# Is a bivariate function that is a polynomial function with respect to each variable necessarily a bivariate polynomial?

Let $\mathbb{F}$ be an uncountable field. Suppose that $f: \mathbb{F}^{2} \rightarrow \mathbb{F}$ satisfies the following two properties:

1. For each $x \in \mathbb{F}$, the function $f(x,\cdot): y \mapsto f(x,y)$ is a polynomial function on $\mathbb{F}$.
2. For each $y \in \mathbb{F}$, the function $f(\cdot,y): x \mapsto f(x,y)$ is a polynomial function on $\mathbb{F}$.

Is it necessarily true that $f$ is a bivariate polynomial function on $\mathbb{F}^{2}$? What if $\mathbb{F}$ is merely countably infinite?

• (Note that this is trivially true for finite fields (because then every function $F\times F\to F$ is a polynomial), and is also easy to prove in the general case if there's an upper bound for the degrees of the polynomial functions in the hypothesis). Oct 11, 2012 at 21:33
• Also note this is false if we replace $\mathbb{F}$ by $\mathbb{Z}_{\geq 0}$: Take $f(x,y) = \binom{x+y}{x}$. Nice question! Oct 12, 2012 at 1:22

Maybe this works for the countably infinite case. Order the rationals (or whatever countably infinite field you have) as $r_1,r_2,\dots$. Let $$f(x,y)=(x-r_1)(y-r_1)+(x-r_1)(x-r_2)(y-r_1)(y-r_2)+\cdots$$ Then if $r$ is any rational, say, $r=r_j$, then $f(r,y)$ is a polynomial of degree $j-1$ in $y$, and similarly for $f(x,r)$. But clearly $f$ is not a polynomial function --- what would be its degree?

• I don't get why $(x,y)\mapsto \displaystyle \sum \limits_{k=1}^{+\infty}\prod\limits_{j=1}^{k}\left(\left(x-r_j\right)\left(y-r_j\right)\right)$ exists at all. Why does, for instance, $\displaystyle \sum \limits_{k=1}^{+\infty}\prod\limits_{j=1}^{k}\left(\left(r_j\right)^2\right)$ converge? Jun 10, 2018 at 22:38
• If $x=r_m$ and $y=r_n$ all terms $\prod\limits_{j=1}^{k}\left(\left(x-r_j\right)\left(y-r_j\right)\right)$ vanish for $k\geq\max(m,n)$. Jun 11, 2018 at 2:45
• It seems that I need a hint concerning the assertion that $f$ is not a polynomial. Jun 15, 2018 at 2:16
• @Jens, note that, for each $k=1,2,\dots$, $f(r_k,y)$ is a polynomial in $y$ of degree $k-1$. This can't happen if $f$ is a polynomial, as in that case the degree of $f(r_k,y)$ as a polynomial in $y$ would be bounded above independent of the choice of $k$. Jun 15, 2018 at 3:04
• Now I see. Years ago in a paper written jointly with Wolfgang Prager we had a different and more complicated example. Jun 15, 2018 at 9:13

As shown is Gerry Myerson’s answer, the answer is NO when $\mathbb F$ is countably infinite.

The answer is YES when $\mathbb F$ is uncountable, however.

Sketch of proof : since there are only countably many degrees, the polynomials will share a common degree on an uncountable set. This bound on the degree allows one to use interpolation, and to retrieve the whole of $f$.

More detailed proof : Denote by $d(x)$ the degree of the univariate polynomial $f(x,.)$ for $x\in {\mathbb F}$ (recall that the degree of the zero polynomial is $-\infty$), and put $U_d=\lbrace x \in {\mathbb F} | d(x)=d\rbrace$ for $d\in \lbrace -\infty \rbrace \cup {\mathbb N}$. Then the $U_d$ form a countable partition of $\mathbb F$, so at least one of the $U_d$, say $U_{n}$, is uncountable.

We may assume that $n>0$, as the cases $n=-\infty$ and $n=0$ are similar and simpler. Let $y_0,y_1, \ldots y_{n}$ be $n+1$ distinct values in $\mathbb F$, this is possible because $\mathbb F$ is uncountable. (if the characteristic of $\mathbb F$ is zero, we can simply take $y_i=i$). Using Lagrange interpolation, let us put

$$L_k(y)=\frac{\prod_{j \neq k}{(x-x_j)}}{\prod_{j \neq k}{(x_k-x_j)}}$$

for $0 \leq k \leq n$. Then one has, for any polynomial $P$ of degree $\leq n$ and any $y\in{\mathbb F}$,

$$P(y)=P(y_0)L_0(y)+P(y_1)L_1(y)+ \ldots +P(y_n)L_n(y)$$

In particular, one has for any $(x,y)\in U_n \times {\mathbb F}$,

$$(1) \ f(x,y)=f(x,y_0)L_0(y)+f(x,y_1)L_1(y)+f(x,y_2)L_2(y)+ \ldots +f(x,y_n)L_n(y)$$ The right-hand side is a fixed bivariate polynomial, let us denote it by $Q(x,y)$. Let $y\in {\mathbb F}$. Then the two univariate polynomials $f(.,y)$ and $Q(.,y)$ coincide on the uncountable set $U_n$, so they must coincide everywhere. Finally $f=Q$ everywhere and we are done.

• Nice insight concerning the motivation behind the proof. Oct 18, 2012 at 16:01