# A functorial definition of a projective curve

Define a functorial scheme as in "Two functorial definitions of schemes".
Here there is one big problem.
We first need to define a projective scheme, but we also need a closed subscheme.
(cf. Projective scheme)
So The definition of a closed subscheme requires a structure sheaf.
However the definition of functorial scheme directly don't have a structure sheaf, therefore a closed subscheme is not defined.
For this reason, like Hartshorne, we also can't define projective morphism as a composition of closed immersion into projective space and natural projection because a closed immersion is not defined.

Is it possible to define a projective curve?
Is there a PDF or something that does that?

Thanks in advance.

• A closed immersion is the same thing as a proper monomorphism, which can definitely be defined from a functor of points perspective (see the `valuative criterion for properness) May 15, 2020 at 12:54
• Thank you very much for answering my questions. It seems that it is possible to define a projective scheme by defining closed immersion in terms of proper monomorphism and closed subscheme in terms of the scheme in which the closed immersion exists. May 16, 2020 at 3:50

## 1 Answer

Every projective curve admits a finite morphism to $$\Bbb P^1_k$$. Conversely, every $$k$$-scheme admitting a surjective finite morphism to $$\Bbb P^1_k$$ is one-dimensional. So if we can describe a surjective finite morphism categorically and figure out how to describe irreducibility in categorical terms, we're done.

As mentioned in the comments, proper morphisms can be defined using category-theoretic terms via the valuative criteria for properness. Surjectivity can also be checked categorically: $$f:X\to Y$$ is surjective iff for every field $$K$$ and $$y\in Y(K)$$, there is a field extension $$L/K$$ and $$x\in X(L)$$ whose image under $$X(L)\to Y(L)$$ is the image of $$y$$ under $$Y(K)\to Y(L)$$. If we can show that quasi-finiteness is a categorical property, then we'll be done as finite is proper + quasi-finite.

In order to check quasi-finiteness of a morphism $$X\to Y$$, it suffices to check that the fiber $$X_y$$ is quasi-finite for every point $$y\in Y$$. This is categorical: for any morphism $$\operatorname{Spec} K\to Y$$ where $$K$$ is a field, form the fiber product $$X_K:= X\times_Y \operatorname{Spec} K$$, and then count $$X_K(K)$$: if it's finite for all choices of $$K$$ and morphism $$\operatorname{Spec} K\to Y$$ then our morphism $$X\to Y$$ is quasi-finite.

Now to figure out irreducibility. We can certainly detect connectedness via categorical data: a $$k$$-scheme $$X$$ is disconnected iff there exists an epimorphism $$X\to \Bbb A^0_k\sqcup \Bbb A^0_k$$. I claim that this is actually enough to detect irreducibility, too: a scheme $$X$$ is irreducible iff for every closed immersion $$Z\to X$$ we have $$X\setminus Z$$ is connected. As closed immersions can be described categorically and describing complements is also categorical, this shows that irreducibility can be defined using categorical data.

In summary, the projective curves over $$k$$ are exactly the irreducible $$k$$-schemes admitting a finite morphism to $$\Bbb P^1_k$$, and all of these conditions are purely categorical.

• Thank you for your answer! I think it's beautiful to be able to define a purely categorical projective scheme. It would be a waste that there is no textbook with this perspective. May 17, 2020 at 12:25
• @undertate You're welcome. Please note that I am not claiming there's no textbook that does this - personally I don't think about the functor of points perspective that much, so it's very possible there is a book that does this but I'm unfamiliar with it. If this answers your question, please consider accepting the answer via clicking the checkmark over by the votes. May 17, 2020 at 19:08
• I understand. I will try to find it.And, I voted. May 18, 2020 at 13:47