# Automorphisms of vector spaces and the complex numbers without choice

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?

• "Every vector space has a basis" implies choice; see math.stackexchange.com/questions/207990/vector-spaces-and-ac Feb 22 at 20:28
• To continue @KevinCasto's comment, just in case it's not clear: and once you have choice, you're in ZFC, so you can cook up many automorphisms of $\mathbb{C}$. Mar 1 at 16:00
• @AndrejBauer Although I don't think it's known that "every vector space over $\mathbb{C}$ has a basis" implies choice. Mar 1 at 16:35
• Oh, I see. So how should I understand @KevinCastos comment? Mar 1 at 16:50
• Ah fair enough, I didn't read Blass' proof closely enough (it relies on function fields over $\mathbb C$ also having this property). Still, it does in any case seem like a result worth mentioning in the context of the question! Mar 1 at 17:20

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.