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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. (So, as Remy says, this argument does not apply to 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. But nothing in the argument assumes that $K$ is a field, so it is quite conceivable that even in such a topology, there are non-fields with the property in question.

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.

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. (So, as Remy says, this argument does not apply to the fpqc topology.)

Indeed, if $\phi : K \to C$ is any homomorphism that is not a cover and $F$ is the sheaf image of $\operatorname{Spec} \phi : \operatorname{Spec} C \to \operatorname{Spec} K$, 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 $\operatorname{Spec} K \to \operatorname{Spec} C$ is either a cover or has $C$ trivial. But nothing in the argument assumes that $K$ is a field, so it is quite conceivable that even in such a topology, there are non-fields with the property in question.

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. (So, as Remy says, this argument does not apply to 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. But nothing in the argument assumes that $K$ is a field, so it is quite conceivable that even in such a topology, there are non-fields with the property in question.

There is no hope of this for any subcanonical topology for which fields are local rings, i.e. rings $A$ such that every covering family of $\operatorname{Spec} A$ contains a split epimorphism. (Fields are local rings for the fpqc topology, so this is true for all coarser topologies, e.g. the Zariski topology.)

Indeed, for any field $K$, we can find a non-trivial subsheaf of $\operatorname{Spec} K$. Let $\phi : K \to L$ be any non-trivial field extension and let $F$ be the sheaf image of $\operatorname{Spec} \phi : \operatorname{Spec} L \to \operatorname{Spec} K$. Then $F$ is a non-trivial subsheaf of $\operatorname{Spec} K$: it is non-empty, since $\phi \in F (L)$, and $\textrm{id}_K \notin F (K)$, because 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 $\phi : K \to L$ – but we assumed any cover of $\operatorname{Spec} K$ splits, so $\operatorname{id}_K \in F (K)$ if and only if $\phi : K \to L$ splits, and a field extension splits if and only if it is trivial.

(Note that in the above argument it is not essential that $L$ be a field. It suffices to consider any homomorphism $K \to C$ that is not a split monomorphism, with $C$ non-trivial.)

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. (So, as Remy says, this argument does not apply to the fpqc topology.)

Indeed, if $\phi : K \to C$ is any homomorphism that is not a cover and $F$ is the sheaf image of $\operatorname{Spec} \phi : \operatorname{Spec} C \to \operatorname{Spec} K$, 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 $\operatorname{Spec} K \to \operatorname{Spec} C$ is either a cover or has $C$ trivial. But nothing in the argument assumes that $K$ is a field, so it is quite conceivable that even in such a topology, there are non-fields with the property in question.