I am trying to find all algebraically closed fields(up to isomorphism). I found that the field of all algebraic numbers over $\mathbb Q$ is algebraically closed and I also know that the field of complex numbers $\mathbb C$ is also algebraically closed. I want to know that is there exist any other algebraically closed field or not. If exist then what are they.
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3$\begingroup$ Every field has an algebraic closure, so there are many such fields. Each finite field $\Bbb F_p$ has its own algebraic closure; function fields $\Bbb Q(t)$ have algebraic closures; etc. $\endgroup$– Greg MartinMay 18, 2022 at 20:09
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1$\begingroup$ To follow on with Greg Martin comment, you can have fields of arbitrary infinite cardinal and from them build algebraic closure of the same arbitrary infinite cardinal $\endgroup$– mathcounterexamples.netMay 18, 2022 at 20:15
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1$\begingroup$ On the other hand, algebraically closed fields of fixed characteristic all behave “roughly similarly” without being isomorphic – their first-order theory being complete, any first-order statement holding in one of them holds in any other. $\endgroup$– AphelliMay 18, 2022 at 20:25
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3$\begingroup$ Under the axiom of choice an algebraically closed field is determined (up to isomorphism) by its characteristic and the cardinality of a transcendental basis (over $\Bbb{Q}$ or $\Bbb{F}_p$). Conversely for each cardinal $c$ there is the algebraic closure of the field of rational functions in $c$ variables. $\endgroup$– reunsMay 18, 2022 at 20:28
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