Let $K$ be a finitely generated field extension of $\mathbb{Q}$, and let $p$ be a prime number. Can $K$ must be embedded into a p-adic local field (i.e. a finite field extension of $\mathbb{Q}_p$) ?
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$\begingroup$ For finite-degree extensions, it is a consequence of Chebatorev density. I don't know about arbitrary finitely generated extensions. $\endgroup$– Kenta SuzukiCommented Oct 25, 2023 at 2:07
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$\begingroup$ @KentaSuzuki Do you mean when K is a number field? If so, then the completion of K along a valuation over p is the desired local field. $\endgroup$– lolipopCommented Oct 25, 2023 at 2:10
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2$\begingroup$ Ah yes, I agree, my previous comment proves there exists an embedding into $\mathbb Q_p$, but that is stronger than what you ask. $\endgroup$– Kenta SuzukiCommented Oct 25, 2023 at 2:13
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$\begingroup$ @KentaSuzuki Ok, thank you for remind me of the Chebatorev density theorem. $\endgroup$– lolipopCommented Oct 25, 2023 at 2:25
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3$\begingroup$ See this paper by Cassels: cambridge.org/core/services/aop-cambridge-core/content/view/…. And the addendum cambridge.org/core/journals/…. Cassels wrote about this in his book Local fields. $\endgroup$– KConradCommented Oct 25, 2023 at 5:23
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2 Answers
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As YCor indicates, the question in the title is different from the one in the body. Are we assuming the prime $p$ is fixed or not? In both cases the answer is "Yes", but if $p$ is not fixed, there are even stronger results. Below $K$ is always a finitely generated field extension $K$ of $\mathbb{Q}$.
Theorem (Cassels): Given any finite subset $0\neq C\subset K$, there are infinitely many primes $p$ such $K$ embeds in $\mathbb{Q}_p$ and the image of $C$ is in $\mathbb{Z}_p^*$.
See the remark by KConrad (which I was not aware of before, thanks!).
Theorem (Breuillard-Gelander, following Tits): For every infinite subset $C\subset K$ there exists a local field $k$ and an embedding $K$ into $k$ such that the image of $C$ is unbounded.
This is Lemma 2.1 in "A topological Tits alternative". It extends the Tits' lemma cited by YCor in his answer.
I am writing this answer to complete the details in the case where the prime $p$ is fixed. In fact, instead of looking at the extension problem from $\mathbb{Q} \to \mathbb{Q}_p$ and $\mathbb{Q}\to K$ to $K\to k$ for some finite extension $\mathbb{Q}_p\to k$, let us replace $\mathbb{Q}$ by any countable field $K_0$ and $\mathbb{Q}_p$ by any local field $k_0$. Then, by induction, we may assume that $K$ is generated over $K_0$ by a single element $t$. Solving the corresponding extension problem now amounts to choosing a finite field extension $k_0\to k$ and fixing an appropriate element in $k$ as the image of $t$. If $t$ is transcendental, take $k=k_0$ and pick any element in it which is transcendental over the image of $K_0$ (which exists by cardinality considerations) as the image of $t$. If $t$ is algebraic, with minimal polynomial $f$, set $k=k_0[x]/(f)$ and send $t$ to the image of $x$ in it. Thus, indeed, the answer to the problem in the question's body is "Yes".
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Lemma 4.1 in: Jacques Tits, Free subgroups in linear groups, J. of Algebra, 20, (1972). DOI link
Let $k$ be a finitely generated field and let $t\in k^*$ be an element of infinite order. Then, there exists a locally compact field $k’$ endowed with an absolute value $\omega$ and a homomorphism $\sigma:k\to k^*$ such that $\omega(\sigma(t))\neq 1$.
This does not answer the question (the one in the body, since it is distinct from the one in the title — the one in the title does not fix $p$).
But the proof is by reduction to the number field case (which you claim to know). Namely, in the present case let $k_0$ be the algebraic closure of $\mathbf{Q}$ in $k$. Then the proof basically consists in extending a $p$-adic embedding from $k_0$ to $k$.