This example is from the book "A problem Seminar" by D.J. Newman:
Let $F$ be an infinite field and let $f: F \times F \to F$ be a function of two variables such that $f(x_0,y)$ is a polynomial in $y$ for every $x_0 \in F$ and $f(x,y_0)$ is a polynomial in $x$ for every $y_0 \in F$. (Of course, being a polynomial for a function $f: F \to F$ means there exists $p(x) \in F[x]$ such that $f(x)=p(x)$ for all $x \in F$.) Now, is $f$ itself necessarily a polynomial?
Surprisingly the answer depends on the cardinality of $F$. It is negative when $F$ is countable and positive when $F $ is uncountable. For countable $F$, enumerate the elements as $a_1, a_2, \dots $ and consider $$ f(x,y)=\sum_{i=1}^{\infty} (x-a_1)(x-a_2)\cdots (x-a_i)(y-a_1)\cdots (y-a_i) $$
It is obvious that $f$ satisfies the condition, and not hard to show that it is not a polynomial.
