Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
(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). – Henning Makholm Oct 11 '12 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! – David Speyer Oct 12 '12 at 1:22
up vote 2 down vote accepted

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.

share|cite|improve this answer
Nice insight concerning the motivation behind the proof. – Haskell Curry Oct 18 '12 at 16:01

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?

share|cite|improve this answer
+1 Nice! ${}{}$ – Henning Makholm Oct 12 '12 at 15:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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