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Let $K$ be a finite extension of the $2$-adic numbers $\mathbb{Q}_2$, and suppose that $-1$ is not square in $K$. Write $K(i)$ for the quadratic extension $K[X]/(X^2 + 1)$, where $i^2 = -1$. Is it true that $K(i)/K$ is always a totally ramified extension? If so, how do you prove it?

My attempt

My first idea was to find an Eisenstein polynomial defining the extension. The element $i$ has minimal polynomial $f(X) = X^2 + 1$ over $K$, so I wanted to find $a \in K$ such that $f(X + a) = x^2 + 2aX + a^2 + 1$ is Eisenstein. This is equivalent to having $v_K(a^2 + 1) = 1$. The existence of such an $a$ is equivalent to there being some unit $u \in \mathcal{O}_K^\times$ such that $a^2 = u\pi_K - 1$, so we need $u\pi_K - 1$ to be square for some unit $u$. I am unsure how to show that such a $u$ always exists.

Another idea is to consider the unique unramified quadratic extension of $K$, and show that it does not contain a square root of $-1$. I know that an unramified extension is always obtained by adjoining an $m^\mathrm{th}$ root of unity for $(m,p) =1 $, but I don't know how to show that $-1$ won't be square in that extension.

Sebastian Monnet
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2 Answers2

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No, try $K=\Bbb{Q}_2(\sqrt{3})$ whose residue field is $\Bbb{F}_2$ as $(x+1)^2-3$ is Eisenstein.

$K(i)=K(\zeta_3)$ so $K(i)/K$ is unramified.

reuns
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Let $L=\mathbb{Q}_2(\mu_{252})$. The action on roots of unity produces an isomorphism $t: G:=Gal(L/\mathbb{Q}_2) \rightarrow Aut_{\mathbb{Q}_2}(\mu_{4}) \times Aut_{\mathbb{Q}_2}(\mu_{63}) \cong \mathbb{Z}/(2) \times \mathbb{Z}/(6)$.

In particular, let $K=L^{t^{-1}(1,3)}$. Then $K \subset L$ is quadratic, $i,\mu_{63} \subset L$ but $\mu_{63} \not\subset K$ and $i \notin K$. Thus $K(i)=K(\mu_{63})$ is unramified over $K$.

Aphelli
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  • I would think this (good!) construction works with $63$ replaced by any $2^{2m}-1$, and the case $m=1$ is the one exhibited in reuns' answer. I wonder why you chose $m=3$. – Torsten Schoeneberg Jul 13 '23 at 14:27