$\DeclareMathOperator\Spec{Spec}\newcommand\Ring{\mathrm{Ring}}\newcommand\op{^\text{op}}\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Sh{Sh}$In the category of schemes the objects of the form $\Spec(K)$ with $K$ a field can be characterized as follows. They are precisely the non-empty schemes which have no proper subobjects in $Sch$.

Consider now the category $\Ring\op$ with its various Grothendieck topologies (Zariski, étale, fpqc, etc.). From those we get various big sheaf topoi which contain the category of schemes as a full subcategory. As a functor, $\Spec(K)$ is the representable presheaf $\Hom_{\Ring}(K,-)$.

Even though $\Spec(K)$ has no proper subschemes when $K$ is a field, it can have non-trivial sub-pre-sheaves. Can it have non-trivial Zariski sub-sheaves? If yes, can we switch to one of the finer topologies to prevent this? Is one of the topologies on $\Ring\op$ fine enough, so that the sheaves of the form $\Spec(K)$ with $K$ a field are precisely the objects with a trivial subobject lattice in $\Sh(\Ring\op,\text{sth})$?


1 Answer 1


There is no hope for this in any subcanonical topology coarser than the fppf topology, or more generally, any subcanonical topology in which morphisms $\operatorname{Spec} C \to \operatorname{Spec} K$ are not automatically covers when $K$ is a field and $C$ is non-trivial. But it is true with the fpqc topology.

Indeed, if $\operatorname{Spec} \phi : \operatorname{Spec} C \to \operatorname{Spec} K$ is any morphism that is not a cover and $F$ is the sheaf image, then $F$ is not the top subsheaf of $\operatorname{Spec} K$; if we further assume that $A$ is not trivial then $F$ is also not the bottom subsheaf of $\operatorname{Spec} K$. (We have $\alpha \in F (A)$ if and only if there exist a cover of $\operatorname{Spec} A$ by affines $\operatorname{Spec} \beta_i : \operatorname{Spec} B_i \to \operatorname{Spec} A$ such that each $\beta_i \circ \alpha : K \to B_i$ factors through $\phi : K \to C$. So $\textrm{id}_K \in F (K)$ if and only if $\operatorname{Spec} \phi : \operatorname{Spec} C \to \operatorname{Spec} K$ is a cover.)

Conversely, by the above argument, in any subcanonical topology such that $\operatorname{Spec} K$ only has the top and bottom subsheaves, it must be that every morphism $\operatorname{Spec} C \to \operatorname{Spec} K$ is either a cover or has $C$ trivial. Nothing in the argument assumes that $K$ is a field, but if we assume that covers are faithfully flat and $K$ is non-trivial, it will follow that $K$ is a field: because $K$ is non-trivial, there exist a field $L$ and a ring homomorphism $K \to L$, and $\operatorname{Spec} L \to \operatorname{Spec} K$ is a cover so $K \to L$ is faithfully flat, hence $K$ is an integral domain with a unique prime ideal, i.e. a field. Thus, with the fpqc topology, $\operatorname{Spec} K$ has only the top and bottom subsheaves if and only if $K$ is a field.

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    $\begingroup$ Sorry, isn't $\phi\colon\operatorname{Spec}A\to\operatorname{Spec}K$ itself an fpqc cover for any nonzero $K$-algebra $A$? Then $h_{\operatorname{Spec}\phi}\colon h_{\operatorname{Spec}A}\to h_{\operatorname{Spec}K}$ is surjective as sheaves. In the Zariski, étale, or fppf cases I agree with you: take $K\to L$ a field extension that is not an isomorphism, not étale, or not finitely presented, respectively. Then $K\to B_i$ cannot factor through $K\to L$ since this forces an injection $L\to B_i$, so if $K\to B_i$ is an isomorphism, étale, or finitely presented, the same goes for $K\to L$. $\endgroup$ Dec 8, 2022 at 16:35
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    $\begingroup$ Oops. I was confused about what local rings in these topologies were. Let me fix that... $\endgroup$
    – Zhen Lin
    Dec 8, 2022 at 22:06
  • $\begingroup$ Really nice! :) $\endgroup$
    – Nico
    Dec 14, 2022 at 21:02

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