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It is known that if $\mathbb{F}, \mathbb{G}$ are two algebraically closed fields with characteristic zero and equal cardinality, then $\mathbb{F}$ and $\mathbb{G}$ are isomorphic as fields. If $\mathbb{F}, \mathbb{G}$ are topological fields, that is, fields with an inherent notion of topology, then there is no reason to expect that the fields isomorphism (which is usually highly non-canonical) will respect the topologies.

Now fix a prime $p$. Then the above theorem implies that $\overline{\mathbb{Q}_p}$ and $\mathbb{C}$ are isomorphic. Let $\sigma : \mathbb{C} \rightarrow \overline{\mathbb{Q}_p}$ be such an isomorphism. Then it is easy to see that $\sigma(q) = q$ for every $q \in \mathbb{Q}$, and hence $\sigma$ must fix every polynomial with rational coefficients and therefore $\sigma(\overline{\mathbb{Q}}) = \overline{\mathbb{Q}}$.

Now suppose that $\tau \in \mathbb{C}$ is transcendental. Then $\sigma(\tau) \in \overline{\mathbb{Q}_p}$. Hence, it must lie in some finite extension of $\mathbb{Q}_p$ of minimal degree, say $K$. We see that the degree of this extension $K$ over $\mathbb{Q}_p$ is equal to the degree of $\sigma(\tau)$ over $\mathbb{Q}_p$.

Is this degree well-defined? That is, does this degree depend on the choice of the isomorphism, or only on the prime $p$?

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    $\begingroup$ The first statement in your question is incorrect - the algebraic closures of $\mathbb Q$ and $\mathbb Q(t)$ are equipotent but not isomorphic. This is only true if you require fields to be uncountable, or make a more precise assumption that the two fields have equal transcendence degree over $\mathbb Q$. $\endgroup$
    – Wojowu
    Commented Mar 25, 2021 at 17:05
  • $\begingroup$ What do you denote by $\overline{\mathbf{Q}_p}$? the algebraic closure or its norm completion? (From the sequel: you seem to mean the former— using such bars for algebraic closures in a norm context is not optimal.) $\endgroup$
    – YCor
    Commented Mar 25, 2021 at 17:19
  • $\begingroup$ The group of automorphisms of $\mathbf{C}$ acts transitively on transcendental elements so it's quite clear this degree is arbitrary. $\endgroup$
    – YCor
    Commented Mar 25, 2021 at 17:23
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    $\begingroup$ @YCor Norm completion of a field $F$ is usually denoted by $\widehat F$, not $\overline F$. Given that $\mathbb Q_p$ is already complete I don't see why you would want to complete it further. $\endgroup$
    – Wojowu
    Commented Mar 25, 2021 at 17:42
  • $\begingroup$ Actually both the algebraic closure and its norm completion are isomorphic as abstract fields to the complex numbers. The sentence in the OP's post suggests that the degree takes finite values, i.e., that the OP considers the algebraic closure, which is dense in $\mathbf{C}_p$. This doesn't affect much the conclusion anyway. (I'm not familiar to usual conventions in that particular area of math, but I'm puzzled that "overline" denotes an algebraic closure in the norm setting.) $\endgroup$
    – YCor
    Commented Mar 25, 2021 at 19:23

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(Assume AC throughout) For any transcendental $\tau\in\mathbb C$ and any $\tau'\in\overline{\mathbb Q_p}$ transcendental over $\mathbb Q$ there is an isomorphism $\sigma$ satisfying $\sigma(\tau)=\tau'$. To see this, include $\tau,\tau'$ in the trascendence bases of respective fields over $\mathbb Q$, and noting they both have cardinality continuum, extend any bijection between them to a field isomorphism of algebraic closures.

Since transcendentals in $\overline{\mathbb Q_p}$ can have arbitrary degree over $\mathbb Q_p$, the answer to your question is no in the strongest possible sense - for any $\tau\in\mathbb C$, the degree of $\sigma(\tau)$ over $\mathbb Q_p$ can be completely arbitrary.

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