Vector spaces "over a field" We often refer to a vector space $V$ "over the field $F$". Every definition I have read defines this as a set of vectors satisfying the axioms, where the scalars in that vector space belong to some field $F$. 
What about the elements of the vectors, though? Must those vectors also belong to $F$? Is it possible, for example, for $V$ to be the set of vectors with complex entries (so, the vector space $\mathbb{C}^n$) but the scalars be in $\mathbb{R}$? So our vector space would be $\mathbb{C}^n$ over $\mathbb{R}$? Is this a correct use of the definition, or must the entries of the vectors also matter? 
 A: Yes, that's possible. It is not uncommon to consider the complex numbers $\mathbb{C}$ as a two dimensional vector space over the real number $\mathbb{R}$. A basis is given by $1$ and $i$, for example. 
All that is needed is that the scalar multiplication is defined in some way. Often this entails that the elements of the vector space are somehow naturally linked to the field over which the space is defined. 
However, I can say $\{q,u,i,d\}$ is a vector space over the field $\mathbb{Z}/2 \mathbb{Z}$, with $q+ x  = x $ for each $x \in \{q,u,i,d\}$, further $u+u = i+i = d + d = q$, and $u+i  = d$, $u+d = i$, $i+d = u$. 
Further $\overline{0} x = q$  and $\overline{1} x = x$ for each $x$ in $\{q,u,i,d\}$.   
A more natural example is proposed by Rob Arthan: For any set whatsoever $X$ the powerset $\Bbb{P}(X)$  becomes a vector space over $\Bbb{Z}/2\Bbb{Z}$ if one defines the sum of two sets to be their symmetric difference so that $0 = \emptyset$ and the action of $\Bbb{Z}/2\Bbb{Z}$ in the only possible one: $0A = \emptyset$ and $1A = A$.
A: A vector in $\mathbb C^n$ over the field $\mathbb R$ has dimension $2n$ but over the field $\mathbb C$ is just $n$.
This means that the case $\mathbb C^1$ has real dimension $2$ but it has dimension $1$ relative to the complex numbers:
If you give a vector $z_0$ different from $0+0i$ on $\mathbb C^1$ and you have  any other vector $z\in\mathbb C^1$ then 
$$z=\mu z_0$$
where $\mu$ is the complex number $\frac{z}{z_0}$.
