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 $X$?

  • $\begingroup$ 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. $\endgroup$ – user38268 Oct 10 '13 at 23:46
  • $\begingroup$ I think the finite type hypothesis insures that if $f$ is defined at $x$, then it is defined at an open neighborhood of $x$. $\endgroup$ – Cantlog Oct 16 '13 at 14:06
  • 1
    $\begingroup$ 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. $\endgroup$ – 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.

Seeing as I didn't actually have a particular point of worry in the proof, but rather in the statement, seeing this alternate statement is enough to quell my concerns. (Not to mention the fact that Vakil's proof is essentially the same as the one I gave.)


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