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Let $F(x,y)$ be a bivariate polynomial, of degree n. Hence:

$F(x,y) = \underset{i+j \leq n}{\sum_{i=0}\sum_{j=0}}a_{ij}x^{i}y^{j}$

Can there exist an upperbound for the number of isolated zeros for $F(x,y)$ ? I understand that if we ask for number of zeros in general, it can be infinite. But by isolated zeros I mean that zeros that are not connected by a curve (on the x-y plane).

I saw an answer to a similar question: Point 2 of this answer that one can bound the number of isolated zeros by $n^{2}$, or , it says some quadratic function of $n$. The answer was not elaborate to the point I could understand.

Can anyone kindly help me verify this ?

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That answer also mentions "Harnack's Theorem." Did you try to find anything about it? –  Gerry Myerson Dec 11 '12 at 0:38
The paper at sciencedirect.com/science/article/pii/S0747717198902723 looks forbidding, but some of the references may give you what you need. –  Gerry Myerson Dec 11 '12 at 0:43
@GerryMyerson I looked at Harnack's Theorem for bivariate polynomials. It seems helpful, bounding the number of zero curves of the bivariate polynomial. Thanks for the reference, the paper seems heavy for me, but from what I understood, in it they are discussing solutions of a polynomial system, whereas I am looking for the number of zeros of a single bivariate polynomial. So, I was wondering it there are infact the same. Thanks again for the reference. –  Pavithran Iyer Dec 11 '12 at 3:09
Well, a single polynomial is a system of $1$ polynomial; conversely, the system $f_1=f_2=\cdots=f_m=0$ has the same (real) zeros as the single equation $f_1^2+f_2^2+\cdots+f_m^2=0$. So it may be possible to use results about the one to get answers about the other. –  Gerry Myerson Dec 11 '12 at 3:16
Note also that if $f(x)$ has degree $n$ and $n$ distinct real zeros, then $(f(x))^2+(f(y))^2=0$ has degree $2n$ and $n^2$ isolated zeros. –  Gerry Myerson Dec 11 '12 at 3:19

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