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I am working on a model theory problem that has very little to do with field theory.

I basically want to know if and why every algebraically closed field of characteristic 0 is uncountable. I cannot think of a reason why this is true, but given the problem I am working on, I suspect it must be. I know many abstract algebra theorems assume uncountability. I also know that the cardinality of the algebraic closure of a field F is $max\{\aleph_0, |F|\}$. Is it true that algebraically closed fields of characteristic 0 are uncountable?

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    $\begingroup$ No, the field of algebraic numbers is countable. $\endgroup$
    – Angina Seng
    Commented Nov 10, 2019 at 9:03

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$\mathbb Q$ has characteristic $0$ and is countable by a famous spiral argument. As you correctly state, the cardinality of the algebraic closure of a field $F$ is $\max\{\aleph_0, |F|\}$, so the cardinality of the algebraic closure of $\mathbb Q$ is $\aleph_0$.

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    $\begingroup$ Well, this is of course also correct, but since there is no finite algebraically closed field (there are irreducible polynomials of any degree over any $\mathbb F_p$) we have that equality always holds. Or was it something else you were doubting? $\endgroup$
    – Levi
    Commented Dec 11, 2019 at 21:59

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