Why do we lose geometric intuition regarding a scheme due to nilpotent elements? I have quite often read that the functor of points allows us to recover some geometric intuition for a scheme which was lost due to the presence of nilpotent elements in the structure sheaf. 
I know this question is somewhat vague, but, regardless, what exactly is it about the nilpotent elements which don't allow us to view a scheme as a collection of points and how does the functor of points approach resolve this issue? 
 A: Consider $X=Spec(k[x]/(x^2)$. There is only one prime ideal of $k[x]/(x^2)$ which is $(x)$. Hence $X$ consists of a single point, just like $Spec(k)$. The only difference between these schemes is that $X$ contains nilpotents in the structure sheaf but $Spec(k)$ does not.
A: I don't know very much algebraic geometry, so my understanding is that the functor of points view is a restatement of our notions of the points of a scheme and of what we think of as geometric intuition.  
If $R$ is a ring with nilradical $J$, then for every ideal $I$ between $(0)$ and $J$, there is no difference between the underlying topological spaces $\textrm{Spec}(R/I)$ and $\textrm{Spec}(R)$.  
More generally, if $X$ is a scheme, and $\mathscr J$ is the quasicoherent sheaf of ideals of $\mathcal O_X$ given by $\mathscr J(U) = \{ s \in \mathcal O_X(U) : s_x \in \textrm{Nil }\mathcal O_{X,x} \textrm{ for all } x \in U \}$, then for any quasicoherent sheaf of ideals $\mathscr I$ between $0$ and $\mathscr J$, there is no difference between the underlying topological spaces of the schemes $(X,\mathcal O_X)$ and $(X, \mathcal O_X/\mathscr I)$.
So it's not enough to view a scheme as a collection of points, i.e. as a topological space.  Instead, what we do is we use the functor of points view to change our definition of what are the points of a scheme.  And from this new definition of what is meant by "points," we are then changing what we mean by "geometric intuition."  And from this new notion of geometric intuition, we do recover everything there is to know about the scheme, by the Yoneda lemma.
Namely, if $Z$ is any scheme, we define the $Z$-points of $X$ to be the set $X(Z) = \textrm{Hom}_{\textrm{Sch}}(Z,X)$.  The functor $X(-): \textrm{Sch}^{\textrm{op}}\rightarrow \textrm{Set}$ then determines $X$ up to isomorphism, in the sense that if $Y$ is a scheme, then any isomorphism of functors $X(-) \rightarrow Y(-)$ (that is, any collection of bijections $X(Z) \rightarrow X(Z)$ for all schemes $Z$, natural in $Z$) determines an isomorphism of schemes $X \rightarrow Y$.
