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?

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, $I\mathcal{O}_L = (\alpha^{1/m})$.
