Vector spaces "over a field"

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 $1A = A$.