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I'm trying to prove the following: Any rational map $f: \mathbb{A}_\mathbb{C}^2 \dashrightarrow \mathbb{A}_\mathbb{C}^1$ which is defined on a cofinite set of $\mathbb{A}_\mathbb{C}^2$ (and thus open for the Zariski topology being $T_1$) must be also defined on the whole $\mathbb{A}_\mathbb{C}^2$ (and thus, is a regular function).

The result isn't true for rational maps $\mathbb{A}_\mathbb{C}^1 \to \mathbb{A}_\mathbb{C}^1$. The rational map given by $(\mathbb{A}_\mathbb{C}^1\setminus\{0\},\frac{1}{x})$ cannot be defined in any open Zariski neighborhood of $0$. Neither is true if the cofiniteness hypothesis of some set of definition of $f$ is abandoned. Both hypothesis, $f$ having two variables and the cofiniteness one, must be exploited.

Two results come to my mind which I think could come handy for this problem, but I'm not sure how exactly. They are the following:

  1. Any non-constant polynomial of $k[x_1,...,x_n]$ with $k$ an algebraically closed field and $n\geq 2$ has an infinite number of roots (see https://math.stackexchange.com/a/355227/394668).
  2. The identity theorem from the complex variable theory: any two holomorphic functions defined on a common open connected subset $D\subset\mathbb{C}$ and that coincide on a subset $S\subset D$ which has an accumulation point on $D$ must be the same function.

I cannot think of any potential proof which uses the hypothesis of the two variables of $f$. I have no clue on how this could be used.

Any help will be appreciated.

Elías Guisado Villalgordo
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1 Answers1

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A rational function $\Bbb A^2\to \Bbb A^1$ can be represented by an element $f$ of the function field of $\Bbb A^2$. The condition that $f$ is defined everywhere except for a finite set means that $f \in k[x,y]_{\mathfrak{p}}$ for all height one primes $\mathfrak{p}\subset k[x,y]$. As $R=\bigcap_{\mathfrak{p} \text{ of ht 1}} R_\mathfrak{p}$ for $R$ a noetherian normal domain, this implies that $f\in k[x,y]$ or that $f$ is actually regular on $\Bbb A^2$.

The result about the intersection of localization of height one primes can be found in Matsumura's Commutative Algebra, Theorem 38, or as "Algebraic Hartog's Lemma" (11.3.11) in Vakil's notes.

If you're looking for a more elementary proof, one option in this case is to write our representative of the rational map as an element $g/h$ for $g,h\in \Bbb C[x,y]$ and $h$ nonzero. As $\Bbb C[x,y]$ is a UFD, we may assume that $g,h$ are coprime. If $h$ is nonconstant, then every point on $V(h)$, a curve in $\Bbb A^2_\Bbb C$ and thus an infinite set gives that $g/h$ is undefined. But we assumed that $g/h$ was undefined in finitely many places, so $h$ must be constant after all and our map is regular.

KReiser
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  • @Ulam More generally, if $X$ is a normal variety over $k$ and $U\subset X$ is a dense open with $codim(X\setminus U)\geq 2$, then a morphism $U\to \mathbb{A}^1$ extends (uniquely) to a morphism $X\to \mathbb{A}^1$. – Ariyan Javanpeykar Nov 11 '20 at 08:23
  • Why does the condition "$f$ is defined everywhere except for a finite set" is equivalent to "$f\in k[x,y]_\mathfrak{p}$ for all height one primes $\mathfrak{p}\subset k[x,y]$"? Also, I've seen on my edition of Matsumura that the result you say is actually Theorem 11.5. The book has no Theorem 38. – Elías Guisado Villalgordo Nov 11 '20 at 10:24
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    Sorry, I made a mistake - I meant to refer to Matsumura's Commutative Algebra and somehow swapped the reference. For the other question, suppose $\mathfrak{p}$ is a height-one prime. In our case, that determines a curve, $V(\mathfrak{p})$, which has infinitely many points. Pick one of them where $f$ is defined: that means that $f\in k[x,y]m$, where $m$ is the maximal ideal corresponding to our point. As $k[x,y]_m\subset k[x,y]{\mathfrak{p}}$ in the field of fractions, we have the answer you were looking for. – KReiser Nov 11 '20 at 10:52
  • I do feel comfortable with everything except for the result "$R=\bigcap_{\mathfrak{p} \text{ of ht 1}} R_\mathfrak{p}$ for $R$ a noetherian normal domain". I've just looked up the definition for a normal domain (which I understand but didn't see before) but I'm not keen with what Matsumura uses on the proof for this fact, I simply can't understand it. I don't know enough commutative algebra yet. Wouldn't there be a more elementary approach for this problem, perhaps? Maybe for the specific case of $k=\mathbb{C}$, instead of giving a general proof for arbitrary $k$. – Elías Guisado Villalgordo Nov 11 '20 at 11:28
  • @Ulam It might help to try to prove the equality $R=\cap_{p} R_p$ where $p$ runs over all primes of height one when $R$ is a field or a principal ideal domain. – Ariyan Javanpeykar Nov 11 '20 at 17:19
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    @Ulam I've added another explanation which is a bit more elementary. – KReiser Nov 12 '20 at 06:36