ag.algebraic geometry - What does the nilpotent cone represent?

Notation

Let $mathfrak g$ be a the Lie algebra of an algebraic group $Gsubseteq GL(V)$ 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 $X=X_n$).

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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, $mathcal N(mathrm{Spec}(A))={Xin mathfrak{g}(mathrm{Spec}(A))| X^N=0$ 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 $1times 1$ 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 $A$?