Algebraic closure with no nontrival automorphism

Question

In Milne's notes on Galois theory, Chapter 7, p.91 he remarked that it is consistent without the axiom of choice that there exists an algebraic closure $L$ of $\mathbb{Q}$ with no nontrivial automorphisms.¹

With Zorn's lemma, we can always extend an automorphism of $\mathbb{Q}(i)$ into $L$, thus always get a nontrivial automorphism of $L$. Without axiom of choice, we cannot do what we did. But it is hard for me to think why this would happen, could anyone give an brief explanation?

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1. Wilfrid Hodges, Läuchli’s algebraic closure of $\mathbb Q$, Math. Proc. Cambridge Philos. Soc. 79 (1976), no. 2, 289--297.

Answer 1

There is no simple explanation. The only thing you can give as an intuition is that Zorn's lemma fails. And its failure is witnessed already in these parts.

The construction is written in a fairly outdated language, but here's what I can speculate from what's in the paper and the things I know about these constructions.

We add a generic copy of the algebraic closure of $\Bbb Q$ to our universe, and then we restrict to a subuniverse of this extension in which every element is essentially defined from finitely many elements of the field as the field structure itself.

So we have a field which is still algebraically closed, as being algebraically closed is witnessed "locally" by finite collections of elements; and it is still of characteristics $0$, so it has a copy of the rationals. We can also show that no subfield is algebraically closed in a fairly straightforward way.

Therefore our generic copy is an algebraic closure of $\Bbb Q$. One can show, however, that it is not countable (an enumeration cannot depend only on finitely many elements). So now $\Bbb Q$ has two non-isomorphic closure. Which is great. Finally, an automorphism cannot be defined by a finite subset of this field, since it would mean it has to be an automorphism of a subfield and we have to make choice outside that subfield.

In a nutshell, we add a copy of $\overline{\Bbb Q}$ and then forget all its subsets which are not definable (in a set theoretic way, but also model theoretic way) from the roots of finitely many polynomials. It is not hard to show that if an automorphism of this field is defined from a polynomial, it has to be more or less an automorphism of that polynomial's splitting field. And we cannot "decide how to extend it" by the relative homogeneity of the [original] algebraic closure.