Field and subfield vector spaces. Here is this example in Golan's "Linear Algebra":



My question is:
Can we reverse i.e. think of $\mathbb{R}$ as a vector space over $\mathbb{C}$ and of $\mathbb{Q}$ as a vector space over $\mathbb{R}$? If no, why? 
 A: In the example you have, the scalar multiplication is simply the field multiplication: If $F\subseteq K$, then we have $f\cdot k=fk$ for all $f\in F$ and $k\in K$, where "$f\cdot k$ denotes the multiple of the "scalar" $f$ with the "vector" $k$, and "$fk$" denotes the product inside the field $K$.
So interesting questions are:

(a) Does $\mathbb{Q}$ have any vector space structure over $\mathbb{R}$, if we only assume the vector space addition is the usual addition of $\mathbb{Q}$?
(b) Does $\mathbb{R}$ have any vector space structure over $\mathbb{C}$, if we only assume the vector space addition is the usual addition of $\mathbb{R}$?

Answers:
(a) No! Any nontrivial vector space over $\mathbb{R}$ has cardinality $\geq$ continuum, and $\mathbb{Q}$ is countable (and nontrivial).
(b) Yes! The additive groups $\mathbb{R}$ and $\mathbb{C}$ are isomorphic, see https://math.stackexchange.com/a/337121/58818.
Let $\phi\colon\mathbb{R}\to\mathbb{C}$ be an additive isomorphism. Define a $\mathbb{C}$-vector space structure on $\mathbb{R}$ as $\lambda \cdot x=\phi^{-1}(\lambda\phi(x))$ for all $\lambda\in\mathbb{C}$ and $x\in\mathbb{R}$, and the product inside the argument of $\phi^{-1}$ is the usual complex product.
This $\mathbb{C}$-vector space has dimension $1$ over $\mathbb{C}$.
