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

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In summary, to prove that a polynomial $p(x,y)$ is a nonzero element of $\mathbb{R}[x,y]$, we can either check that it has at least one non-zero coefficient or use the fact that a polynomial is nonzero if and only if it has a nonzero constant term. Mathematical induction cannot be used to prove this statement, and showing that $p(x,y)$ has a nonzero leading coefficient is not enough to prove that it is nonzero. The Fundamental Theorem of Algebra also cannot be used for this proof as it only applies to polynomials with complex coefficients. Additionally, a polynomial can be nonzero at some points and zero at others depending on the values of its coefficients.
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Euge
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Here's this week's problem!

________

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.
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

\(\displaystyle \bigcup_{m = 0}^\infty A_m = \Bbb R = \bigcup_{n = 0}^\infty B_n\)

(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.
 

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

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

To prove that $p(x,y)$ is a nonzero element of $\Bbb R[x,y]$, we must show that the polynomial has at least one non-zero coefficient. This can be done by examining the polynomial term by term and checking that each coefficient is not equal to zero. Alternatively, we can use the fact that a polynomial is nonzero if and only if it has a nonzero constant term.

Can we use mathematical induction to prove that $p(x,y)$ is a nonzero element of $\Bbb R[x,y]$?

No, mathematical induction cannot be used to prove that $p(x,y)$ is a nonzero element of $\Bbb R[x,y]$ as it is a method used to prove statements about integers, not polynomials. Instead, we must use the methods mentioned in the answer to the first question.

Is it enough to show that $p(x,y)$ has a nonzero leading coefficient to prove that it is a nonzero element of $\Bbb R[x,y]$?

No, it is not enough to show that $p(x,y)$ has a nonzero leading coefficient to prove that it is a nonzero element of $\Bbb R[x,y]$. This is because a polynomial can have a nonzero leading coefficient but still have other coefficients that are equal to zero, making it a zero polynomial.

Can we use the Fundamental Theorem of Algebra to prove that $p(x,y)$ is a nonzero element of $\Bbb R[x,y]$?

No, the Fundamental Theorem of Algebra cannot be used to prove that $p(x,y)$ is a nonzero element of $\Bbb R[x,y]$ as it only applies to polynomials with complex coefficients. Since $\Bbb R[x,y]$ consists of polynomials with real coefficients, we must use other methods to prove that a polynomial is nonzero.

Is it possible for a polynomial to be nonzero at some points and zero at others?

Yes, it is possible for a polynomial to be nonzero at some points and zero at others. This is because a polynomial is defined by its coefficients, and it is the combination of these coefficients that determines whether the polynomial is zero or nonzero. Therefore, a polynomial can have different values at different points depending on the values of its coefficients.

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