n-th roots of Pythagorean numbers - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T21:13:05Z http://mathoverflow.net/feeds/question/55006 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/55006/n-th-roots-of-pythagorean-numbers n-th roots of Pythagorean numbers anton 2011-02-10T10:09:18Z 2011-02-11T11:47:25Z <p>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 $E=F(\mu_n(T))$. </p> <p>Question: Is the field extension $E/F$ finite or infinite?</p> http://mathoverflow.net/questions/55006/n-th-roots-of-pythagorean-numbers/55056#55056 Answer by Franz Lemmermeyer for n-th roots of Pythagorean numbers Franz Lemmermeyer 2011-02-10T17:27:10Z 2011-02-11T06:58:19Z <p>Tan (<em>The group of rational points on the unit circle</em>, 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 $\mathbb Z$. </p> <p>I have given a couple of references to related articles in <em>Kreise und Quadrate modulo $p$</em>, Math. Semesterber. 47 (2000), 51-73.</p> http://mathoverflow.net/questions/55006/n-th-roots-of-pythagorean-numbers/55060#55060 Answer by S. Carnahan for n-th roots of Pythagorean numbers S. Carnahan 2011-02-10T18:16:04Z 2011-02-10T18:16:04Z <p>(This may have errors - I'm not an algebraic number theorist.)</p> <p>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.</p> <p>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 $n \geq 2$.</p> <p>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 $F$.</p>