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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?

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  • $\begingroup$ Since $1$ is in $T$, the n-th roots of unity $\mu_n(1)$ are in $E$. Thus $E$ contains the field of all roots of unity, hence is infinite. $\endgroup$
    – Franz Lemmermeyer
    Feb 10, 2011 at 12:13
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    $\begingroup$ oops - n is supposed to be fixed. The answer is still that $E$ is infinite because the group of rational points on the unit circle is not finitely generated. I'll provide references later. $\endgroup$
    – Franz Lemmermeyer
    Feb 10, 2011 at 12:20
  • $\begingroup$ @Franz: If you could show that the group $T$ is an direct sum of infinitely many copies of $\mathbb Z$, that would prove that the extension is infinite. Otherwise it could be that $T$ is not finitely generated, but contains the $n$-th roots nevertheless, at least all but finitely many. $\endgroup$
    – user1688
    Feb 10, 2011 at 12:59
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    $\begingroup$ For every prime $p$ in $\mathbb{Z}$ that splits in $\mathbb{Z}[i]$ as $q\bar{q}$, $T$ contains $q/\bar{q}$. All such elements generate an infinite dimensional free subgroup in $T$. $\endgroup$
    – Yaakov Baruch
    Feb 10, 2011 at 14:28

2 Answers 2

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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.

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    $\begingroup$ @Franz : I guess you mean infinitely many copies of Z. Recently I came across the following unpublished note by Elkies : math.harvard.edu/~elkies/Misc/hilbert.pdf There he shows that the mentioned result is a nice application of Hilbert's Theorem 90. $\endgroup$
    – François Brunault
    Feb 10, 2011 at 20:58
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(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 $F$.

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  • $\begingroup$ First error! I should have said that in realizing the description of $\mathbb{Q}(i)^\times$, we fixed a choice of generators for the splittings of the primes $p$ congruent to 1 mod 4. Otherwise, the number $\frac{a+ib}{a-ib}$ is only a well-defined primitive element in the copy of $\mathbb{Z} \oplus \mathbb{Z}$ after we quotient by the copy of $\mathbb{Z}/4\mathbb{Z}$ that makes the 4th roots of unity. $\endgroup$
    – S. Carnahan ♦
    Feb 10, 2011 at 18:31

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