In Zermelo-Fraenkel set theory without the Axiom of Choice (AC), it is consistent to say that there are models in which:


1. there are vector spaces without a basis;
2. the field of complex numbers $\mathbb{C}$ only has two automorphisms (identity and complex conjugation).


Before I even ask my main question, I want to ask two "pre-questions" (as I am not a logician):


* is my formulation "in Zermelo-Fraenkel set theory without the Axiom
of Choice (AC), it is consistent to say that there are models in
which ..." formally correct? (And if not, what is a formal correct
statement ?);
* for the second statement above, what is a precise reference in which I can find this statement?


Now for my main question: is it also true that it is consistent to say that there are models of ZF set theory without AC, in which every vector space over $\mathbb{C}$ *has* a base (or even stronger: in which *dimension* is well defined), and in which $\mathbb{C}$ also has precisely two field automorphisms?



Automorphisms of vector spaces and the complex numbers without choice
This is not a full answer, but it is too long to be a comment.


Let $B(F)$ for field $F$ be the statement "every vector space over $F$ has a basis" and let $AL19(F)$ be the statement "for every vector space $V$ over $F$, every generating subset of $V$ contains a basis", $AL20(F)$ means "for every vector space $V$ over $F$, every independent subset of $V$ is contained in a basis".


In 2012 Paul Howard and Eleftherios Tachtsis said in [1] that both:


* There exists a field $F$ such that "$B(F)\implies AC$"
* There exists a field $F$ such that "$B(F)\;\not\!\!\!\implies AC$"


Are open, in particular *if the answer of your question is positive, then it is an open problem*.


On the other hand, Paul Howard had proven in [2] that $(∃F\ s.t.\ AL19(F))⇒AC$, in particular, the strengthening of $B(ℂ)$ to $AL19(ℂ)$ does imply AC and hence imply that there are wild automorphisms for $ℂ$.


Similarly, both [1] and [2] claim that in [3,4] it was proven that $(∃F\ s.t.\ AL20(F))⇒MC$ (which over ZF implies $AC$), and hence the dual strengthenin of $B(ℂ)$ to $AL20(ℂ)$ also imply that there are wild automorphisms for $ℂ$ (although from quick glance over [3,4] I couldn't see this result, a proof of this result, together with the result of the previous paragraph can be found in [5]).



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> [1] *Howard, Paul; Tachtsis, Eleftherios*, [**On vector spaces over specific fields without choice**](https://doi.org/10.1002/malq.201200049), Math. Log. Q. 59, No. 3, 128-146 (2013). [ZBL1278.03082](https://zbmath.org/?q=an:1278.03082).
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> [2] *Howard, Paul*, [**Bases, spanning sets, and the axiom of choice**](https://doi.org/10.1002/malq.200610043), Math. Log. Q. 53, No. 3, 247-254 (2007). [ZBL1121.03064](https://zbmath.org/?q=an:1121.03064).
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> [3] *Armbrust, M. K.*, [**An algebraic equivalent of a multiple choice axiom**](https://doi.org/10.4064/fm-74-2-145-146), Fundam. Math. 74, 145-146 (1972). [ZBL0234.04011](https://zbmath.org/?q=an:0234.04011).
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> [4] *Bleicher, M. N.*, [**Some theorems on vector spaces and the axiom of choice**](https://doi.org/10.4064/fm-54-1-95-107), Fundam. Math. 54, 95-107 (1964). [ZBL0118.25503](https://zbmath.org/?q=an:0118.25503).
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> [5] Rubin, H., & Rubin, J. E. (1985). Algebraic Forms. In [*Equivalents of the axiom of choice, II*](https://www.elsevier.com/books/equivalents-of-the-axiom-of-choice-ii/rubin/978-0-444-87708-6) (p. 122). North-Holland.
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