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Let $p$ be a prime number, $\mathbb{Q}_p$ the $p$-adic number field. We fix an algebraic closure $\Omega$ of $\mathbb{Q}_p$. Any algebraic extension of $\mathbb{Q}_p$ is assumed to be a subfield of $\Omega$. Let $n$ be a positive rational integer.

Is the number of finite extensions of $\mathbb{Q}_p$ of degree $n$ finite? If yes, is there an algorithm to construct all of them?

The motivation is as follows. Let $p$ be an odd prime number. I came up with the following result using Hensel's lemma.

The number of quadratic extensions of $\mathbb{Q}_p$ is $3$. They are $\mathbb{Q}_p(\sqrt a)$, $\mathbb{Q}_p(\sqrt{ap})$, $\mathbb{Q}_p(\sqrt p)$, where $a$ is a quadratic non-residue rational integer mod $p$. $\mathbb{Q}_p(\sqrt a)$ (resp. $\mathbb{Q}_p(\sqrt{ap})$) does not depend on the choice of $a$.

Makoto Kato
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1 Answers1

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Maybe the best way to count the quadratic extensions of any $p$-adic field $K$ is to use Kummer theory, which says that each such extension is described by a nontrivial element of $K^*/{K^*}^2$. In general, Kummer theory lets you count the cyclic extensions of $K$ degree $n$, as long as the characteristic doesn’t divide $n$, and the $n$-th roots of unity are in $K$. So you can count the cyclic cubic extensions of $\Bbb Q_p$ when $p\equiv1\pmod6$.

Lubin
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