Does the algebraic closure of the field $\mathbb{Q}_p$ of $p$-adic numbers exist in ZF without the axiom of choice? If so, how to prove it?
I know that algebraic closures of countable fields can be constructed without the axiom of choice and that it is the uncountable case where choice-free proofs could be a problem. But what about the specific example of $\mathbb{Q}_p$?
The typical construction of $\mathbb{C}_p$ goes through the following steps:
You are probably worried about the first step, but there's actually also a subtle issue in the third step, which is that the standard definition of the completion of a metric space (i.e., the collection of Cauchy sequences modulo distance $0$) isn't actually 'correct' in situations with very little choice. Specifically, there can be metric spaces $X$ and $Y$ with $X$ dense in $Y$ such that not every element of $Y$ is the limit of a sequence of elements of $X$. So with that in mind, we'll say that a metric space $X$ is complete if the image of any isometric embedding of $X$ into another metric space is closed. $\mathsf{ZF}$ proves that every metric spaces has a unique completion in the typical sense. (There is a way to explicitly construct this completion in terms of 'Cauchy filters' instead of Cauchy sequences.)
To actually answer your question, $\mathsf{ZF}$ does prove that for each prime $p$, there is a valued field $\mathbb{C}_p$ extending $\mathbb{Q}_p$ with the properties that
(I don't know whether we can prove that this $\mathbb{C}_p$ is unique in any way.)
Here is a sketch of a terrible proof of this:
Fix a model $V$ of $\mathsf{ZF}$. Let $\mathbb{Q}_p^V$ be $V$'s copy of the $p$-adic numbers. Every model of $\mathsf{ZF}$ has an inner model $L$ of $\mathsf{ZFC}$. Let $\mathbb{Q}_p^L$ be $L$'s copy of the $p$-adic numbers. Since the integers are dense in $\mathbb{Q}_p$, there is a canonical way of thinking of $\mathbb{Q}_p^L$ as a sub-field of $\mathbb{Q}_p^V$. Perform the normal algebraic closure argument inside $L$ to get $\overline{\mathbb{Q}}_p^L$. Let $\mathbb{C}_p^V$ be the metric completion of $\overline{\mathbb{Q}}_p^L$ computed in $V$. Argue that $\mathbb{C}_p^V$ has the required properties: Note that $\mathbb{Q}_p^V$ can be identified with a certain subset of $\mathbb{C}_p^V$. Let $F$ be a proper sub-field of $\mathbb{C}_p^V$ containing $\mathbb{Q}_p^V$ (and therefore also $\mathbb{Q}_p^L$). If $F$ is dense in $\mathbb{C}_p^V$, then $F$ is not metrically complete. If $F$ is not dense in $\mathbb{C}_p^V$, then there is an element $a$ of $\overline{\mathbb{Q}}_p^L$ and a rational $\varepsilon >0$ such that the ball $B_{\leq \varepsilon}(a)$ is disjoint from $F$, but this implies that $F$ is not algebraically closed. Finally we need to check that $\mathbb{C}_p^V$ is algebraically closed. This follows from Hensel's lemma and approximating elements of $\mathbb{Q}_p^V$ with elements of $\mathbb{Q}_p^L$. (Technically we also need to check that $\mathbb{C}_p^V$ has a sensible field structure extending $\overline{\mathbb{Q}}_p^L$, but this is pretty easy.) $\square$
A more honest proof of this would be to build the algebraic closure of some countable dense sub-field of $\mathbb{Q}_p$, such as just the rationals, and then extend the valuation and take the metric completion.
A more difficult question is whether you can always form a spherically complete, algebraically closed field extension of $\mathbb{Q}_p$, since such fields (sometimes denoted $\Omega_p$) are usually constructed using an ultraproduct.
Here is a more "algebraic" answer. Start with the ring of algebraic integers $R=\mathcal O_\overline{\mathbb Q}$. This ring is countable, so we can use the usual constructions to find a maximal ideal $P$ in $R$ which contains $p$. We can use this $P$ to define an extension $v_P$ of the $p$-adic valuation to all of $\overline{\mathbb Q}$ - for a finite Galois extension $K/\mathbb Q$, $P_K:=P\cap K$ is a prime ideal in a Dedekind domain $R\cap K$ so induces a $P_K$-adic valuation $v_{P_K}$ on $K$, and for any $a\in K$ we can define $v_P(a)=v_{P_K}(a)/v_{P_K}(p)$. Standard results in algebraic number theory give that this is independent of $K$ and so gives us the desired function on $\overline{\mathbb Q}$.
We can then use $v$ to define a norm via $|a|_P=p^{-v_P(a)}$, and a metric via $d(a,b)=|a-b|_P$. We can now form the metric completion $C$ in the usual way. The usual argument using Krasner's lemma (which goes through just fine in ZF) implies that $C$ is algebraically closed. Since $\mathbb Q$ embeds into $\overline{\mathbb Q}$, this extends to an embedding $\mathbb Q_p\hookrightarrow C$. The algebraic closure of $\mathbb Q_p$ is clearly dense, so $C$ is also its metric completion. The resulting field $C$ can thus most reasonably be called $\mathbb C_p$, for instance it satisfies the properties James Hanson specifies in their answer.
I cannot answer the parenthetical question they posed in their answer about those properties guaranteeing any sort of uniqueness, but let me briefly mention that while my construction depends on the choice of the ideal $P$, the resulting choices are (noncanonically) isomorphic - any two choices $P,P'$ are related by an automorphism of $\overline{\mathbb Q}$ (this is true for finite Galois extensions by a standard argument, and since $\overline{\mathbb Q}$ is countable we can combine those in ZF) and this automorphism will take $v_P,|\cdot|_P$ to $v_{P'},|\cdot|_{P'}$ and induces isomorphisms between resulting $C$'s.
