# Is $\mathbb{C}$ still algebraically closed in ZF-models without the Axiom of Choice?

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?

Yes, $$\mathbb{C}$$ is still algebraically closed. You can prove this by going through any of the usual proofs (e.g. via Liouville's theorem) and seeing that choice is never used.

Alternately, you can use a nuke: Shoenfield's absoluteness theorem says that in a precise sense choice cannot be relevant to theorems of this simplicity.$$^1$$ There is absolutely no reason to use Shoenfield here ... except that it's really funny.

$$^1$$Technically, Shoenfield showed that every $$\Pi^1_2$$ sentence true in $$L$$ is true is true in $$V$$, under the assumption that $$V\models\mathsf{ZF}$$. Since $$L\models\mathsf{ZFC}$$ and "$$\mathbb{C}$$ is algebraically closed" is $$\Pi^1_2$$, this gives the desired result. Moreover, Shoenfield is completely constructive: it gives a totally explicit (if rather unsatisfying) way to transform a $$\mathsf{ZFC}$$-proof of a $$\Pi^1_2$$ sentence into a $$\mathsf{ZF}$$-proof. It's also worth noting that Shoenfield is actually even broader than this - e.g. it also lets us remove the continuum hypothesis from proofs of $$\Pi^1_2$$ sentences, and many more hypotheses besides.

• I see your nuke, and I raise you a thermonuclear one. ;-) Mar 31, 2021 at 22:08

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 $$V$$.

• I think this is actually less of a nuke than mine. But +1 certainly! Mar 31, 2021 at 22:13
• Fine. Assume two supercompact cardinals; collapse the first to force $\sf DC$, then collapse the second which will well-order the continuum. Now use generic absoluteness of the Chang model. Better? :-D Mar 31, 2021 at 22:18
• There we go! That's wonderful. Mar 31, 2021 at 22:22