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The Question:

Let $K=\Bbb Q(\sqrt{-17})$ and $L=K(\sqrt{17})$. How to show that $\sqrt{-17}\mathcal O_K$ is unramified in $L$?

Context:

This is actually the last step I need to show that the Hilbert class field of $K=\Bbb Q(\sqrt{-17})$ is $K\left(\sqrt{\frac{1+\sqrt{17}}{2}}\right)$, following exercise 5.25 of Cox's "Primes of the form $x^2+ny^2$". For this, I need in particular to show that all primes are unramified in the extension.

The main tool used by Cox is the following fact:

Theorem: Let $L/K$ be a Galois extension with $L=K(\alpha)$ and $\alpha \in \mathcal O_L$, and let $f(x)$ be the minimal polynomial of $\alpha$ over $K$. Consider a prime ideal $\mathfrak p$ in $\mathcal O_K$. If $f$ is seperable mod $\mathfrak p$ (i.e., in $(\mathcal O_K/\mathfrak p)[x]$), then $\mathfrak p$ is unramified in $L$.

The strategy is to create a tower of quadratic extensions: $K \subset K(\sqrt{17}) \subset K\left(\sqrt{\frac{1+\sqrt{17}}{2}}\right)$, and we must show that both extensions are unramified. The second one I've solved, but for the first one, this is my progress:

Let $\mathfrak p$ be a prime in $O_K$. Then divide into cases: If $17\notin\mathfrak p$, write $K(\sqrt{17}) = K\left(\frac{-1+\sqrt{17}}{2}\right)$, so let $\alpha = \frac{-1+\sqrt{17}}{2}$ in the theorem. The minimal polynomial is $f(x) = x^2+x-4$ with discriminant $17\not\in\mathfrak p$, so $f$ is separable mod $\mathfrak p$, so $\mathfrak p$ is unramified.

But if $17\in\mathfrak p$, I'm stuck. Maybe there is just a clever choice of $\alpha$ that works again? But $\mathcal O_L$ has four degrees of freedom, and I got lost. I can proceed a bit concretely. Evidently, $17\mathcal O_K = (\sqrt{-17}\mathcal O_K)^2$, so we know that in fact $\mathfrak p = \sqrt{-17}\mathcal O_K$ (hence the main question). I did calculate in sage that actually $\sqrt{-17}\mathcal O_L$ factors into two principal ideals: $$ \sqrt{-17}\mathcal O_L = \left(\left(\frac 12 - \frac{3}{34} \sqrt{-17}\right)\sqrt{17} + \frac12\sqrt{-17} - \frac52)\right) \cdot \left(\left(\frac 12 + \frac{3}{34} \sqrt{-17}\right)\sqrt{17} - \frac12\sqrt{-17} - \frac52\right). $$ That looks.... complicated.

Milten
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1 Answers1

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Instead of $\alpha=\frac{-1+\sqrt{17}}{2}$ choose $\alpha'=\sqrt{-1}$ as a generator of $L/K$. We may do this, since we obviously have $K(\alpha)=K(\alpha')$.

So now we just need to check separability of the minimal polynomial $f'=X^2+1$ of $\alpha'$ over $\mathcal{O}_K/\sqrt{-17} \cong \mathbb{F}_{17}$, which is trivial since $X^2+1$ factors over that field as $(X+4)(X-4)$.

R.P.
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