Is the intersection of all p-adic fields equal to Q? Fix an algebraic closure $\overline{\mathbb Q}$ of the field $\mathbb Q$ of rational numbers. For a prime $p$ let $K_p$ the field of all algebraic elements in ${\mathbb Q}_p$.


*

*Question: Is $K_p$ normal over $\mathbb Q$?


If so, it defines a unique subfield of $\overline{\mathbb Q}$ which we might want to write as ${\mathbb Q}_p\cap \overline{\mathbb Q}$. Then the second question is:


*Is $\bigcap_p{\mathbb Q}_p\cap \overline{\mathbb Q}=\mathbb Q?$

 A: *

*No:  $x^3-2$ factors linear $\times$ quadratic over ${\mathbb Q}_5$.

*Yes: any algebraic extension of ${\mathbb Q}$ is ramified at some prime.
A: *

*No: the field $\mathbb{Q}_5$ has a cube root of $2$, but does not contain a square root of $-3$. (It's easy to form such examples for all primes.)

*An alternate version of the question is as follows: Given an algebraic number $\alpha$ such that there is an inclusion $\mathbb{Q}(\alpha) \rightarrow \mathbb{Q}_p$ for all $p$, is $\alpha$ necessarily rational? The answer is yes. Let $f(x) \in \mathbb{Q}[x]$ be an irreducible polynomial with $\alpha$ as a root, and assume the degree of $f(x)$ is $> 1$. Let $G$ be the Galois group of the splitting field $F$ of $f(x)$. The group $G$ acts transitively on the roots of $f(x)$. By a theorem of Jordan, there exists an element $\sigma \in G$ which has no fixed points. If $p$ is a prime such that the corresponding Frobenius element of $p$ in $F$ is (the conjugacy class of) $\sigma$, then there will be no inclusion $\mathbb{Q}(\alpha) \rightarrow \mathbb{Q}_p$. 
A replacement concept would be to take the elements in $\overline{\mathbb{Q}}$ which land in $\mathbb{Q}_p$ for any embedding of $\overline{\mathbb{Q}} \rightarrow \overline{\mathbb{Q}}_p$. The corresponding elements $K_p$ do form a field which is Galois over $\mathbb{Q}$. 
