# Characterizing the domain of definition of a rational map in terms of the function field homomorphism

The following is an exercise in Liu (exercise 3.3.13(b), page 111):

Let $f$ be a dominant rational map $X\dashrightarrow Y$ where $X$ and $Y$ are integral schemes of finite type over a locally Noetherian scheme $S$. Show that a point $x\in X$ lies in the domain of definition if and only if there is a point $y\in Y$ such that $\mathscr{O}_{Y,y}$ is dominated by $\mathscr{O}_{X,x}$ under the field extension $K(Y)\to K(X)$.

This reduces in a straightforward way to a the affine case, where it becomes a statement in commutative algebra. The proof of that statement, as far as I can tell, is just as straightforward. The thing that has me worried is that I believe this holds with no Noetherianness condition on $S$ ($S$ can be an arbitrary scheme) and no finite type condition on $X$. Those conditions, of course, are pretty mild, so it's not surprising that Liu would include them anyways, but I want to make sure that I'm not missing a major obstruction.

Here's my statement/idea of proof of that local case:

Let $A$ and $B$ be $R$-algebras, with $A$ finite type, and suppose we have a homomorphism $f:A\to B_\beta$ for some $\beta\in B$. Then for any prime $\mathfrak p$ of B such that $f(A)\subseteq B_\beta \cap B_\mathfrak p$, there is $\beta'\notin \mathfrak p$ such that $f(A)\subseteq B_\beta\cap B_{\beta'}$.

The proof is shorter than the statement. Choose $R$-generators $x_1,\dots,x_n$ for $A$. Then $f$ is given by $f(x_i) = f_i/\beta^n$. The condition $f(A)\subseteq B_\mathfrak p$ means that we have $t_i\notin\mathfrak p$ so that $t_if(x_i)\in B$. If we let $\beta'=\prod t_i$, then $f(A)\subseteq B_\mathfrak p$, Q.E.D.

A note to relate this to global case: having an affine patch of $Y$ whose coordinate ring maps into the local ring is just as good as having a point whose local ring is dominated: we can just pull back the maximal ideal of the local ring, and then localize.

Certainly knowing $A$ can be described by only finitely many relations on the $x_i$ comes only if $S$ is locally Noetherian, but is there any reason we benefit from that? And is there any reason why we would need some sort of finiteness conditions on $X$?

• I would really love to look at this question now (as I am reading Liu's book too ) but unfortunately I'm swamped with stuff for the next 12 hours approximately. I will look at this after that. – user38268 Oct 10 '13 at 23:46
• I think the finite type hypothesis insures that if $f$ is defined at $x$, then it is defined at an open neighborhood of $x$. – Cantlog Oct 16 '13 at 14:06
• Understood: finiteness hypotheses are a common requirement for stalk-local-to-affine-local transitions, since they can be used to get a guarantee that you only need to invert finitely many things to get into the stalk. In this case however, "being defined on some subset/point of $X$" is the same as saying that our function field map puts a coordinate ring on $Y$ into the corresponding subring of $K(X)$. In practice, this only really has to do with finiteness on $Y$, it seems. – Xander Flood Oct 17 '13 at 14:33

In the June 11, 2013 edition of Vakil's FOAG, proposition 6.5.7 is a closely related (read: equivalent) statement we have this statement without f.t. assumptions on $X$, but with $S$ replaced by a field, rather than an arbitrary locally Noetherian scheme.