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Suppose that$ f(x, y)$ given by $\sum_{i=0}^{a}\sum_{j=0}^{b}c_{i,j}x^iy^j$ is a polynomial in two variables with real coefficients such that among its coefficients there is a non-zero one. Prove that there is a point $(x_0, y_0) ∈ R^2$ such that $f(x_0, y_0)\neq 0$.

So basically this is a non-zero polynomial i.e. a polynomial with at least one non-zero coefficient. I do not understand how can one make such a statement like I do not understand intuition. Moreover, I am not getting any properties like this on the Wikipedia page. I think I am lacking some real analysis basics. Could anyone please provide any hints or direct me towards something helpful for this question?

Logo
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  • Did you mean $x^iy^j?$ – zhw. May 28 '21 at 18:08
  • @zhw. oh yeah sorry. fixing it. – Logo May 28 '21 at 18:11
  • How would you prove the analogous statement for a polynomial in one variable using techniques from analysis? – saulspatz May 28 '21 at 19:02
  • @saulspatz to be honest I'm not completely sure. I'm thinking of using derivative and double derivative techniques to see if I can come up some solutions. I read somewhere that such type of polynomial can be expressed in the form of product of its derivatives. – Logo May 28 '21 at 19:11
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    In the one-variable case, you could just argue that it goes to $\infty$ in absolute value as $x\to\infty$ so it can't be identically $0$. It's more complicated in the two-variable case, obviously, but I think you can make it work. – saulspatz May 28 '21 at 19:15

2 Answers2

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Hint: $\dfrac{\partial^{i+j}f}{\partial x^i\partial y^j}= i!j!c_{ij}.$

zhw.
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  • even I was also thinking about using derivative but didn't know how it will b useful. – Logo May 28 '21 at 19:14
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Another hint, without calculus: $f$ can be written as a polynomial in $y$ with coefficients that are polynomials in $x$, in other words $f(x,y) = \sum_j g_j(x)\,y^j$ where $g_j(x) = \sum_i c_{ij} x^i$.

Suppose the non-zero coefficient is $c_{pq}\ne0$, then $g_q(x)$ cannot be the zero polynomial, so there exists an $x_0$ such that $g_q(x_0)\ne0$.

It follows that the polynomial $P(y)=f(x_0,y)=\sum_j g_j(x_0)\,y^j$ cannot be the zero polynomial, so there exists an $y_0$ such that $P(y_0) \ne 0 \iff f(x_0,y_0)\ne0$.

dxiv
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