# $X(C) = \{\text{nilpotent elements in$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.

• What is the space $C$? – Joppy Oct 29 '17 at 4:19
• @Joppy $X$ is an $R$-functor for a ring $R$, and it is applied on $C$ where $C$ is an $R$-algebra. – Y.X. Oct 29 '17 at 4:21

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 $k[t]/(t^N)$).
• Maybe I have some misunderstandness. But may I please ask if this $Y$ is equivalent to taking $R$ to be $k$, so if $Y$ is an $R$-scheme for all $R$, then it must be a $k$ scheme? If so, I think if we take an $k$-algebra $C$ such like $k[x]/\langle x^2\rangle$. Then $0$ is not the only nilpotent so $Y(C)$ is not a single point. – Y.X. Oct 29 '17 at 8:16
• Yes, $Y$ is equivalent to taking $R$ to be $k$. When I refer to "points", I mean in the sense of points of Spec of a ring being prime ideals, not in the sense of the functor of points. – Eric Wofsey Oct 29 '17 at 8:23
• It seems that you are talking about some intrinsic notion of a point of a functor without mention of an $k$-algebra $C$. It seems unfamiliar to me. If possible, may I please ask what is the "points" here refers to, and how does it related to "prime ideals"？ – Y.X. Oct 29 '17 at 8:36
• Oh, it seems you have never seen that definition of schemes at all then. Instead of defining schemes as functors, you can define them as certain locally ringed spaces. If $A$ is a commutative ring, then $\operatorname{Spec} A$ is the set of all prime ideals of $A$ equipped with a certain topology and sheaf of rings. General schemes are then locally ringed spaces that are locally isomorphic to spaces of the form $\operatorname{Spec} A$. – Eric Wofsey Oct 29 '17 at 8:39