Properties of Norm of Prime Ideals

Theorem

Let \(I \trianglelefteq \mathcal{O}_\mathbb{K}\) be a prime ideal of a number ring. Then \(N(I) = p^m\) for some prime number \(p\) and \(m \leq [\mathbb{K}:\mathbb{Q}]\), and such a \(p\) is the only prime in \(I\).

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 \(I \mid \langle N(I) \rangle\). 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

\[ I \mid \langle N(I) \rangle = \langle p_1 \rangle^{e_1} \dots \langle p_r\rangle^{e_r}.\]

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

\[\begin{align*} p, q \in I &\implies \langle p, q \rangle \subseteq I \\ &\implies \langle \gcd(p, q) \rangle \subseteq I \\ &\implies \langle 1 \rangle \subseteq I \\ &\implies 1 \in I \\ &\implies I = \mathcal{O}_\mathbb{K} \\ \end{align*}\]

however this contradicts the fact that \(I\) is prime by definition.

Theorem

Let \(I \trianglelefteq \mathcal{O}_\mathbb{K}\). If \(N(I)\) is a prime number then \(I\) is prime.

Proof

For any ideal \(I\) we have a unique factorisation into prime ideals

\[ I = \mathfrak{p}_1^{e_1} \dots \mathfrak{p}_r^{e_r}\]

where \(e_i > 0\).

By multiplicativity of the norm we can deduce that

\[ N(I) = N(\mathfrak{p}_1)^{e_1} \dots N(\mathfrak{p}_r)^{e_r}.\]

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 \(j \neq i\). This would mean

\[ N(I) = N(\mathfrak{p}_i).\]

However

\[ N(\mathfrak{p}_j) = 1 \implies |\mathcal{O}_\mathbb{K}/\mathfrak{p}_j| = 1 \implies \mathfrak{p}_j = \mathcal{O}_\mathbb{K}\]

and this contradicts the fact that \(p_j\) is prime.

Therefore we have that there are no such \(j\), and in particular

\[ I = \mathfrak{p}_1\]

which is prime by definition.