3
$\begingroup$

Let $f(z_1,z_2,\ldots,z_n)$ be a function on $\mathbf{C}^n$ such that for all $i$, the restriction $$ [z_i\mapsto f(z_1,z_2,\ldots,z_n)] $$ is a "rational function".

(added: to be precise here one should allow $(z_2,z_3,\ldots,z_n)$ to avoid a closed exceptional variety $E\subseteq\mathbf{C}^{n-1}$, nowhere dense in $\mathbf{C}^{n-1}$ such that for every point $P\in E$ and every open polydisc neighborhood $U$ of $P$, $U\backslash E$ is connected --- see Sakai's paper mentioned below)

Then I would expect $$ [(z_1,\ldots,z_n)\mapsto f(z_1,\ldots,z_n)] $$ to be rational as well. There should be somewhere in the literature an elementary proof of this fact....

Note that if we replace in the above statement the word "rational" by "holomorphic" then the result holds true (this is the well-known result due to Hartogs) or if we replace it by "meromorphic" it is again true (due to Sakai 1957).

Are there elementary proofs of Hartogs' and Sakai theorems that only use the usual basics which are covered in a first course in one complex variable ?

$\endgroup$
3
$\begingroup$

Let us prove the desired result for $n=2$. We have $$f(x,y)=\frac{\sum_{i=0}^m a_i(y)x^i}{\sum_{i=0}^k b_i(y)x^i}=r_x(y),\tag{1}$$ where the $a_i$'s and $b_i$'s are some functions and, for each $x$, $r_x$ is a rational function. We want to show that $f$ is a rational function. Without loss of generality (wlog), $b_0=1$. It is then enough to verify the claim that the $a_i$'s and $b_i$'s are rational functions.

Let us do this by induction on the (total) degree $d:=m+k$ of $f$ in $x$. If $d=0$, then the claim is obvious. Suppose now that $d=m+k\ge1$. Wlog, $m\ge k$ (or take the reciprocal of $f$). Let $$g(x,y):=\frac{r_x(y)-r_0(y)}x=\frac{f(x,y)-r_0(y)}x =\frac{\sum_{j=0}^{m-1}c_j(y)x^j}{\sum_{i=0}^k b_i(y)x^i},$$ where $c_j(y):=a_{j+1}(y)-b_{j+1}(y)r_0(y)$, with $b_i(y):=0$ for $i>k$. Then $g(x,y)$ is of degree $<d=m+k$ in $x$ and is rational in $x$ and in $y$. So, by induction, all the $b_i$'s and all the $c_j$'s are rational functions, and hence all the $a_i$'s are rational functions, just as claimed.


As Wojowu noted, the above argument tacitly assumes that $m=m_y$ and $k=k_y$ do not depend on $y$. Also, (for uncountable fields) Wojowu showed how to fix this argument. His reasoning can now be used to prove the desired result for any $n\ge2$. This can be done by induction on $n$, as sketched below.

As noted by the OP, the function $f$ is meromorphic and hence defined on a nonempty open subset $E$ of $\mathbb C^n$. Let $x:=z_1$ and $y:=z_2,\dots,z_n$. For natural $d$, let $S_d:=\{y\colon\exists(x,y)\in E, D_y(f)\le d\}$, where $D_y(f)=m_y+k_y$ and $m_y=m,k_y=k$ with $m,k$ as in (1). The sets $S_d$ are closed in the open set $U:=\{y\colon\exists(x,y)\in E\}\subseteq\mathbb C^{n-1}$ and $\bigcup_d S_d=U$. So, by the Baire category theorem, for some natural $p$ the set $S_p$ contains a nonempty open ball $B$. Fixing now $z_3,\dots,z_n$ and using the above argument, we see that, for each $i$, $a_i(y)=a_i(z_2,z_3,\dots,z_n)$ is rational in $z_2$ (that is, in $z_2\in\{t\colon (t,z_3,\dots,z_n)\in B\}$). Similarly, $a_i(y)=a_i(z_2,\dots,z_n)$ is rational in $z_j$ for each $j\in\{2,\dots,n\}$. So, by induction on $n$, $a_i(y)=a_i(z_2,\dots,z_n)$ is rational in $z_2,\dots,z_n$, for each $i$. Similarly, $b_i(y)=b_i(z_2,\dots,z_n)$ is rational in $z_2,\dots,z_n$, for each $i$. Thus, $f$ is rational.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ This answer assumes that the degrees $m,k$ are independent of $y$, which is not a priori clear. Indeed, the result is false if $F$ is a countably infinite field because of this. This proof can be salvaged by using a cardinality argument to pick out an infinite set of $y$'s on which $m,k$ are indeed bounded and going from there $\endgroup$ – Wojowu Oct 6 at 8:37
  • $\begingroup$ Thanks @Iosif for the nice inductive argument. I also just realized that my original statement is a bit sloppy since it may well happen that for a fixed $z_2$, $[z_1\mapsto f(z_1,z_2)]=\infty$ which is not rational per se (e.g. $1/z_2$). The same issue arises in more variables....it is not completely clear to me how to correct that. May be $(z_2,\ldots,z_{n} )$ should be allowed to avoid a finite union of analytic subvarieties in $\mathbf{C}^{n-1}$.... $\endgroup$ – Hugo Chapdelaine Oct 6 at 11:35
  • $\begingroup$ @Iosif, should you not multiply $r_0(y)$ by $x^{m-k}$ if you want the leading term in $x$ to cancel ? $\endgroup$ – Hugo Chapdelaine Oct 6 at 12:49
  • $\begingroup$ @HugoChapdelaine : I am canceling the term free of $x$, rather than the leading term in $x$. $\endgroup$ – Iosif Pinelis Oct 6 at 13:06
  • $\begingroup$ Ok got it since after all you want your numerator to be a again a polynomial in $x$ after having divided it by $x$. $\endgroup$ – Hugo Chapdelaine Oct 6 at 13:14
4
$\begingroup$

This is an extended comment to Iosif's answer, explaining why it's not true over arbitrary fields, and how to patch the argument.

Indeed, the statement is not true if $F$ is a countably infinite field. For let $a_1,a_2,\dots$ be an enumeration of all elements of $F$. Consider the function $$f(x,y)=\sum_{n=1}^\infty\prod_{i=1}^n(x-a_i)(y-a_i)$$ (which makes sense for any $x,y\in F$ since all but finitely many terms will be zero). For any fixed $y=a_k$, all the terms with $n\geq k$ will vanish, so $f(x,a_k)$ is a polynomial, same for fixed $x$. However, $f(x,y)$ itself is not rational because the number of roots of $f(x,a_k)$ is unbounded, while for a rational function it'd be bounded by the degree of the numerator.

However, if $F$ is uncountable, the argument can be fixed. Indeed, for each $d\in\mathbb N$, let $S_d$ be the set of $y$ for which $f(x,y)$, as a rational function of $x$, has numerator and denominator of degree at most $d$. Since the union of all $S_d$ is all of $F$, which is uncountable, one of the $S_d$ has to be infinite (even uncountable). From there we can essentially repeat Iosif's argument to deduce $f$ restricted to $F\times S_d$ agrees with a rational function $g:F^2\to F$. Lastly we note that for any $x$, $f(x,y)$ and $g(x,y)$ agree for $y\in S_d$, which means, since both are rational, that they agree everywhere, so $f$ is rational.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks @Wojowu for this great example! $\endgroup$ – Hugo Chapdelaine Oct 6 at 11:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.