Does the category of Schemes admit a (single) generator (or a cogenerator)?

What if we restrict to the category of schemes of finite type over a field $k$?


The category of schemes does not admit a generating family, see MO/2015.

But the category of schemes which are (locally) of finite type over some ring $k$ has a generating family: First of all, the category of affine schemes of finite type is a large generating family. Next, one observes that this subcategory is essentially small: Up to isomorphism there is only a set of $k$-algebras of finite type. In fact, they are all isomorphic to $k[x_1,x_2,\dotsc]/I$ for some ideal $I$ such that the quotient is of finite type. Taking the disjoint union of all these affine schemes, we obtain a generating object if we work with schemes locally of finite type.

I doubt that the category of schemes of finite type has a generating object.

  • $\begingroup$ I think the user is asking for a generating object rather than a generating family. $\endgroup$ – Zhen Lin Sep 6 '13 at 8:34
  • $\begingroup$ There seems to be something wrong in your second paragraph, but I'm not sure what. Taking the disjoint union of the objects in a generating family doesn't yield a generating object in general. $\endgroup$ – Zhen Lin Sep 6 '13 at 8:51
  • $\begingroup$ Consider the category $\mathbf{Set}^2$: it has no generator, but it does have a generating family. (A generator in a cocomplete locally small topos is necessarily a dense generator, but the only categories that have a dense generator are full subcategories of $\mathbf{Set}$.) $\endgroup$ – Zhen Lin Sep 6 '13 at 9:02
  • $\begingroup$ Okay, it is true in pointed categories. Unfortunately schemes are not pointed ... $\endgroup$ – Martin Brandenburg Sep 6 '13 at 9:05
  • $\begingroup$ Indeed. The obstruction is basically the existence of empty hom-sets. $\endgroup$ – Zhen Lin Sep 6 '13 at 9:07

First question: no. Consider morphisms $X \to \operatorname{Spec} \mathbb{F}_p$ for fixed $X$ and different $p$. Thus there is always a non-empty scheme $Y$ such that there are no morphisms $X \to Y$. However, if $X$ were a generator, then there are no morphisms $X \to Y$ if and only if the unique morphism $\emptyset \to Y$ is an epimorphism, which is clearly impossible for $Y$ non-empty.

Second question: probably no. Geometrically speaking, if $X$ is a generator, then all objects can be "covered" by copies of $X$. When there is only one kind of "point", we could take $X$ to be that point, but the whole raison d'être for scheme theory is the need for "fat points", e.g. nilpotent thickenings like $\operatorname{Spec} k[x]/(x^n)$ or more generally spectra of local $k$-algebras.

So, consider the family of schemes $Y_n = \operatorname{Spec} k[x]/(x^n)$. These are $k$-schemes of finite type and they are all distinct, and we have a chain of monomorphisms $Y_0 \to Y_1 \to Y_2 \to \cdots$. However, any morphism $X \to Y_n$ must factor through the "affine envelope" $X \to \operatorname{Spec} \Gamma (X, \mathscr{O}_X)$. If $X$ is a generator, then the resulting chain $$\mathrm{Hom}(X, Y_0) \to \mathrm{Hom}(X, Y_1) \to \mathrm{Hom}(X, Y_2) \to \cdots$$ must be strictly increasing, but that can only happen if $\Gamma (X, \mathscr{O}_X)$ has nilpotents of unbounded nilpotency index. If $\Gamma (X, \mathscr{O}_X)$ is noetherian, that is impossible. Unfortunately it is not guaranteed that $\Gamma (X, \mathscr{O}_X)$ is noetherian even when $X$ is a $k$-scheme of finite type, but at the very least we can conclude that any generator for the category $k$-schemes of finite type has to be very strange.

If you really want a generator, then I suggest looking at the category of varieties over an algebraically closed field. It is also worth pointing out that the category of affine schemes has a cogenerator, because the category of commutative rings has a generator (namely $\mathbb{Z} [x]$). Geometrically this corresponds to the fact that every affine scheme can be embedded in a cartesian power of the affine line.

  • $\begingroup$ "since $X$ is of finite type over $k$, so is $\operatorname{Spec} \Gamma(X, \mathscr{O}_X)$." Unfortunately this isn't true in general. See Ravi Vakil's "An example of a nice variety whose ring of global sections is not finitely generated", math.stanford.edu/~vakil/files/nonfg.pdf $\endgroup$ – Martin Brandenburg Sep 6 '13 at 9:35
  • $\begingroup$ Right. I have replaced the argument with something more heuristic for now. $\endgroup$ – Zhen Lin Sep 6 '13 at 9:39

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