What does the nilpotent cone represent? Notation
Let $\mathfrak g$ be a the Lie algebra of an algebraic group $G\subseteq GL(V)$ over a(n algebraically closed) field $k$ (I'm actually thinking $G=GL_n$, so $\mathfrak g=\mathfrak{gl}_n$). Then any element $X$ of $\mathfrak g$ can be uniquely written as the sum of a semi-simple (diagonalizable) element $X_s$ and a nilpotent element $X_n$ of $\mathfrak g$, where $X_s$ and $X_n$ are polynomials in $X$. The nilpotent cone $\mathcal N$ is the subset of nilpotent elements of $\mathfrak g$ (elements $X$ such that $X=X_n$).

People often talk about the nilpotent cone as having the structure of a subvariety of $\mathfrak g$, regarded as an affine space, but usually don't say what the scheme structure really is. To really understand a scheme, I'd like to know what its functor of points is. That is, I don't just want to know what a nilpotent matrix is, I want to know what a family of nilpotent matrices is (i.e. what a map from an arbitrary scheme $T$ to $\mathcal N$ is). Since any scheme is covered by affine schemes, it's enough to understand what an $A$-valued point (a map $\mathrm{Spec}(A)\to \mathcal N$) is for any $k$-algebra $A$. So my question is

What functor should $\mathcal N$ represent?

A guess
Well, an $A$-point of $\mathfrak g$ is "an element of $\mathfrak g$ with entries in $A$" (again, I'm really thinking $\mathfrak g=\mathfrak{gl}_n$, so just think "a matrix with entries in $A$"), so I would expect that such an $A$-point happens to be in $\mathcal N$ exactly when the given matrix is nilpotent. That is, $\mathcal N(\mathrm{Spec}(A))=\{X\in \mathfrak{g}(\mathrm{Spec}(A))| X^N=0$ for some $N\}$.
However, this is wrong. That functor isn't even an algebraic space, even for the nilpotent cone of $\mathfrak{gl}_1$. If it were, the identity map on it would correspond to a nilpotent regular function $f$ (a nilpotent $1\times 1$ matrix), and this would be the universal nilpotent regular function; every other nilpotent regular function anywhere else would be a pullback of this one. But whatever the degree of nilpotence of this function (say $f^{17}=0$), there are some nilpotent regular functions which cannot be a pullback of it (something with nilpotence degree bigger than 17). If this version of the nilpotent cone were representable, you can show that the $\mathfrak{gl}_1$ version would be too.
Another guess
I think the answer might be that an $A$ point of $\mathcal N$ is a matrix ($A$ point of $\mathfrak g$) so that all the coefficients of the characteristic polynomial vanish. This is a scheme and it has the right field-valued points, but why should this be the nilpotent cone? What is the meaning of having all coefficients of the characteristic polynomial vanish for a matrix with entries in $A$?
 A: The ideal which defines the nilpotent cone is generated by the homogeneous elements of positive degree in $\mathbb{C}[\mathfrak g]^G$. In the case of $GL_n$, this ideal is generated by the functions $\mathrm{tr}\;X^k$ for $k$ up to the dimension, and also by the coefficients of the characteristic polynomial. See, for example, [Springer, T. A. Invariant theory. Lecture Notes in Mathematics, Vol. 585. Springer-Verlag, Berlin-New York, 1977. iv+112 pp. MR0447428]
A: Another way of saying Mariano's answer is that the nilcone is the scheme-theoretic fiber of the projection map $\mathfrak{g}\to \mathfrak{g}// G$  where $\mathfrak{g}// G$ is the categorical quotient of the adjoint action.  This is usually how it comes up anyways.
Yet another way to say this is that the nilcone is the classical limit of the quotient of the universal enveloping algebra modulo a maximal central ideal.  This might sound complicated, but I certainly can't think of a more important appearance of the nilcone than this fact.
Well, there's also its appearance as the affinization of $T^*G/B$, which also gives the reduced scheme structure.
