Sanity check: prime ideals in number fields Work in $\mathbb{Q}(\zeta_m)/\mathbb{Q}$. If $P$ is a prime ideal of norm $N(P)=p \hspace{1mm}\not \vert m$, does it follow that $p \equiv 1 \text{ mod }m$? I am sure this is not true, but am bad at generating counterexamples.
 A: There are no counterexamples, indeed if the norm of an ideal $P$ in $\mathbb  Q(\zeta_m)$ is $p$ and $p \not\vert m$ then $p \equiv 1 \pmod{m}$. 
Here is the argument, the hypothesis implies that $\zeta_m$ is congruent to an integer $\mod P$, otherwise not all the integers $k-\zeta_m (k=0,\dots,p-1)$ would be incongruent $\mod P$, and so $P$ would divide the difference of two of them, say $$P \vert (a-\zeta_m)-(b-\zeta_m)=a-b = k <p $$ 
so
    $$ p = N(P) \vert  k^{\varphi(m)} $$
but $(p, k)=1$, a contradiction. 
So $\zeta_m$ is congruent to an integer $\pmod{P}$ say $t$, then
    $$ \Phi_m(t) = \prod_k' (t-\zeta_m^k) \equiv 0 \pmod{P} $$
where $\Phi_m(x)$ is the $m$-th cyclotomic polynomial and the product is over all the integers $k$ coprime with $m$ and smaller than $m$. As the left hand side is an integer we have also
    $$ \Phi_m(t) \equiv 0 \pmod{p} $$
but $\Phi_m(x)$ is a divisor of the polynomial $x^m-1$, this implies that $t^m\equiv 1 \pmod{p}$ that is the order of $t$ is a divisor of $m$. We are going to show that it is exactly equal to $m$. 
Suppose otherwise that the order of $t$ is $n$, with $n\vert m$, $n<m$, we use the the fact that 
   $$ x^m-1 = \prod_{d\vert m} \Phi_d(x) = \Phi_m(x) \times (x^n-1) \times Q(x) $$
for some polynomial $Q$, but then $x^m-1$ has at least a double root at $t$ mod $p$, and this means that it's derivative is zero at $t$ say
  $$ m t^{m-1} \equiv 0 \pmod{p} $$
a contradiction. 
This means that the order of $t$ mod $p$ is $m$ and so $m \vert p-1$ ie
   $$ p \equiv 1 \pmod{m} $$ 
