf Q$ implies irreducibility of $x^4 - a$ over $\mathbf Q$, and
Claim: If $x^4 - a$ is irreducible over $\mathbf Q$ then $\sqrt[4]{a}$ is in a cyclotomic field if and only if $a = -b^2$ for some integer $b$.
Since positive $a$ don't fit the conclusion of the claim, for $a \in \mathbf Z^+$ with $x^n - a$ irreducible over $\mathbf Q$, $\sqrt[n]{a}$ is not in a cyclotomic field if $n > 2$.
Proof of claim:
As in the argument above, if $\sqrt[4]{a}$ is in a cyclotomic field then $\mathbf Q(\sqrt[4]{a})$ is Galois over $\mathbf Q$.  The ratio of different 4th roots of $a$ are the 4th roots of unity, so if $\mathbf Q(\sqrt[4]{a})$  is Galois over $\mathbf Q$ then $\mathbf Q(i)$ is inside $\mathbf Q(\sqrt[4]{a})$. Let's look at $\mathbf Q(\sqrt[4]{a})$ as a quadratic extension of $\mathbf Q(i)$.
The number $\sqrt[4]{a}$ has a quadratic minimal polynomial over $\mathbf Q(i)$. What could that minimal polynomial be? It must be a factor of $x^4 - a$ (the minimal polynomial of $\sqrt[4]{a}$ over $\mathbf Q$), so let's just write down all possible monic quadratic factors of $x^4 - a$ with $\sqrt[4]{a}$ as a root and see which ones could have all coefficients in $\mathbf Q(i)$.
By unique factorization in $\mathbf Q(\sqrt[4]{a})[x]$, where $x^4 - a$ splits completely, there are only three monic quadratic factors of $x^4 - a$ with $\sqrt[4]{a}$ as a root:
$$
(x-\sqrt[4]{a})(x+\sqrt[4]{a}) = x^2 - \sqrt{a}, 
$$
$$
(x-\sqrt[4]{a})(x-i\sqrt[4]{a}) = x^2 - (1+i)\sqrt[4]{a}x + i\sqrt{a}, 
$$
$$
(x-\sqrt[4]{a})(x+i\sqrt[4]{a}) = x^2 - (1-i)\sqrt[4]{a}x - i\sqrt{a}.
$$
The second and third choices have linear coefficient $-(1\pm i)\sqrt[4]{a}$, and if this is in $\mathbf Q(i)$ then so is $\sqrt[4]{a}$, which is impossible since $\sqrt[4]{a}$ has degree $4$ over $\mathbf Q$. Thus the minimal polynomial of $\sqrt[4]{a}$ over $\mathbf Q(i)$ must be $x^2 - \sqrt{a}$, so
$\sqrt{a} \in \mathbf Q(i)$, which implies $\mathbf Q(\sqrt{a}) = \mathbf Q(i)$.
From $\mathbf Q(\sqrt{a}) = \mathbf Q(i)$, $a = -b^2$ for some integer $b$.
Conversely, suppose $a = -b^2$. Then the roots of $x^4 - a = x^4 + b^2$ are $\pm\sqrt{bi}$ and $\pm\sqrt{-bi}$, which are in cyclotomic fields since $\sqrt{b}$ is in a cyclotomic field and $\sqrt{i}$ is an $8$th root of unity.  In fact, $\mathbf Q(\sqrt{bi}) = \mathbf Q(\sqrt{2b},i)$ is a composite of two different quadratic extensions of $\mathbf Q$: $i = (\sqrt{bi})^2/b$ is in $\mathbf Q(\sqrt{bi})$, the number $\alpha = (1-i)\sqrt{bi}$ in $\mathbf Q(\sqrt{bi})$ satisfies $\alpha^2 = -2i(bi) = 2b$, and $x^2 - 2b$ is irreducible over $\mathbf Q$ since otherwise $2b = c^2$ for some integer $c$ and that would reveal $x^4 - a$ to be reducible over $\mathbf Q$: $a = -b^2 = -c^4/4$, so
$$
x^4 - a = x^4 + \frac{c^4}{4} = (x^2 + cx + c^2/2)(x^2 - cx + c^2/2).
$$
From irreducibility of $x^2 - 2b$, $\mathbf Q(\alpha) = \mathbf Q(\sqrt{2b})$ is a quadratic subfield of $\mathbf Q(\sqrt[4]{a})$ and it is not the quadratic subfield $\mathbf Q(i) = \mathbf Q(\sqrt{-1})$ since $-2b$ is not a perfect square: if $-2b$ were a perfect square then (by unique factorization in $\mathbf Z$) $b = -2d^2$, so $a = -b^2 = -4d^4$, but then $x^4 - a = x^4 + 4d^4$ would be reducible over $\mathbf Q$:
$$
x^4 - a = x^4 + 4d^4 = (x^2 + 2dx + 2d^2)(x^2 - 2dx + 2d^2).
$$
Remark. By a similar argument to the case $n = 4$,
if $n = 2^k \geq 4$ and $x^n - a$ is irreducible over
$\mathbf Q$ with $\sqrt[n]{a}$ in a cyclotomic field then $\mathbf Q(\sqrt[n]{a})$ is quadratic over $\mathbf Q(\zeta_{n}) = \mathbf Q(\zeta_{2^k})$ and $\mathbf Q(\sqrt[n/2]{a}) = \mathbf Q(\zeta_n)$. Must $a = -b^{n/2} = -b^{2^{k-1}}$ for some integer $b$?
