$X(C) = \{\text{nilpotent elements in $C$}\}$ is not a scheme I am try to prove that 
$X(C) = \{\text{nilpotent elements in C}\}$ is not a scheme.
I have proved that it is local, so what makes it is not a scheme must be the open cover stuff. And I think it is covered by $X_n(C) = \{x|\text{$x^n=0$ in C}\}$. So I think it is covered by an infinite union of affine schemes. Therefore, I think the problem is that the cover is not open.
But even if I could prove this obvious cover is not open. It is still not sufficient to prove that $X$ has no open affine cover. And I have no idea for finding some contradiction assuming $X$ has an open affine cover.
So how to show that $X$ has no open affine cover? Thanks for any help.
 A: Here's one way to prove it.  Take your favorite field $k$ and let's base-change to $k$: supposing $X$ is a scheme, consider the scheme $Y=X\times \operatorname{Spec} k$ as a scheme over $\operatorname{Spec} k$.  For a $k$-algebra $C$, then, $Y(C)$ is the set of nilpotent elements of $C$.  As you say, this is covered by the affine schemes $Y_n=\operatorname{Spec} k[t]/(t^n)$ of elements $x$ such that $x^n=0$.  Now the key observation is that each $Y_n$ has only one point, and so when you consider $Y$ as the nested union of the $Y_n$, this means $Y$ has only one point as well.  In particular, $Y$ must be affine, so it must be $\operatorname{Spec} A$ for some $k$-algebra $A$.  Then $A$ would be the universal $k$-algebra with a nilpotent element: there is an element $a\in A$ such that for any nilpotent element $x$ in any $k$-algebra $C$, there is a unique homomorphism $A\to C$ sending $a$ to $x$.  This is impossible, since we must have $a^n=0$ for some particular $n$ and then $a$ cannot be mapped to $x$ if $x^n\neq 0$ but $x^N=0$ for some $N>n$ (e.g., $x=t$ in $k[t]/(t^N)$).
