n-th roots of Pythagorean numbers Let $F$ be the field ${\mathbb Q}(i)\subset \mathbb C$ and let $T\subset F$ be the set of all elements of complex absolute value 1.
Let $n$ be a natural number $\ge 2$ and let $\mu_n(T)\subset\mathbb C$ be the set of all $n$-th roots of elements of $T$.
Finally, let $E=F(\mu_n(T))$. 
Question: Is the field extension $E/F$ finite or infinite?
 A: Tan (The group of rational points on the unit circle, Math. Mag. 69 (1996), 163-171)
proved that the group of rational points on the unit circle modulo torsion is isomorphic to infinitely many copies of $\mathbb Z$. 
I have given a couple of references to related articles in Kreise und Quadrate modulo $p$,  Math. Semesterber. 47 (2000), 51-73.
A: (This may have errors - I'm not an algebraic number theorist.)
We have a complete description of the multiplicative structure of $F = \mathbb{Q}(i)$.  It is: 
$$\mathbb{Q}(i)^\times \cong \left( \bigoplus_{p \cong 1 \mod 4} (\mathbb{Z} \oplus \mathbb{Z}) \right) \oplus \left( \bigoplus_{\text{other } p} \mathbb{Z} \right) \oplus \mathbb{Z}/4\mathbb{Z}.$$
Note that there are no $n$-divisible subgroups, except the 4th roots of unity (when $n$ is odd).  This is a good sign of infinite degree.
Each prime $p$ congruent to 1 mod 4 can be written as product of primes $(a+ib)(a-ib)$, with $a$ and $b$ unique up to obvious symmetries.  We find that $\frac{a+ib}{a-ib} \in T$, and is a primitive element in the copy of $\mathbb{Z} \oplus \mathbb{Z}$ in the big sum corresponding to $p$.  In particular, it is not an $n$th power for $n \geq 2$.
We can now construct a sequence of fields $F=F_0 \subset F_1 \subset \dots$, where $F_k$ is given by starting with $F_{k-1}$, and adjoining an $n$th root of the number $\frac{a+ib}{a-ib}$ corresponding to some prime congruent to 1 mod 4 over which $F_{k-1}$ is unramified.  Since finite extensions are ramified over finitely many primes, and adjoining the $n$th root creates ramification over $p$, we have strict containment at each step, and the chain does not terminate after finitely many steps.  The union of the chain is an infinite degree extension that is contained in $E$, so $E$ has infinite degree over $F$.
