Unramified p-adic field extension

Let $\beta$ be an algebraic integer and $p$ be a prime number. Let $\mathbb{Q}_p(\beta)$ be an unramified extension of degree $2^{n}$ over $\mathbb{Q}_p$. Does it imply that $\mathbb{Q}(\beta)$ is an extension of degree $2^n$ over $\mathbb{Q}$ ?

Solution 1:

As in @mathmo123's example: high-degree algebraic extensions of $\mathbb Q$ can (and do) "collapse" quite a lot as extensions of $\mathbb Q_p$.

For example, for every finite extension $k$ of $\mathbb Q$ there are infinitely-many primes $p$ in $\mathbb Z$ which split completely (or else the zeta function of $k$ would not have a pole at $s=1$). For any one of these, there is a corresponding imbedding $k\to\mathbb Q_p$.

EDIT: And, yes, more restricted forms of the question still have affirmative answers. For example, for fixed prime $p$, to make degree $p^n$ extensions $k$ of $\mathbb Q$ that collapse to $\mathbb Q_p$, it suffices to find a monic polynomial of degree $p^n$ in $\mathbb Z[x]$ which factors into distinct linear factors mod a sufficiently large power of $p$, but is irreducible mod $q$ for some auxiliary prime $q$. The irreducibility mod $q$ assures that adjoining a root to $\mathbb Q$ gives a degree $p$ extension, and the complete splitting mod $p$ assures that the prime $p$ splits completely in the extension, so that all the local-field extensions are trivial.

For example, with $n=1$, we could require $f$ congruent mod $p$ to $x^p-x$ and mod $q$ to $x^p-q$. This requirement is easy to meet.

Certainly with $p^n$ and $n>1$, the condition about distinct roots has to be mod $p^N$ with $N>1$, just on counting considerations.