Sorry if this is standard, but I couldn't find a reference.
Background:
- Let $F: \textbf{Aff}^\text{op} \to \textbf{Set}$ be a functor. If $R$ is a ring (commutative with $1$) and $I \subseteq R$ is an ideal, one has the subfunctor of $\text{Spec}(R)$ given by
$$S \mapsto \{ \varphi \in Hom(R , S) : \varphi(I) = 0 \}.$$
A subfunctor $G$ of $F$ is closed if for every ring $R$ and every morphism $\text{Spec}(R) \to F$ the inverse image of $G$ is a subfunctor of the form above.
- The intersection of a family of subfunctors of $F$ is defined in the naive way.
Actual Question: What's the definition of the closure of a subfunctor? Does the naive construction (intersection of all closed subfunctors which contain $G$) work?
If it simplifies matters, I'm particularly interested in the case when $F$ is a sheaf in the fpqc topology on $\textbf{Aff}$ (but NOT necessarily representable by a scheme).
In addition to answers, any references would be much appreciated! :) Thanks so much!!
The answer is "yes"! The closure of a subfunctor is defined to be the intersection of all closed subfunctors containing it. See, for example, Demazure and Gabriel's excellent book: "Introduction to Algebraic Geometry and Algebraic Groups".