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