Extending functors between K-algebras to schemes

Assume we have $$K$$ and $$L$$ (comm.) rings, and we have a functor $$F$$ from the category of $$K$$-Algebras to the category of $$L$$-Algebras (I work only with commutative rings). What conditions need to satisfy this functor in order to "extend" to a functor $$F^*$$ form the category of $$\operatorname{Spec}(L)$$-Schemes to the category of $$\operatorname{Spec}(K)$$-Schemes?

It is clear that the extension, if it exists, is unique, by Yoneda's lemma.

On the other hand, I am not sure any functor will work; it seems to me we need that the functor sends "open affine immersions" to "open affine immersions", i.e. flat morphisms $$g:R\to S$$ of finite presentation which are epimorphism, to the same type of morphisms, in order to be able to "patch" the image of an affine covering of a $$\operatorname{Spec}(L)$$-scheme to get a $$\operatorname{Spec}(K)$$-scheme. But I don't see how to do it if the intersections are not affine, and one needs to get an affine covering of the intersections, and so on. Moreover, I don't know if it is sufficient to show this for principal open schemes, so for maps of the type $$g:R\to R[1/a]$$ for $$a\in R$$.

I am interested, for example, in the functor sending any ring $$R$$ to its total ring of fractions $$Q(R)$$. I have not seen this mentioned in the scheme world. But such a general result, if it exists in the literature, could be useful to get e.g. the base change functor, or the reduced subcheme functor, etc.

• It is not necessary that the functor preserve open embeddings or open coverings or intersections – after all you could start with an arbitrarily bad functor of schemes that happens to preserve affine schemes. Commented Jun 19, 2021 at 11:18
• I voted to close since the question needs more details. What do you mean by extending exactly? For example, do you require that the extension has some properties? (If not, the question is a bit meaningless.) Also, I am very confused about the last paragraph, there is a well-known sheaf $K_X$ of meromorphic functions for any scheme $X$. Commented Jun 21, 2021 at 13:56
• May be I was not sufficiently clear on my question and my example, which is not the sheaf of meromorphic functions. Given any scheme $X$, I want a scheme $Q(X)$ and a morphism $Q(X) \to X$ which, restricted to affine schemes is the "total ring of fractions". Commented Jun 22, 2021 at 11:26
• Yes, it's $\mathrm{Spec}_X(K_X)$. Again, what do you mean by extending? (I am even more confused why you accepted the answer below, which does not address the question as how I understand it so far.) Commented Jun 27, 2021 at 16:57
• @Martin Brandenburg I must say that after your comments and the answer I realize that there could be functors from the category of S-schemes to elsewhere that coincide on affine schemes, being different. So one needs to ask for some property of the functor, like being a Zariski sheaf. This is why I accepted the answer. Commented Jun 27, 2021 at 18:22

I would prefer to reverse the question: instead of starting with a functor defined on affine schemes and trying to extend it to all schemes, you can consider a functor defined on all schemes and ask whether it can be uniquely reconstructed from its restriction to affine schemes. Also, the fact that the target category happened to be schemes in your question is not really important. So we may choose any functor $$F \colon \mathrm{Sch}^{op} \to C$$, where $$C$$ is any category, and ask if it is determined by its restriction to $$\mathrm{Aff}^{op} \to C$$.
The natural condition that implies this is that $$F$$ is a sheaf for the Zariski topology on $$\mathrm{Sch}$$, i.e. if $$X = \bigcup U_i$$ is a Zariski open cover of a scheme $$X$$ then $$F(X)$$ is the equalizer of the two maps $$\prod_i F(U_i) \to \prod_{i,j} F(U_i \cap U_j)$$. Let's assume that $$C$$ has all limits, so the products and equalizer exist. In this case there's an equivalence of categories between functors $$\mathrm{Sch}^{op} \to C$$ satisfying the sheaf condition for Zariski open covers, and functors $$\mathrm{Aff}^{op} \to C$$ satisfying the same sheaf condition. Moreover, as suggested in your question, both are also equivalent to functors $$\mathrm{Aff}^{op} \to C$$ satisfying the sheaf condition for open covers by principal open subsets.
• Re: the last sentence. More precisely, the equivalence between sheaves on $\mathrm{Sch}$ and sheaves on $\mathrm{Aff}$ comes from the comparison lemma. The fact that we only need to check the sheaf condition on $\mathrm{Aff}$ for covers by principal open subsets is because these two pretopologies on $\mathrm{Aff}$ generate the same Grothendieck topology. Commented Jun 22, 2021 at 11:23