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?

  • $\begingroup$ 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. $\endgroup$ – 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.


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