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For each integer $n$ I am looking for a real-valued polynomial in two variables, $A_n(x,y)$, such that $A_n(x,y) = 0$ defines a curve with precisely $n$ closed components in the plane $\mathbb{R}^2$. Could someone point me in the direction of a theorem or result that might partially/fully solve this problem?

I should point out that the curves must not be concentric, i.e. one should not contain another in its interior.

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    $\begingroup$ The zero scheme of $\prod_{i=1}^n ( (x-3i)^2+y^2-1)$ is such a plane curve. There are many others. $\endgroup$
    – Jason Starr
    Commented Apr 29, 2017 at 22:27
  • $\begingroup$ How does one prove this? $\endgroup$
    – jess
    Commented Apr 29, 2017 at 22:29
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    $\begingroup$ How does one prove what? How does one prove that a product of real numbers equals zero if and only if one of the factors equals zero? Is that what you are asking? $\endgroup$
    – Jason Starr
    Commented Apr 29, 2017 at 22:35
  • $\begingroup$ No, sorry, I rushed. Thanks for pointing that out! $\endgroup$
    – jess
    Commented Apr 29, 2017 at 22:36

2 Answers 2

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Take any polynomial $P$ of degree $n$ with $n$ distinct roots. The curve $|P(z)|^2=\epsilon$, where $z=x+iy$, is algebraic and has $n$ components when $\epsilon$ is small enough. This curve is called a lemniscate.

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Here is @JasonStarr's polynomial for $n=5$: $$ \left((x-3)^2+y^2-1\right) \left((x-6)^2+y^2-1\right) \left((x-9)^2+y^2-1\right) \left((x-12)^2+y^2-1\right) \left((x-15)^2+y^2-1\right) $$ And here is a plot of its zeros:

            JasonStarr

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  • $\begingroup$ What is this plot good for? Visualizing a complicated example ...? $\endgroup$
    – Peter Mueller
    Commented Apr 30, 2017 at 14:44

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