Can we make the real numbers a $\mathbb{C}$-module? 
Can we make the real numbers into a $\mathbb{C}$-module with unity? 

We should take the real numbers as a group and the complex numbers as our ring. Please tell me what is the operation?
 A: Yes, we can! 
$\mathbb R$ is a $\mathbb C$-module iff there exists a ring homomorphism $\mathbb C\to\operatorname{End}(\mathbb R)$. 
A group endomorphism of $\mathbb R$ is a $\mathbb Q$-linear map. Now let's define such a linear map which satisfies the condition $(f\circ f)(x)=-x$ for all $x\in\mathbb R$. Let $(x_i)_{i\in I}$ be a $\mathbb Q$-basis of $\mathbb R$ and write $I$ as a disjoint union of subsets made of two elements. For such a subset, say $\{u,v\}$, define $f(x_u)=x_v$ and $f(x_v)=-x_u$. Now check that $(f\circ f)(x_i)=-x_i$ for all $i\in I$.
Then define a ring homomorphism $\phi:\mathbb C\to\operatorname{End}(\mathbb R)$ by $\phi(a+bi)=\phi_a+\phi_b\circ f$, where $\phi_a(x)=ax$, respectively $\phi_b(x)=bx$ for all $x\in\mathbb R$. 
A: As an abelian group, $\mathbb{R}$ is uniquely divisible (that is, for each $x$ and nonzero integer $n$, there is a unique $y$ so that $x = n \cdot y$), so $\mathbb{R}$ is isomorphic to a vector space over $\mathbb{Q}$. (there's a more obvious proof that $\mathbb{R}$ is a rational vector space -- the point is that "uniquely divisible abelian group" means the same thing as "rational vector space", so I will talk about the latter instead)
In fact, (I'm surely assuming axiom of choice), $\mathbb{R}$ is a $\mathfrak{c}$-dimensional rational vector space, where $\mathfrak{c}$ is the cardinality of $\mathbb{R}$.
$\mathbb{C}$ is also a $\mathfrak{c}$-dimensional rational vector space. Therefore, $\mathbb{C} \cong \mathbb{R}$ as rational vector spaces, and thus as abelian groups.
Thus, your question is equivalent to the follwing:


Can we make the complex numbers into a $\mathbb{C}$-module with unity? 

We should take the complex numbers as a group and the complex numbers as our ring. Please tell me what is the operation?
