We define a scheme as in "Two functorial definitions of schemes". That is, for the category $\textbf {Psh}=\operatorname{Fun}(\textbf{Ring},\textbf{Set})$ with any Grothendieck topology we define a scheme to be such an object $X$ in $\textbf{Psh}$ which has an open coverings with functors $h_R$ and for which $h_X$:$\textbf{Psh}^{op}$$\textbf{→Set}$ is a sheaf in the Grothendieck topology. I want to define a generic point. A point of a scheme $X$ is defined as for some ring $R$ an element $x \in X(R)$. Although a generic point usually is defined as $\overline{\{x\}}$, in the above definition, we have not thought an ordinary topology space, whence we don't know a closure. Can we define a generic point?
A generic point of a functorial scheme
algebraic-geometrycategory-theory
Best Answer
Yes we can, and it is quite elegant. I will go through the setup first to make sure we talk about the same thing. We work in the category $\text{Sh}(\text{Aff,Zar})$ of sheaves on the Zariski site. I will call its objects Zariski-local functor. Since $\text{Aff}$ is the opposite of $\text{cRing}$, a Zariski-local functor is just a functor $\text{cRing}\to \text{Set}$ which satisfies a gluing condition. (If we want the category of Zariski-local functors to be a topos, then we have to play around with universes or chose a smaller site of definition.) We write $\text{Spec} A$ for the representable functor $\text{Hom}_{\text{cRing}}(A,-)$. The Zariski topology is subcanonical and $\text{Spec} A$ is Zariski local.
The open subobjects of $\text{Spec}A$ are by definition those of the form $D(\mathfrak a)$ where $\mathfrak a$ is an ideal of $A$. The functor $D(\mathfrak a)$ is defined through the following formula. $$D(\mathfrak a)R=\{\phi:A\to R\,|\,\text{$\phi(\mathfrak a)$ generates the unit ideal in $R$ }\}$$ One can check that all open subfunctors are actually joins of basic open subfunctors of the from $D(f)$. Now a subobject $U\hookrightarrow X$ of a Zariski-local functor $X$ is defined to be open iff each pullback to an affine scheme is open in that scheme. It is the finest topology on $X$ such that each test map $\text{Spec} R\to X$ is continuous. A scheme is a Zariski-local functor which can be covered by open affines.
Functions
We write $\mathbb A$ for the forgetful functor. It is a ring object internal to the category of schemes, and it is called the affine line. Note that $\mathbb A = \text{Spec }\mathbb Z[x]$ naturally. We write $\mathcal OX$ for the set of morphisms $X\to \mathbb A$. The ring structure of $\mathbb A$ turns $\mathcal OX$ into a ring. There is a fundamental adjunction $\text{Hom}(X,\text{Spec}A) = \text{Hom}(A,\mathcal OX)$ which is very easy to describe in the functorial approach. We get unit and counit morphisms of the form $X\to \text{Spec }\mathcal OX$ and $A\to \mathcal O \,\text{Spec}A$. The latter map is always an isomorphism and the former is an isomorphism if and only if $X$ is affine.
Points and primes
To associate a topological space $|X|$ to a scheme one proceeds as follows. A point $p$ of $X$ is an equivalence class of field valued generalised point $\text{Spec}K\to X$, where two such field valued points are equivalent if there is a third field valued generalised point which factors through both. One can show that all representatives of a class $p$ are contained in exactly the same subschemes of $X$, so the equivalence relation makes sense intuitively. The topology of $X$ puts a topology onto the set $|X|$ of such equivalence classes. If $X=\text{Spec} A$ is affine, then one discovers that $|\text{Spec}A|$ is precisely the set of prime ideals of $A$ with its usually topology. More importantly one can show that each class $p$ of $|X|$ has a canonical representative $\text{Spec }\kappa(p)\to X$. It has the universal property that it is the least representative in $p$. Any other representative of $p$ factors uniquely through it (i.e. is $\text{Spec }\kappa (p)\to X$ precomposed with a field extension if you so like). The morphism $\text{Spec }\kappa(p)\to X$ is monic. In case of an affine functor the morphism $\text{Spec } \kappa(p)\to \text{Spec }A$ is precisely the one you think it is.
