Do there exist non-rational algebraic numbers that belong to $\mathbb Q_p$ for all prime $p$? If yes, can one characterize them? I spent several days for the first question, and I found nothing. The second one looks even more diffuclt and surely out of my skills.
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10$\begingroup$ See here $\endgroup$– WojowuCommented Nov 11, 2021 at 22:17
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4$\begingroup$ For some $n$ and some $p$, there may be a solution of $X^3=n$ in $\mathbb Q_p$. But there is no way to say whether a solution on $\mathbb Q_p$ and a solution in $\mathbb Q_q$ are the same or not. So you need to explain how you identify non-rationals in one $\mathbb Q_p$ with those in another. $\endgroup$– Gerald EdgarCommented Nov 11, 2021 at 22:24
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3$\begingroup$ One meaningful question of the same spirit: is there an irreducible polynomial of degree $\ge 2$ in $\mathbf{Q}[t]$ that has a root in $\mathbf{Q}_p$ for every $p$? there's none of degree $2$. $\endgroup$– YCorCommented Nov 11, 2021 at 22:38
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7$\begingroup$ @Wojowu thanks. So, equivalently, the only algebraic extension of $\mathbf{Q}$ that embeds in $\mathbf{Q}_p$ for all primes $p$ [or even for a density 1 set of primes] is $\mathbf{Q}$ itself. $\endgroup$– YCorCommented Nov 11, 2021 at 22:56
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4$\begingroup$ Please let us know if the answer Wojowu linked to doesn't answer your question. $\endgroup$– Will SawinCommented Nov 11, 2021 at 23:15
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1 Answer
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(cw answer based on Wojowu's link)
The only algebraic extension of $\mathbf{Q}$ that embeds into $\mathbf{Q}_p$ for all $p$ (or even for a density 1 set of primes) is $\mathbf{Q}$ itself.
For if $P\in\mathbf{Q}[t]$ is an irreducible polynomial of degree $\ge 2$, the set of primes $p$ for which $P$ has no root in $\mathbf{Q}_p$ has positive density: see this answer, which also applies to an arbitrary number field (=finite extension of $\mathbf{Q}$), where primes now mean primes of the number field.
The above statement roughly states that "the only algebraic number that belong to the intersection of all $\mathbf{Q}_p$ are rationals", which seems to answer your question. As already mentioned in comments, this statement doesn't really make sense, since there is no natural way to identify algebraic elements in different completions. An artificial way for it to makes sense is to mod out $\mathbf{Q}_p$ by the equivalence relation "having the same minimal polynomial (monic or zero) over $\mathbf{Q}$".