# Cyclotomic Fields over Q and prime ideals

Q 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. $k_{p^r}$=Q(x) where x is exp(2$\pi$i/$p^r$). [$k_{p^r}$:Q]=$p^{r-1}(p-1)$.

Question: Does q remain a prime in the integer ring of $k_{p^r}$?

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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 $K$.

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To match the notation here with the notation of the question, set $n := p^r$ and $p:=q$. Short version of answer: Yes iff $\phi(p^r)/f = 1$. –  S. Carnahan Jun 14 '10 at 15:28
Thanks, Scott. I had to run so I just quickly wrote up the statement. –  Rob Harron Jun 14 '10 at 17:42

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.

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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 inert) are those congruent to 3 mod 4.

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Ha. Beat you by one minute. –  Gerry Myerson Jun 14 '10 at 12:59
Yup, you did :) –  Alberto García-Raboso Jun 14 '10 at 13:11

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.

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Could you elaborate on the congruence condition? The Galois group for $k_{p^r}$/**Q** is a cyclic group. Do you mean the condition that q does not divide p-1? –  7-adic Jun 14 '10 at 13:06
The behaviour of a prime q depends only on the order of the cyclic subgroup it generates in that (multiplicative) group. The generators of the whole group mark out certain residue classes, and so the q are those that lie in certain arithmetic progressions. –  Charles Matthews Jun 14 '10 at 14:08

Yes if multiplicative order of $q$ modulo $p^r$ equals to $(p-1)p^{r-1}$.

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This seems to echo Rob Harron's answer (all the more since there is no '$n$' in the question; cf. what Scott Carnahan wrote in his comment). –  Todd Trimble Jan 12 '14 at 14:00
Thx for your comment. I have editted. –  Oguz Yayla Jan 17 '14 at 19:54