# Subsheaves of Spec K, K a field


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 $$\Sh(\Ring\op,\text{sth})$$?

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.
• 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 $K\to L$. Dec 8, 2022 at 16:35