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 ?



On a variation of Hartogs' separate analyticity theorem
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.


