TITLE: Is $\mathbb{C}$ still algebraically closed in ZF-models without the Axiom of Choice?
QUESTION [5 upvotes]: Is $\mathbb{C}$ still algebraically closed in ZF-models without the Axiom of Choice (AC) ?
I know several results about algebraic closures of fields in models with AC become really delicate in ZF-models without AC.
So this must be one of the basic questions, I guess (but I cannot find a clear reference or definite proof online).
Side question: do finite fields have "unique" algebraic closures (i. e. up to isomorphism) in ZF-models without AC?

REPLY [6 votes]: Noah gave a good answer. Yes, it's algebraically closed. The usual proof uses the topology, and that topology is complete without using choice.
But since we're taking out nukes, allow me to throw a hydrogen bomb of a proof.
Given a polynomial $p(x)$ it can be written as $c_nx^n+\dots c_1x+c_0$, where each $c_i$ is a complex number. That means that it is a polynomial in $L[\vec c]$ which is the smallest model of $\sf ZF$ which contains all these complex numbers, and it happens to be a model of $\sf ZFC$ because complex numbers can be seen as subsets of $\omega$, i.e. the natural numbers. Moreover, the encoding of the complex numbers is robust enough that $L[\vec c]$ agrees with $V$, the full universe of $\sf ZF$, about the arithmetic of $\Bbb C$.
So in $L[\vec c]$ we have $p(x)$ and it has a root, $c$. But this means that $c_nc^n+\dots c_1c+c_0=0$, but those are all complex numbers and the arithmetic is robust (as we said before) that $p(c)=0$ remains true in $V$.