By factoring each of \(\mathfrak {a},\mathfrak {b}\) into prime ideals, it suffices to prove that \(N(\mathfrak {p}\mathfrak {b})=N(\mathfrak {p})N(\mathfrak {b})\) where \(\mathfrak {p}\) is a prime ideal and \(\mathfrak {b}\) is any proper ideal.

Now, by group theory \(\mathfrak {b}/\mathfrak {p}\mathfrak {b}\) is a subgroup of \(\mathcal{O}_K/\mathfrak {p}\mathfrak {b}\) and the quotient is \(\mathcal{O}_K/\mathfrak {b}\). This means

So we need to show \(|\mathfrak {b}/\mathfrak {p}\mathfrak {b}|=|\mathcal{O}_K/\mathfrak {p}|\).

Since \(\mathfrak {p}\neq \mathcal{O}_K\) and (fractional) ideals form a group we have \(\mathfrak {p}\mathfrak {b}\neq \mathfrak {b}\) (otherwise we could multiply through by \(\mathfrak {b}^{-1}\) giving a contradiction), so let \(x \in \mathfrak {b}\backslash \mathfrak {b}\mathfrak {p}\). Let us consider the map

given by \(a \mapsto ax + \mathfrak {p}\mathfrak {b}\). The kernel of this map contains \(\mathfrak {p}\) and the map is non-zero by our choice of \(x\), so the kernel is an ideal containing \(\mathfrak {p}\). But \(\mathfrak {p}\) is maximal, so the kernel must be \(\mathfrak {p}\). Therefore we have an injective map

Now, to see that this map is also surjective we just note that by unique factorization, there cannot be any ideals strictly between \(\mathfrak {b}\) and \(\mathfrak {p}\mathfrak {b}\). Since otherwise, we would have \(\mathfrak {b}\mathfrak {p}\subset \mathfrak {c}\subset \mathfrak {b}\) which means \(\mathfrak {b}\mid \mathfrak {c}\) and \(\mathfrak {b}\mid \mathfrak {b}\mathfrak {p}\) so by uniqueness of factorization we would have either \(\mathfrak {c}=\mathfrak {b}\) or \(\mathfrak {c}=\mathfrak {p}\mathfrak {b}\).

Therefore \(\mathfrak {b}=(x)+\mathfrak {p}\mathfrak {b}\) giving surjectivity.

Clearly, if \(\mathfrak {a}=\mathfrak {b}\) they have the same norm. So lets check the other direction.

First note that since \(\mathfrak {a}\subset \mathfrak {b}\subset \mathcal{O}_K\), the tower law (which works for groups) gives us that \([\mathcal{O}_K:\mathfrak {a}]=[\mathcal{O}_K:\mathfrak {b}][\mathfrak {b}:\mathfrak {a}]\). But \([\mathcal{O}_K:\mathfrak {a}]=N(\mathfrak {a})=N(\mathfrak {b})=[\mathcal{O}_K:\mathfrak {b}]\) therefore \([\mathfrak {b}:\mathfrak {a}]=1\) giving the result.

If \(\mathfrak {a}\) is not prime then \(\mathfrak {a}=\mathfrak {p}\mathfrak {b}\) with \(\mathfrak {p}\) a prime ideal and \(\mathfrak {b}\neq (1)\). Then by Proposition 3.4.2 we have

which means \(N(\mathfrak {a})\) could not have been prime, which is a contradiction.

For the converse we note that since \(\mathfrak {p}\) is prime it is therefore maximal. Therefore \(\mathcal{O}_K/\mathfrak {p}\) is a field, which we know is finite. Moreover, we know every finite field has size \(p^f\) for some prime \(p\) and natural number \(f\), which gives the result.

Let \(e_1,\dots ,e_n\) be an integral basis of \(\mathcal{O}_K\), then clearly \(\alpha e_1,\dots ,\alpha e_n\) is an integral basis of \((\alpha )\). On the other hand, since \((\alpha ) \subset \mathcal{O}_K \cong \mathbb {Z}^n\) we can make sure to choose the \(e_i\) such that there exist \(m_i \in \mathbb {Z}\) such that \(m_1e_1,\dots ,m_ne_n\) is an integral basis of \((\alpha )\). Alternatively, this follows from the structure theorem for finitely generated abelian groups. Then

and therefore \(N((\alpha ))=\prod _i |m_i|\).

Next we need to relate this number to \(N_{K/\mathbb {Q}}(\alpha )\) in some way. For this, let us compare the three bases for \(K\) over \(\mathbb {Q}\) that we have written down: \(\{ e_1,\dots ,\epsilon _n\} \), \(\{ m_1e_1,\dots ,m_ne_n\} \) and \(\{ \alpha e_1,\dots ,\alpha e_n\} \).

Now, lets see look at the diagram summarising the associated change of basis matrices.

The top arrow corresponds to the action of \(A_\alpha \). The arrow on the left corresponds to the identity matrix \(I\). The arrow along the bottom corresponds to a change of basis matrix with is diagonal which diagonal entries \(m_i\), lets call it \(D\). Lastly, whatever the matrix corresponding to the arrow going upwards on the right is, it represents a change of basis matrix between two integral bases for \((\alpha )\), therefore similarly to what we have seen before, it must correspond to an invertible matrix with integer coefficients, which we will denote by \(N\). Note that this also means, \(\det (N)=\pm 1\).

Now, from the diagram it is clear that going along the top of the diagram is the same as first going down, then right and then up. This means this is a commutative diagram and therefore we have \(A_\alpha =NDI\) therefore

Taking absolute values now gives the result.

Since \(\alpha \in \mathfrak {a}\) we have \((\alpha ) \subset \mathfrak {a}\). Then by Lemma 3.4.3 and Proposition 3.4.5 gives the result.

Recall that by Lagrange’s Theorem, every element in \(\mathcal{O}_K/\mathfrak {a}\) has order at most \(a\). Therefore, if we look at the element \(1+\mathfrak {a}\in \mathcal{O}_K/\mathfrak {a}\) we have \(a\cdot (1+\mathfrak {a})=0 +\mathfrak {a}\) therefore \(a \in \mathfrak {a}\).

If \(\mathfrak {a}\) is an ideal with \(N(\mathfrak {a})=a\) then by Lemma 3.4.8 we have \(a \in \mathfrak {a}\) which means \(\mathfrak {a}\mid (a)\). But by uniqueness of factorization, \(a\) only has finitely many factors, which gives the result.