Can the dimension of a quotient space be 0?

Question

I am trying to think of examples of finite vector spaces and subspaces to observe the relationship for the corresponding quotient space: $\dim(V)=\dim(W)+\dim(V/W)=\dim(W)+\textrm{codim}(W)$. I'm considering $V=\mathbb{Z}_5$ and $W=\mathbb{Z}_2$ (not entirely sure if $W$ is in fact a subspace of $V$ here, however), both of which have dim 1 so by the relation $\dim(V/W)=0$. I know that we can extend a basis of $W$ to one of $V$ since a basis of $W$ is already linearly independent in $V$. Then if we consider the ordered set of those adjoined elements of the basis, the image of that set in the homomorphism $\phi:v\mapsto [v]$ is an ordered set which forms a basis for $V/W$. I tried constructing a basis for $V/W$ like this: I think $\{1\}$ is a basis for both $V$ and $W$ (matches the dim as well), so there are no other vectors to construct a basis for $V/W$ from. It seems like that supports $\dim(V/W)=0$, but I can't think of an intuitive explanation of this. Any insights?