Automorphisms of vector spaces and the complex numbers without choice

Question

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?

Answer 1

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]).

[1] Howard, Paul; Tachtsis, Eleftherios, On vector spaces over specific fields without choice, Math. Log. Q. 59, No. 3, 128-146 (2013). ZBL1278.03082.

[2] Howard, Paul, Bases, spanning sets, and the axiom of choice, Math. Log. Q. 53, No. 3, 247-254 (2007). ZBL1121.03064.

[3] Armbrust, M. K., An algebraic equivalent of a multiple choice axiom, Fundam. Math. 74, 145-146 (1972). ZBL0234.04011.

[4] Bleicher, M. N., Some theorems on vector spaces and the axiom of choice, Fundam. Math. 54, 95-107 (1964). ZBL0118.25503.

[5] Rubin, H., & Rubin, J. E. (1985). Algebraic Forms. In Equivalents of the axiom of choice, II (p. 122). North-Holland.