Uniformly bounded separate rationality implies rationality

For $n=0$ the function is constant. For $n=1$ the hypothesis is the conclusion. For $n\ge 2$, put $d=\lfloor D\rfloor$; since polynomial degrees are integers, every degree bound in the hypothesis is then at most $d$. Proceed by induction. Choose $2d+1$ distinct constants $a_i$ in the first coordinate such that the slices $U_i=U\cap \{z_1=a_i\}$ are nonempty Zariski open and inherit the coordinate-line hypotheses in the remaining variables. Require also that none of the slices is identically a polar slice of $f$, and that the generic first-coordinate restrictions are finite at all nodes. These additional choices are possible locally: represent $f=g\mathrm{/}h$ near a point where $h\mathrm{\ne }0$ and choose the nodes in that first-coordinate neighborhood; analytic uniqueness then prevents an entire chosen slice from being polar. Such constants exist: each exceptional algebraic set is defined by finitely many nonzero polynomials; a polynomial becomes identically zero after substituting $z_1=a$ for at most finitely many $a$, unless it was zero to start with. For the direction $z_1$, exclude also those $z^{\mathrm{\prime }}$ where the line is exceptional or any node lies outside $U$. The algebraic exclusions leave a nonempty Zariski open set, and the finitely many meromorphic pole exclusions still leave a nonempty analytic open subset.

Induction makes $f_i(z^{\mathrm{\prime }})=f(a_i,z^{\mathrm{\prime }})$ rational functions. Over $L=\mathbb{C}(z^{\mathrm{\prime }})$, solve the $2d+1$ homogeneous linear equations $P(a_i)=f_{i}Q(a_i)$ for the $2d+2$ coefficients of two polynomials of degrees at most $d$. There is a nonzero solution. It has $Q\mathrm{\ne }0$, because otherwise $P$ has $2d+1>d$ distinct roots and is zero, contradicting nonzeroness; this also covers $d=0$.

For generic fixed $z^{\mathrm{\prime }}$, write the actual line restriction as $P_{\ast }\mathrm{/}{Q}_{\ast }$ with degrees at most $d$ and $Q_{\ast }(a_i)\mathrm{\ne }0$. The polynomial $P{Q}_{\ast }-P_{\ast }Q$ has degree at most $2d$ and vanishes at every $a_i$, hence is zero. Possible zeros of $Q$ at nodes cause no problem for this cross-multiplied identity. Thus $f=P\mathrm{/}Q$ on a nonempty open set. Clear the rational coefficients to write this as equality of meromorphic functions $f=P_0\mathrm{/}{Q}_0$ on a nonempty open subset of $U$, with polynomials $P_0,Q_0$. Let $r=f-P_0\mathrm{/}{Q}_0$, and let $A$ be the set of points of $U$ near which $r$ is the zero meromorphic function. This set is open and nonempty. If $p\in \bar{A}$, choose a connected ball $B\subset U$ around $p$ on which $r=g\mathrm{/}h$ for holomorphic functions $g,h$ with $h$ not identically zero. The open set $B\cap A$ contains a point where $h\mathrm{\ne }0$, so $g$ vanishes on a nonempty open subset of $B$. The holomorphic identity theorem [F1] gives $g=0$ on $B$, whence $r=0$ meromorphically on $B$ and $p\in A$. Thus $A$ is also closed; connectedness from [F1] gives $A=U$. This proves the meromorphic identity propagation needed here, and hence the induction and the assertion.