Germs and local rings
Forming germs in the functor of points approach is very easy. Say you have a scheme $X$ and a point $p$ of $|X|$. Then you just form the intersection of all open subspaces of $X$ which contain $p$ and call the result the germ at $p$. Let us denote that space by $X_p$. Explicitly $$X_p(R) = \{r\in X(R)\,|\, \text{the associated $\text{Spec} R\to X$ factors through the same opens as $\text{Spec }\kappa(p)\to X$}\} $$ Clearly the germ at a point $p$ is a subobject $X_p\hookrightarrow X$. To get the usual formula of $X_p$ as a spec of a local ring you can proceed as follows. Note that we have formed $X_p$ as a limit over a diagram of open subfunctors of $X$. Since the open affines form a basis, we would have gotten the same $X_p$ if we would have taken the intersection over those. Now limits in $\text{Aff}$ are the same as limits in the sheaf topos, so we could have taken the limit in $\text{Aff}$ instead. But this is the same as taking the colimit over the associated rings in $\text{cRing }$, and you end up with the usual construction. In particular $X_p$ is affine and the natural map $X_p \to \text{Spec }\mathcal OX_p$ is an isomorphism. To use the standard algebraic geometry notation we write $\mathcal O_{X,p}$ for the ring $\mathcal OX_p$. The ring $\mathcal O_{X,p}$ is local, which can be seen by showing it for affines first and then noting that we get the same $X_p$ if we first pass to an open affine which contains $p$.
The germ $X_p\to X$ at a point also has a universal property. It states that whenever $(R,\mathfrak m)$ is a local ring and $\text{Spec }R\to X$ a morphism such that $\text{Spec }\kappa(\mathfrak m)\to \text{Spec }R\to X$ is a representative of $p$, then $\text{Spec }R \to X$ factors (necessarily uniquely) through the germ $X_p\to X$. In this sense $X_p$ is the larges local subobject around $p$.
Generic points, specialisation and generalization
A generic point of a scheme is a point which is dense in $X$. In the functor of points approach we can define it as follows. A generic point of a scheme $X$ is a field valued point $\text{Spec }K\to X$ which factors through every non-empty open subscheme of $X$. If we are being extra careful then we could replace $\text{Spec }K\to X$ by its canonical best representative $\text{Spec }\kappa(p)\to X$ through which it factors.
One can show that a scheme has a generic point if and only if it is irreducible and that the generic point is unique (up to the equivalence of field valued points which we described when we constructed $|X|$).
Thus it makes sense to speak of THE generic point $\text{Spec }\kappa(\eta)\to X$ of an irreducible scheme also in the functor of points approach. We just take the best representative from the class of the generic point in $|X|$.
One can show that a point $p$ is more general than a point $q$ if and only if $\text{Spec }\kappa(p)\to X$ lies in the germ $X_q\to X$ of $q$. So here is a second definition of a generic point. A generic point of $X$ is a point $p$ of $|X|$ (or its canonical representative which lies in every germ of the space $X$.
Sorry, this was a lot of info-dumping. If you need more explanation to one of the steps, write me a comment. Also your question is two years old, but maybe you are still interested.
Edit. Here is how you can see that $Spec$ and $\mathcal O$ are adjoint. Assume you are given a morphism $\phi: X\to Spec(A)$ of functors on $Ring$. How can you produce a corresponding map $\phi^\#:A\to \mathcal O(X)$ of rings. Well, given $a\in A$, you need to define a morphism $\phi^\#(a):X\to \mathbb A$ of functors. Such a morphism is a natural transformation, which means that you have to defnine its component $\phi^\# a\, R: X(R)\to \mathbb A(R)$ for each ring $R$. The functor $\mathbb A$ is the forgetful functor, and so $\mathbb A(R)= R$ as a set. This means you need to find an element $\phi^\# a R r\in R$ given an element $a\in A$, a ring $R$ and an element $r\in X(R)$. What you have is $\phi: X\to Spec(A)$. You can easily see that your only option is the following definition of $\phi^\#$: $$\phi^\# a\, R\, r = \phi\, R \, r\, a$$ You can check that it makes sense. In a similar way, you can figure out how to get a transformation $X\to Spec(A)$ from a ring morphism $A\to \mathcal O(X)$.