There's a question in my college's module -
Let $V$ be the vector space of ordered pairs of complex numbers over the real field $\mathbb{R}$. Show that $V$ is of dimension 4.
How can vector space of complex numbers be over the field $\mathbb{R}$, shouldn't be over the field $\mathbb{C}$?
You can consider the complex numbers $(\mathbb{C}, +)$ as an abelian group under addition, and define the action $\mathbb{R} \times \mathbb{C} \to \mathbb{C}$ by $(x,z) \mapsto xz$ to obtain a vector space over the real numbers. When $\mathbb{C}$ is considered as a vector space over $\mathbb{R}$, it has $\{1,i\}$ as a basis and hence dimension 2. It is then easy to show that $\mathbb{C}^2$ has dimension 4 (taking, say, $\{(1,0), (i,0), (0,1), (0,i)\}$ as a basis).