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Intersection of $ \Bbb{R}$ and $ \Bbb{Q}_p$

For distinct prime $p$ and $q$,intersection of $ \Bbb{Q}_p$ and $ \Bbb{Q}_q$ is just $ \Bbb{Q}$, but what about $ \Bbb{Q}_p$ and $ \Bbb{R}$ ?

$ \Bbb{R}$ is often said to be completion at infinity, so maybe the intersection should be $ \Bbb{Q}$ as the same as other $ \Bbb{Q}_q$, but how to prove formally this ?

Thank you.

Poitou-Tate
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    "For distinct prime $p$ and $q$, intersection of $\mathbf Q_p$ and $\mathbf Q_q$ is just $\mathbf Q$": oh, how do you prove that? (Note $x^2 - 3$ has two roots in $\mathbf Q_{11}$ and $\mathbf Q_{13}$ and no root in $\mathbf Q$. Does that have any significance?) Unless two fields already live inside a common larger field, what does their intersection even mean? – KCd May 04 '22 at 03:45
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    $\mathbb Q_{5}$ and $\mathbb Q_{13}$ both contain $\sqrt{-1},$ so what do you mean by the intersection of two entirely disjoint fields? It is true that $\mathbb Q_p$ contains a subfield isomorphic to $\mathbb Q,$ but it also contains many subfields isomorphic to $\mathbb Q(x).$ – Thomas Andrews May 04 '22 at 04:00
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    Something that may be more natural is the intersection of all the $\sigma(\Bbb{Q}_p)$ and $\rho(\Bbb{Q}_q)$ with $\sigma,\rho$ ranging over the embeddings into $\Bbb{C}$. – reuns May 04 '22 at 04:03
  • Oh, these comments are very new to me, thank you so much. Then, does there exists some real number except for rational number, which is also p adic number ? I want to know an example. – Poitou-Tate May 04 '22 at 07:37
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    @Iwasawa that's like asking "Does there exit some element of $S_n$, except the identity, that is also an element of $\mathrm{GL}_2(\mathbb R)$". The question just doesn't mean anything, until you somehow find a way to view $\mathbb R$ and $\mathbb Q_p$ as living inside some common object $X$. Only then does it make sense to ask if some element $x\in X$ is an element of $\mathbb R$ and $\mathbb Q_p$. – Mathmo123 May 04 '22 at 10:02
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    Ther is a $\sqrt 2$ in $\mathbb Q_7$ and in $\mathbb R,$ but they aren’t the “same” in any traditional sense of the word. Do you know what a real number is? A $p$-adic number? In fact, there is no real number which is a $p$-adic number and visa versa. There is no real number which is a rational number, technically, but there is a unique way to embed the rationals in the reals, so we often call the image of that embedding as rational numbers. – Thomas Andrews May 04 '22 at 14:04
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    Here is a related theorem using all real and $p$-adic fields: if $f(x) \in \mathbf Q[x]$ is irreducible and has a root in $\mathbf R$ and every $\mathbf Q_p$ then $f(x)$ is linear. This makes precise the idea that an algebraic number that (somehow) lives in $\mathbf R$ and every $\mathbf Q_p$ is rational. An abstract algebraic number $\alpha$ has a minimal polynomial $f(x)$ in $\mathbf Q[x]$, and $f(x)$ is irreducible in $\mathbf Q[x]$. If $\alpha$ were to make sense in all completions of $\mathbf Q$ then $f(x)$ should have a root in all these fields, so $\deg f = 1$: $\alpha$ is rational. – KCd May 05 '22 at 14:50

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