Cyclotomic Fields over Q and prime ideals - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T13:16:56Z http://mathoverflow.net/feeds/question/28113 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/28113/cyclotomic-fields-over-q-and-prime-ideals Cyclotomic Fields over Q and prime ideals 7-adic 2010-06-14T12:36:40Z 2010-06-14T14:02:31Z <p><strong>Q</strong> is the rational number field. p is a prime number. q is a prime number other than p. $k_{p^r}$ is a cyclotomic field. $k_{p^r}$=<strong>Q</strong>(x) where x is exp(2$\pi$i/$p^r$). [$k_{p^r}$:<strong>Q</strong>]=$p^{r-1}(p-1)$.</p> <p>Question: Does q remain a prime in the integer ring of $k_{p^r}$?</p> http://mathoverflow.net/questions/28113/cyclotomic-fields-over-q-and-prime-ideals/28115#28115 Answer by Gerry Myerson for Cyclotomic Fields over Q and prime ideals Gerry Myerson 2010-06-14T12:51:26Z 2010-06-14T12:51:26Z <p>Not necessarily. Simplest case, $p=2$, $r=2$, so the integer ring is ${\bf Z}[i]$, and $q$ stays prime if and only if it's 3 (mod 4). Any algebraic number theory text should tell you lots about this question, e.g., Marcus, Number Fields. </p> http://mathoverflow.net/questions/28113/cyclotomic-fields-over-q-and-prime-ideals/28116#28116 Answer by Alberto García-Raboso for Cyclotomic Fields over Q and prime ideals Alberto García-Raboso 2010-06-14T12:52:01Z 2010-06-14T12:52:01Z <p>No. E.g., $p=2$, $r=2$ gives you the extension $\mathbb{Q}(i)$ with ring of integers $\mathbb{Z}[i]$. The prime 2 ramifies in this extension, and those congruent to 1 mod 4 split into two distinct prime ideals of $\mathbb{Z}[i]$. The only ones that remain prime (the terminology is <em>inert</em>) are those congruent to 3 mod 4.</p> http://mathoverflow.net/questions/28113/cyclotomic-fields-over-q-and-prime-ideals/28117#28117 Answer by Charles Matthews for Cyclotomic Fields over Q and prime ideals Charles Matthews 2010-06-14T12:54:15Z 2010-06-14T12:54:15Z <p>Only if it satisfies a congruence condition. The Frobenius for q in the Galois group is the determining condition, and inert primes are those for which it is a generator.</p> http://mathoverflow.net/questions/28113/cyclotomic-fields-over-q-and-prime-ideals/28121#28121 Answer by Rob Harron for Cyclotomic Fields over Q and prime ideals Rob Harron 2010-06-14T14:02:31Z 2010-06-14T14:02:31Z <p>Theorem I.2.13 of Washington's book on cyclotomic fields says the following: $K$ is the $n$th cyclotomic field and $p\nmid n$, let $f$ be the smallest positive integer such that $p^f\equiv 1 (\mathrm{mod}~n)$. Then $p$ splits into $\phi(n)/f$ distinct primes in $K$.</p>