# Every ideal of an algebraic number field can be principal in a suitable finite extension field

Let $K$ be an algebraic number field. Let $I$ be a non-zero ideal of the ring of integers $\mathcal{O}_K$ in $K$. By class field theory, there exists a finite extension(the Hilbert class field) $L$ of $K$ such that $I\mathcal{O}_L$ is principal. Can we prove this without using class field theory?

• I noticed that someone serially upvoted for my questions and answers including this one. While I appreciate them, I would like to point out that serial upvotes are automatically reversed by the system. – Makoto Kato Nov 27 '13 at 7:12

Okay, I don't know if you will be satisfied with this. But here is one way:

Let $I$ be a fractional ideal. Then by finiteness of ideal class group, there exists $m \in \mathbb{N}$ such that $I^m = (\alpha)$ for some $\alpha \in K^*$. Let $L = K(a^{1/m})$. I claim that $I\mathcal{O}_L$ is principal. In fact, I claim that $I\mathcal{O}_L = (\alpha^{1/m})$.

Well, $(I\mathcal{O}_L)^m = I^m\mathcal{O}_L = (\alpha)\mathcal{O}_L = (\alpha)$. Clearly, $(\alpha^{1/m})^m = (\alpha)$. Now it is easy to see that if $I, J$ are fractional ideals such that $I^m = J^m$, then $I = J$. Thus, $I\mathcal{O}_L = (\alpha^{1/m})$.

Here is a nice generalisation of the result you ask for. We claim that there is a finite extension of $$K$$ such that every ideal of $$\mathcal{O}_K$$ is principal in $$\mathcal{O}_L$$.

Proof: Suppose that $$|Cl_K| = n$$ and write $$[I_1],\ldots,[I_n]$$ for the elements of $$Cl_K$$. For each $$1 \leq k \leq n$$ choose a representative $$J_k \in [I_k]$$. Also for each $$k$$ there exists an integer $$m_k$$ and an element $$\alpha_k \in \mathcal{O}_K$$ so that $$J_k^{m_k} = (\alpha_k)$$. By Rankeya's answer above, the ideals $$J_1,\ldots,J_k$$ all become principal in the ring of integers of $$L = K(\sqrt[m_1]{\alpha_1},\ldots, \sqrt[m_n]{\alpha_n}).$$

It now remains to prove why for any ideal $$I\subset\mathcal{O}_K$$ with $$I \simeq J_1$$ say, $$I$$ also becomes principal in $$\mathcal{O}_L$$. If $$I \simeq J_1$$ then there is $$x,y \in \mathcal{O}_K$$ so that $$xI = yJ_1$$. Then $$\begin{eqnarray*} x^{m_1}I^{m_1} &=& y^{m_1}J_1^{m_1}\\ &=& (y^{m_1}\alpha_1) \end{eqnarray*}$$

and thus $$xI\mathcal{O}_L = y\sqrt[m_1]{\alpha_1}\mathcal{O}_L$$. Now there is a $$z \in I\mathcal{O}_L$$ so that $$xz = y\sqrt[m_1]{\alpha_1}$$. We claim that $$I= (z)$$. Clearly $$(z) \subseteq I\mathcal{O}_L$$. For the reverse inclusion take any $$w \in I$$. Then $$xw = y\sqrt[m_1]{\alpha_1}v = xzv$$ for some $$v \in \mathcal{O}_L$$. Since $$\mathcal{O}_L$$ is a domain this implies $$zv = w$$ and so $$I \subseteq (z)$$. Thus $$I$$ is principal in $$\mathcal{O}_L$$ which completes the problem.