Units in cyclotomic fields - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T19:30:08Z http://mathoverflow.net/feeds/question/71885 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71885/units-in-cyclotomic-fields Units in cyclotomic fields expmat 2011-08-02T13:56:15Z 2011-08-02T18:29:07Z <p>Let $q$ and $r$ be distinct prime numbers. I noticed (computing a few cases) that $\zeta_{2q} + \zeta_{2q}^{-1} + \zeta_{2r} + \zeta_{2r}^{-1}$ is a unit (in $\mathbb{Z}[\zeta_{2qr}]$, say). Is this always true? Why is that?</p> http://mathoverflow.net/questions/71885/units-in-cyclotomic-fields/71890#71890 Answer by David Speyer for Units in cyclotomic fields David Speyer 2011-08-02T15:27:16Z 2011-08-02T15:51:02Z <p>I assume you want $q$ and $r$ to be odd primes. Also, note that I will be using the notation that $\zeta_m$ means an arbitrary primitive $m$-th root of unity (but the same one every time it appears in an equation), and will be proving the statement in that generality.</p> <p><strong>Lemma:</strong> For any odd $m>1$ and any $\zeta_m$, the number $\zeta_m+1$ is a unit.</p> <p><strong>Proof:</strong> Let $r$ be such that $m | 2^r-1$. We'll abbreviate $\zeta_m$ to $\zeta$. </p> <p>Then $\zeta^{2^r} = \zeta$ so <code>$$1 = \left( \frac{\zeta^{2}-1}{\zeta -1} \right) \left( \frac{\zeta^{4}-1}{\zeta^{2} -1} \right) \cdots \left( \frac{\zeta^{2^r}-1}{\zeta^{2^{r-1}} -1} \right)=$$</code> <code>$$\left( \zeta+1 \right) \left( \zeta^{2} + 1 \right) \cdots \left( \zeta^{2^{r-1}} +1 \right),$$</code> exhibiting an explicit inverse for $\zeta+1$.</p> <p>Let $\eta$ be a primitive $2qr$ root of unity. Then your proposed unit is $\eta^{r}+\eta^{-r} + \eta^q + \eta^{-q}$ and factors as <code>$$\eta^r (1+\eta^{q-r})(1+\eta^{-q-r}).$$</code> Since $q$ and $r$ are odd and relatively prime, $\eta^{q-r}$ and $\eta^{q+r}$ are primitive $qr$-th roots of unity and we are done by the lemma.</p> http://mathoverflow.net/questions/71885/units-in-cyclotomic-fields/71894#71894 Answer by Tom Goodwillie for Units in cyclotomic fields Tom Goodwillie 2011-08-02T16:22:12Z 2011-08-02T18:29:07Z <p>The answer is also yes if one of the primes, say $r$, is $2$, because then $\zeta_{2r}+\zeta_{2r}^{-1}=0$ and $\zeta_{2q}+\zeta_{2q}^{-1}=\zeta_{2q}(1+\zeta_{2q}^{-2})$ is a unit (as $\zeta_{2q}^{-2}$ is a primitive $q$th root of $1$ and $q$ is an odd prime).</p> <p>Edit: </p> <p>(1) Note that if both primes are odd then $\zeta_q+\zeta_q^{-1}+\zeta_r+\zeta_r^{-1}$ is also a unit. Indeed, $-\zeta_q$ is a primitive $2q$th root of $1$ (this relies on $q$ being odd), so let's call it $\zeta_{2q}$, and likewise for $r$. Then $-(\zeta_q+\zeta_q^{-1}+\zeta_r+\zeta_r^{-1})=\zeta_{2q}+\zeta_{2q}^{-1}+\zeta_{2r}+\zeta_{2r}^{-1}$, and we know that the latter is a unit. </p> <p>(2) Note also that if one of the primes, say $r$, is $2$ then $\zeta_q+\zeta_q^{-1}+\zeta_r+\zeta_r^{-1}$ is not a unit: it is equal to $\zeta_q+\zeta_q^{-1}-2$, so it is a unit times the square of $\zeta_q-1$, but the latter (unlike $\zeta_q +1$) is not a unit because it goes to $0$ under the unique ring homomorphism $\mathbb Z [\zeta_q]\to \mathbb Z/q$, which takes $\zeta_q$ to $1$.</p>