In a given model the axiom is either true or false. When I say “dropped” I mean “false” in that model. It isn’t true that in all models where choice fails we have fields with no algebraic closure.
What about ZF; which is ZFC without the axiom of choice, and notably not ZF¬C, which is ZF with the negation of choice.
In a given model of ZF either choice is true or it is false. In the absence of the Axiom of Choice you can’t say that not all fields have algebraic closures. What you can say is that there is a model in which there are fields with no algebraic closure.