# Embedding $\mathbb Q^c$ into $\mathbb Q^c_p$

Let $p$ be a prime number and $\mathbb Q_p$ the $p$-adic completion of $\mathbb Q$. Let $\mathbb Q^c$ be the algebraic closure of $\mathbb Q$ in $\mathbb C$. Is there an embedding $$j: \mathbb Q^c \hookrightarrow \mathbb Q_p^c$$ where $\mathbb Q^c_p$ is the algebraic closure of $\mathbb Q_p$?

• There is not "the" algebraic closure of $\mathbb{Q}_p$ (or in fact of any given field) and there won't be a "natural" (what does this mean?) embedding. – Martin Brandenburg Dec 22 '13 at 21:00
• The obstacle to 'naturality' in this context is that there is arbitrariness inherent in choosing which root of a polynomial in the target field to send a root of it in the source field. This phenomenon is broader than just $\overline{\Bbb Q}\hookrightarrow\overline{\Bbb Q_p}$. The whole idea of Galois symmetry is based on the observation that conjugate roots are indistinguishable from each other purely abstractly and algebraically. – anon Dec 22 '13 at 21:00
• I'm still considering $\mathbb Q^c$ as the algebraic closure of $\mathbb Q$ in $\mathbb C$ but I've deleted the word 'natural'. Is my question still not well posed? – Josh F. Dec 22 '13 at 21:13

## 1 Answer

Well yes. More generally, if $E \subseteq F$ are fields, then the algebraic closure of $F$ contains a copy of the algebraic closure of $E$. Specifically, let $F^c$ be the algebraic closure of $F$, and let $E'$ be the subfield of $F^c$ consisting of all roots of all polynomials with coefficients in $E$. Then $E'$ is algebraically closed, and is isomorphic to the algebraic closure of $E$.

Your question initially contained the word "natural". This word is a bit dicey — the subfield $E'$ is determined in an entirely natural way, but if you have a specific algebraic closure $E^c$ for $E$ in mind, there won't be a natural isomorphism $E^c \to E'$. This is because, in general, there isn't a natural isomorphism between two different algebraic closures of the same field.

In particular, there is a copy of $\mathbb{Q}^c$ sitting inside of the complex numbers $\mathbb{C}$, and there is also a copy of $\mathbb{Q}^c$ sitting inside of $\mathbb{Q}_p^c$. These copies are isomorphic, but there isn't a single "best" isomorphism between them. For example, $\mathbb{Q}_p^c$ will always contain two different square roots of $2$, but there's not an obvious way to label one of these as the "positive" square root and the other as "negative" square root.

• So you’ll use Axiom of Choice to get your map of the one algebraically closed field into the other. So it looks as if the task can’t be completed without AC. – Lubin Dec 22 '13 at 21:47
• @Lubin Sure, but the usual construction of the algebraic closure of $\mathbb{Q}_p$ uses Zorn's lemma, so we're already in AC territory. See this MathOverflow question for a discussion of the role of AC in the construction of algebraic closures. – Jim Belk Dec 22 '13 at 21:53