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Background

Yet another homework inspired question: A scheme is reduced if no section of the structure sheaf is nilpotent. To every scheme $X$ there is a scheme $X_{red}$ and a morphism $i: X_{red} \rightarrow X$ such that every morphism from a reduced scheme into $X$ factors through $X_{red}$. Hartshorne Ex. 2.2.6 guides you through the construction of this scheme. Basically you leave the topological space alone, but you mod out the nilpotents in the structure sheaf. This gives you a presheaf, which you then sheafify to get the structure sheaf of $X_{red}$.

I have been trying to run through all of the constructions in Hartshorne from a functor of points perspective in addition to the "standard" approach, so naturally I was interested in seeing this construction as well.

From this perspective $X$ is a functor $CRing \rightarrow Sets$, namely $Hom(Spec(-),X)$ if you were using the standard definition of schemes. The functor from $F: CRing \rightarrow CRing$ taking $A$ to $A/nil(A)$ seems relevant here, so it seems natural to ask if $X \circ F: CRing \rightarrow Sets$ is the reduced scheme associated to $X$. I won't spell out the details, but this actually turns out to be true (I think at least!). This brings me to my questions.

Questions

Did I mess up, or does the construction above pan out?

Is there a nice characterization of the functors $CRing \rightarrow Cring$ which will give a scheme when composed with any scheme $CRing \rightarrow Sets$?

Even if there is no simple characterization of all such functors, is there a large class of such functors which is nice?

Do you have any other examples of standard constructions in algebraic geometry which are of this form?

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I would suggest you read Demazure-Gabriel for a much better description of this approach, as it doesn't really make sense without a description of the global Zariski site. –  Harry Gindi Feb 3 '10 at 12:08
    
Yoneda's lemma doesn't make sense without a site? The fact that schemes are sheaves on the Zariski site is important, but it is not important for posing my question. –  Steven Gubkin Feb 3 '10 at 14:15
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3 Answers 3

up vote 6 down vote accepted

Here is a nontrivial example I like. Let $W:\mathrm{Rings}\to\mathrm{Rings}$ denote the Witt vector functor of some fixed finite length. (You can consider the $p$-typical Witt vectors, for some prime $p$, but everything works with the other standard flavors.) Then the functor $W_*(-)=-\circ W$ is an endofunctor of the category of functors $\mathrm{Rings}\to\mathrm{Sets}$, and it takes schemes to schemes. The scheme $W_*(X)=X\circ W$ is the so-called arithmetic jet space of $X$, extensively studied by Buium in the case of $p$-adic formal schemes. The fiber over $p$ is the Greenberg transform.

There is a standard method for proving $W_*$ takes schemes to schemes (or rather for proving almost that), though it probably doesn't work for every functor $F:\mathrm{Rings}\to\mathrm{Rings}$ such that $F_*$ takes schemes to schemes. First you show that the category of sheaves of sets on the category of affine schemes w.r.t. the etale topology is stable under $W_*$. This is true since $W$ takes etale covers of rings to the same and also takes cocartesian squares of etale rings maps to the same. (These properties of $W$ are not obvious.) Then you show $W_*$ takes sheaf epimorphisms to the same, and etale maps of sheaves to the same. (These properties are much easier.) Therefore any etale equivalence relation on an affine scheme is sent to an etale equivalence relation on an affine scheme, and the quotient of the first is sent to the quotient of the second. Therefore the category of quasi-compact quasi-separated algebraic spaces is stable under $W_*$. I have no doubt you could find reasonable abstract properties on the endofunctor $W$ of Rings that allow this argument to go through.

Showing $W_*$ takes schemes to schemes is a bit subtler. You have to deal with quasi-compactness issues (as a right adjoint, $W_*$ doesn't necessarily behave well w.r.t. disjoint unions) and also the fact that it's harder to tell whether a functor is represented by a scheme than by an algebraic space. But as I said, in the Witt vector example above, it is true.

Presumably the same argument works, and is much easier, for the functor $F$ defined by $F(R)=R[t]/(t^{n+1})$. Then $F_*(X)$ should be the usual jet space functor of length $n$. The case $n=1$ should give the total space of the tangent bundle, at least when $X$ is smooth.

Edit: I'm reminded below that this example is just a particular case of the representability of the Weil restriction of scalars for a finite flat map $A\to B$. There you consider the endofunctor of the category of $A$-algebras given by $F(R)=B\otimes_A R$. In particular, it's reasonable to view $W_*$ as a generalized Weil resitrction of scalars.

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How about a 'disconnected' version of your last paragraph: $F(R)=R\times R$? Works just as well, $F_*(X)=X^2$. It seems that if one works with schemes over a field and $A$ is any finite-dimensional algebra, $F(R)=R\otimes A$ works. Then $F_*(X)=Hom(Spec(A),X)$ (which needs to be defined as a scheme first, of course). –  t3suji Feb 3 '10 at 14:03
    
Great answer! This gives me a lot to think about. –  Steven Gubkin Feb 3 '10 at 14:18
    
@t3suji: Good point! I had forgotten about that. I'll add a few words about that. –  JBorger Feb 3 '10 at 21:02
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Actually $X$ is identified with the covriant functor $CRing \to Sets$ $A\mapsto Hom(Spec(A),X)$.

The functor $X\circ F$ is usually called the de Rham space of $X$. This is not the reduced scheme associated to $X$ because you mod out by nilpotents on the source instead of $X$: for $X=Sepc(B)$ you get $Hom(Spec(A);Spec(B)\circ F) = Hom(B,A/nil(A))$ instead $Hom(Spec(A);Spec(B)_red) = Hom(B/nil(B),A)$.

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Thanks for the catch about covariance, I edited my question. I am still not convinced by the second part of your answer. How does it match up with the definition here: ncatlab.org/nlab/show/de+Rham+space ? I will have to think some more. –  Steven Gubkin Jan 28 '10 at 13:23
    
OK, I see how my functor should not be X_red. I still don't know about the connection to the de Rham space. Is the de Rham Space a scheme? –  Steven Gubkin Jan 28 '10 at 14:04
    
You should look at mathoverflow.net/questions/10556/…. –  YBL Jan 28 '10 at 16:53
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It seems to me that you want to get the functor $X \mapsto X_{red}$ (schemes) from a "functor of points"-approach out of the functor $A \to A_{red}$ (rings). This is not possible in a simple formal way (see VBL's answer), but the concept may be generalized as follows:

Let $P$ be a full subcategory of $Ring$, which is local in the sense that for every partition of unity of a ring $A=(f_1,...,f_n)$, all the localizations $A_{f_i}$ belong to $P$ if and only if $A$ belongs to $P$. Now you can endow $P$ with the Zariski topology and define $P$-schemes as sheaves on $P$, which are covered by representable ones. Let $Sch_P$ denote the category of $P$-schemes.

My point is: If $P$ is a reflective subcategory of $Ring$, then also $Sch_P$ is a reflective subcategory of $Sch$. If $F : Ring \to P$ is a retraction, then we may extend this retraction on representable functors and then, by gluing, to $Sch \to Sch_P$.

Of course, you could also formulate this construction in the language of affine schemes or locally ringed spaces. I don't think that this matters here. The functor of points approach does not help you to extend a retraction $Ring \to P$ to $Sch \to Sch_P$ in a formal way, gluing is necessary. In particular, the construction of the reduced scheme structure does have nothing to do with precomposing endofunctors of rings.

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