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?


2 Answers 2


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.

  • $\begingroup$ I see your nuke, and I raise you a thermonuclear one. ;-) $\endgroup$
    – Asaf Karagila
    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 $V$.

  • 2
    $\begingroup$ I think this is actually less of a nuke than mine. But +1 certainly! $\endgroup$ Mar 31, 2021 at 22:13
  • $\begingroup$ 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 $\endgroup$
    – Asaf Karagila
    Mar 31, 2021 at 22:18
  • $\begingroup$ There we go! That's wonderful. $\endgroup$ Mar 31, 2021 at 22:22

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