MHB How do we prove that $p(x,y)$ is a nonzero element of $\Bbb R[x,y]$?

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Here's this week's problem!

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Problem: Suppose $p : \Bbb R^2 \to \Bbb R$ is a function of two variables $x$ and $y$ such that for every $x$, $p(x,y)$ is a nonzero polynomial in $y$, and for every $y$, $p(x,y)$ is a nonzero polynomial in $x$. Show that $p(x,y)$ is a nonzero element of $\Bbb R[x,y]$.
________Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

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No one solved this week's problem. You can find my solution below.

Spoiler

Let $A_m = \{x \in \Bbb R\, |\, \text{deg}_x\, p(x,y) = m\}$ and $B_n = \{y\in \Bbb R\, |\, \text{deg}_y\, p(x,y) = n\}$, for $m, n = 0, 1, 2,\ldots.$ Then

$$\bigcup_{m = 0}^\infty A_m = \Bbb R = \bigcup_{n = 0}^\infty B_n$$

(the nonzero polynomial assumptions ensure that $\text{deg}_x\, p$ and $\text{deg}_y\, p$ are finite for all $x$ and $y$). Since $\Bbb R$ is a complete metric space, by the Baire category theorem, there exist $m$ and $n$ for which $A_m$ and $B_n$ have nonempty interior. If $k \ge \max\{m,n\}$, then there is an infinite subset $J$ of $\Bbb R$ such that for all $x,y\in \Bbb R$, the functions $p_x : a \mapsto p(x,a)$ and $p^y : b \mapsto p(b,y)$ are polynomials of degree not exceeding $k$. Choose $k+1$ points of $J$, say $a_0,\ldots, a_k$. By bilinear interpolation there exists a polynomial $q$ such that $q(a_i, a_j) = p(a_i,a_j)$ for all $i$ and $j$. Given $j$, $p^{a_j}$ and $q^{a_j}$ are polynomials of degree $\le k$ that agree on $k+1$ points. So $p$ agrees with $q$ on each section $\Bbb R \times \{a_j\}$. For $x\in J$, the functions $p_x$ and $q_x$ are also polynomials of degree $\le k$ that agree on $k + 1$ points. Now it follows that $p$ agrees with $q$ on $J \times \Bbb R$. Since for every $y\in \Bbb R$, $p^y$ and $q^y$ are polynomials agreeing on the infinite set $J$, we must have $p = q$ on all of $\Bbb R^2$. Moreover, $p(x,y)$ has finite degree.