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Mar 25, 2021 at 19:23 comment added YCor 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.)
Mar 25, 2021 at 17:42 comment added Wojowu @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.
Mar 25, 2021 at 17:23 comment added YCor The group of automorphisms of $\mathbf{C}$ acts transitively on transcendental elements so it's quite clear this degree is arbitrary.
Mar 25, 2021 at 17:19 comment added YCor 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.)
Mar 25, 2021 at 17:05 comment added Wojowu 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 $\mathbb Q$.
Mar 25, 2021 at 17:03 answer added Wojowu timeline score: 9
Mar 25, 2021 at 16:55 history asked Stanley Yao Xiao CC BY-SA 4.0