Properties of Norm of Prime Ideals

The above tells you that the norm of prime ideals is a power of a prime number, so an interesting question is does the converse hold. The answer is no, but we can recover a weak converse.

Proof

First note that $N(I)\in I$, and therefore $I\mid ⟨N(I)⟩$. Now, considering the factorisation of $N(I)$ in the integers given by $N(I)=p_1^{e_1}\dots p_r^{e_r}$ we have that

This implies that $p_1,\dots ,p_r\in I$. We now will show that $r=1$, and hence only one prime can be in $I$. Suppose there are two distinct primes $p,q\in I$. Then

however this contradicts the fact that $I$ is prime by definition.

Proof

For any ideal $I$ we have a unique factorisation into prime ideals

where $e_i>0$.

By multiplicativity of the norm we can deduce that

Given that $N(I)$ is a prime number, the only way that $N(I)$ can be equal to such a product is if $e_i=1$ for some $i\in \{1,\dots ,r\}$ and $N({\mathfrak{p}}_j)=1$ for each $j\ne i$. This would mean

However

and this contradicts the fact that $p_j$ is prime.

Therefore we have that there are no such $j$, and in particular

which is prime by definition.