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For all complex polynomials in 2 variables I know, the set of zeros looks like a union of curves.

Wrong: I can get a circle with $x^2+y^2-1$ or two vertical lines with $(x-1)(x-2)$?.

Can I get isolated sets of point? In particular, can I find a polynomial in 2 variables which is zero only at a single point?

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    $x^2+y^2\phantom{}$?2012-07-07
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    @J.M. Thanks. I missed that. What about 2 isolated points?2012-07-07
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    Oh, probably something like $(x^2+y^2)((x-1)^2+y^2)$2012-07-07
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    Yes, in general, if you want something that touches the $x$-$y$ plane in some set of isolated points, you try assembling such a polynomial from sums of squares...2012-07-07
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    @J.M. Actually, I was trying to think if there are pairs of polynomials for which the set of common roots is not the set of roots of any single polynomial.2012-07-07
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    What do you mean by "complex polynomials", since you seem to not be allowing $x,y \in \Bbb C$?2012-07-07
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    @BrandonCarter: I got completely confused. I did intend to work with polynomials over $\mathbb{C}$. My examples are wrong.2012-07-07
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    @J.M. I got confused, but anyway, $x^2+y^2$ is satisfied by $(xi,x)$ for any $x \in \mathbb{R}$.2012-07-07
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    Clearly I missed the adjective "complex". You're right, of course.2012-07-07
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    @J.M.: So, to prove what I want to prove (see a previous comment), I'd like to take 2 complex polynomials $(x-1)^2+y^2-1$ and $(x+1)^2+y^2-1$. The only common root is $(0,0)$. So, if there's no complex polynomial with $(0,0)$ as the only root then I'm done.2012-07-07
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    @Derek: There can't be any single complex polynomial $p(x,y)$ with $(0,0)$ as the only root, because for example $p(x,1)=0$ must have a solution due to the Fundamental Theorem of Algebra.2012-07-07
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    @HenningMakholm: Got it. Thanks. The number 1 could have been any constant, right?2012-07-07
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    @Derek The number 1 could be a generic constant but it is possible to have a finite number of exceptions: the polynomial could be $(y-1)(y-2)(y-3)x + 1$, so then $p(x,1)=0$ has no solutions, nor does $p(x,2)=0$ or $p(x,3)=0$.2012-07-07
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    Perhaps someone would like to write up an answer as an answer? Derek, you can do it if you now understand the situation.2012-07-08

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