Scheme of finite type over a field $K$ v.s. $K$-scheme

I'm lost in some definitions about schemes. I have some trouble about two definitions of a scheme of finite type over $K$, for an alg.closed field $K$.

Version 1 (Hartshorn) : a scheme of finite type over $K$ is a scheme $X$ together with a morphism $X \to K$, where $X$ is a scheme (a locally ringed space $(X, \mathcal{O})$ with a cover of spectra of rings) and for me $K$ is the (ridiculous ?) scheme $\text{Spec} K = \{ (0) \}$ with the sheaf sending $\{(0)\}$ to $K$. So the morphism is a continuous function $f \colon X \to \{\ast\}$ irrelevant since only one possible, and a finite morphism of sheaves, that is reduced to a single ring hm $f^{\sharp} \colon K \to \mathcal{O}_X(X)$.

Version 2 : a scheme of finite type over $K$ is a scheme with a finite cover of spectra of finitely generated $K$-algebras.

For example, let's assume $X$ is affine, so $X = \text{Spec} R$ for some ring $R$, so in the first version it is the data of $\text{Spec} R$ with its topology and the sheaf associated, and we add a finite morphism $K \to \mathcal{O}_X(X) = R$. In the second version, it is a scheme of the form $\text{Spec}(A)$ for some finitely generated $K$-algebra $A$.

Oh, actually i think this makes the bridge between the two notions... Well... Sorry. I have however another question : While studying algebraic groups in the Borel, he considers $K$-schemes, which are almost schemes of finite type over $K$, in the sense that the topological space is not the whole spectrum $\text{Spec} A$ but only $\text{max} A = \text{Spec}_K A$ of maximal ideals of the finitely generated $K$-algebra $A$. So clearly the topological space contains "less" points, what does it change ? Why does he do that ? There is a bijection between $\text{max} A$ and $\text{Hom}_K (A,K)$, but what does it bring along ? Because we lose the functoriality (inverse image of maximal ideal is not maximal) and the result is not a scheme anymore...

Sorry for this not linear question, I hope it's understandable, or I'll edit or delete.. Thanks for any hint or piece of information !! Bogdan

P.S. Actually,

• I wouldn't call $\operatorname{Spec} K$ ridiculous. The map of sheaves carries real information, as you yourself say! – Dylan Moreland Nov 13 '11 at 21:41
• I'll try to write more later, but I think Borel often works over an algebraically closed field in that book. In that case, one does have functoriality for $\operatorname{maxSpec}$ applied to f.g. $K$-algebras. See Chapter 3 of Milne's notes for a clean presentation of this. – Dylan Moreland Nov 13 '11 at 21:53
• Thanks ! Yeah i'm looking there. He also does at some point the switch to only considering max ideals. I'll try to understand why, can't see it for now – Bogdan Nov 13 '11 at 22:03
• I guess the point is that it feels simpler. The points of your space correspond to the points you would plot if you could plot a variety over an algebraically closed field. And moreover you get a category anti-equivalent to (maybe I need more adjectives here) f.g. $k$-algebras (so you could say that the maxSpec with its sheaf knows everything about the ring) and this is what you wanted. – Dylan Moreland Nov 13 '11 at 22:09
• Haha yeah actually i've just seen this equivalence of categories. Too lazy to work all the adjectives out for the moment though, thanks for the help :) – Bogdan Nov 14 '11 at 21:46

Technically Borel can get away with that approach because for a scheme $X$ of finite type over a field $K$, the subset of closed points $X_{cl} \subset X$ is very dense in $X$.
This means that the restriction map $Open(X) \to Open (X_{cl}): U \mapsto U\cap X_{cl}$ is bijective.
The reason for that is that a finitely generated algebra $A$ over $K$ is a Jacobson ring, meaning that every prime ideal in $A$ is the intersection of the maximal ideals which contain it.
And for Jacobson rings we actually have functoriality: given a morphism $A\to B$ between two Jacobson rings, the inverse image of a maximal ideal of $B$ is a maximal ideal of $A$.