Can a rational map $X\leadsto Y$ be defined as a scheme morphism $Z\to Y$ for some $Z$?

Question

Let $X=\operatorname{Spec}(R)$ be an integral scheme with generic point $\eta$ and let $Y$ be a separated scheme.

A rational map $X\leadsto Y$ is a certain equivalence class and it is represented by an actual morphism of schemes $U\to Y$ where $U$ is a dense open subscheme of $X$. I chose the above properties for $X$ and $Y$ since two morphisms of schemes from a reduced scheme to a separated scheme which agree on a dense open subscheme of the domain are the same.

Since $\operatorname{Spec}(\mathcal{O}_{X,\eta})$ is the inverse limit over all the open subschemes $U$ of $X$, every rational map $X\leadsto Y$ should define an actual morphism $\operatorname{Spec}(\mathcal{O}_{X,\eta})\to Y$ of schemes.

Is the data of a rational map $X\leadsto Y$ the same as an actual morphism $$\operatorname{Spec}(\mathcal{O}_{X,\eta})\to Y$$ of schemes?

I think that this is not true (see the example in the comments by Zhen Lin): Every element $f\in \operatorname{Spec}(\mathcal{O}_{X,\eta})$ comes from an dense open subscheme $U$ of $X$ but there is probably no common $U$ for all such elements.

Then, a rational map $X\leadsto Y$ may perhaps be seen as a generalization of a scheme morphism $X\to Y$ but it is more special than a scheme morphism $\operatorname{Spec}(\mathcal{O}_{X,\eta})\to Y$. Is this intuition correct?
In that case, I wonder about this question:

Is there a scheme $Z$ together with scheme morphisms $\operatorname{Spec}(\mathcal{O}_{X,\eta})\to Z\to X$ such that a rational map $X\leadsto Y$ can be defined as an actual scheme morphism from $Z$ to $Y$?

Answer 1

If you want to interpret rational maps in terms of fields, you have to reverse your arrows!

a) Suppose we have a rational map $f:X\leadsto Y$ between, say, two integral schemes over a field $k$.
If the map is dominant it induces a morphism of $k$-extensions in the reverse direction between the stalks at the generic points i.e. the rational function fields: $$f^*:\mathcal O_{Y,\eta}=Rat(Y)\to \mathcal O_{X,\xi}=Rat(X)$$
This way we obtain in particular an equivalence between the category of $k$-algebraic varieties and dominant morphisms and the category of finitely generated extensions of $k$.

b) So that solves the problem: given a question on rational maps, translate it into a question of field extensions and the result drops out, right?
Completely wrong: the algebraic problem will be so difficult that it probably cannot be solved at all!
Actually, it is the other way round: problems on transcendental extensions of fields have often be solved by expressing them into an algebraic geometry question which can be solved.
A spectacular example is Lüroth's problem: is a subfield of the purely transcendental extension $k\subset k(X_1,\cdots,X_n)$ also purely transcendental ?
The case $n=1$ was solved by Lüroth and is often taught in undergraduate courses.
The case $n=2$ was solved by algebraic geomety : the answer is "yes" in zero characteristic (Castelnuovo) and "no" in positive characteristic (Zariski).
Finally in dimensions $\geq 3$, the answer is "no", even over $\mathbb C$: this was proved independently by the three pairs of redoubtable algebraic geometers Artin-Mumford, Clemens-Griffiths and Manin-Ivskovskih, using three completely different geometric methods.