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Is every algebraic curve in $\mathbb{R}^2$($\mathbb{C}^2$), the set of critical points of a polynomial in $\mathbb{R}[x,y]$($\mathbb{C}[x,y]$)?

In particular what is a real (complex) polynomial whose critical set is the circle (cylinder) $x^2+y^2=1$?

Added: For the real case we have the perfect answer by Matt F. What about the complex case? Is there an algebraic complex function or even a holomorphic entire function whose critical set is the cylinder $x^2+y^2=1$?

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    $\begingroup$ There is no such polynomial in the real case. Any polynomial $p$ which is critical on the circle must be constant on the circle, so suppose $p=0$ there. If $p$ has positive points inside the circle, then it has a maximum inside the circle, which is a criticial point. If $p$ has negative points inside the circle, then it has a minimum inside the circle, which is a critical point. And if $p$ is constant inside the circle, then all the points inside the circle are critical points. In every case the set of critical points is more than the circle. $\endgroup$
    – user44143
    Commented May 13, 2019 at 12:30
  • $\begingroup$ @MattF. Thank you for your perfect answer in real case. $\endgroup$
    – Ali Taghavi
    Commented May 13, 2019 at 18:53
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    $\begingroup$ Comments on the complex case: If $f$ is critical on $x^2+y^2=1$ then $f$ is constant, and we may as well assume the constant value is $0$. Then $f$ vanishes to order $2$ on the circle, so $f = (x^2+y^2-1)^2 g$. For generic $g$, the critical points of $(x^2+y^2-1)^2 g$ will be the circle and finitely many other points. I see no reason the finitely many other points can't lie on the circle $\endgroup$
    – David E Speyer
    Commented May 14, 2019 at 0:30
  • $\begingroup$ @DavidESpeyer, you can also reformulate that with a Nullstellensatz. If $h=x^2+y^2-1$, $f=h^ag$ where $h\nmid g$, then the condition is $\exists b\ h^b \in (hg_x+2agx, hg_y+2agy)$. $\endgroup$
    – user44143
    Commented May 14, 2019 at 12:20

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