Is $\mathbb{R}$ a $\mathbb{C}$-module without AC? Assuming ZFC. We can make $(\mathbb{R},+)$ into a nontrivial (scalar multiplication is not identically zero) $\mathbb{C}$-module.
Now my questions are?
0.Is it consistent with $ZF$ that $\mathbb{R}$ is not a $\mathbb{C}$-module?
1.Does $AD$ (Axiom of determinacy) implies that $\mathbb{R}$  is not a  $\mathbb{C}$-module?
We know $AD$ leads many amazing theorems about reals, both algebraic and analytic. How does Solovay model affect Algebraic construction of reals? More precisely and especially
2.If the answer of question 1 is yes. Is it possible that $\mathbb{R}$ is not a  $\mathbb{C}$-module in Solovay model?
Any suggestion and reference is appreciated.
 A: If all sets of reals have the Baire Property (as holds under $\sf AD$ and in Solovay's model, and in Shelah's model whose consistency strength does not require any large cardinals), then the answer is negative. Let us denote by $\sf BP$ this principle.
To see that, first we use the fact that under $\sf BP$ if $\varphi\colon\Bbb R\to\Bbb Q$ is linear, then it is continuous (in general any homomorphism from a Polish group to a normed group is continuous under $\sf BP$). This means that there are no discontinuous solutions for Cauchy's functional equation: $f(x+y)=f(x)+f(y)$ with $f\colon\Bbb R\to\Bbb R$.
But if $\Bbb R$ is a $\Bbb C$-module, then we can reinterpret it over $\Bbb R$ as a module of dimension $\geq 1$, but since $\Bbb C$ has dimension $2$, this gives us a nontrivial decomposition of $\Bbb R$ as a vector space over $\Bbb Q$ and therefore it gives us a discontinuous solution to Cauchy's functional equation (to see that this is the case, just note that continuous solutions are zero or have trivial kernel, here the projection on one of the direct summands is neither zero nor has a trivial kernel).
Another option, which again appeals to Pettis' theorem is to note that if $\Bbb R$ is a $\Bbb C$-module, then $z\mapsto z\cdot 1_\Bbb R$ is a group homomorphism, which is therefore continuous, but therefore the range of this function is a connected subgroup of $\Bbb R$ which has to be $\Bbb R$ itself. Now the previous paragraph holds, since $\Bbb R$ is a $2$-dimensional vector space over itself and we can find a discontinuous endomorphism as before.

Some references:

*

*The fact that $\sf BP$ implies that any homomorphism from a Polish group into a normed group appears in Kecrhis' book about descriptive set theory. It's a consequence of Pettis' theorem.


*Shelah constructed a model of $\sf ZF+DC+BP$ from a model of $\sf ZFC$. Thus eliminating the need for an inaccessible cardinal to get the Baire property. He also showed that in order to get Lebesgue measurability of all sets of reals in $\sf ZF+DC$ one has to have that there is an inaccessible cardinal in an inner model.

Saharon Shelah, Can you take Solovay’s inaccessible away?, Israel J. Math. 48 (1984), no. 1, 1--47.

