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Let $I \subset \mathcal{O}_{\mathbb{C}^2,0}$ be the ideal generated by $z_1^2-z_2^3+z_1$ and $z_1^4-2z_1z_2^3+z_1^2$. Describe $\sqrt{I}$. I know $\sqrt{I}:=\{f \in \mathcal{O}_{\mathbb{C}^2,0}: f^k \in I, \space \text{for some $k$}\}$. So how so I find the $f \in \mathcal{O}_{\mathbb{C}^2,0}$ such that $f^k \in I$??

I tried factoring out a $z_1$ from the second expression but the first one has no factoring. So I get $z_1(z_1^3-2z_2^3+z_1)$ for the second and $z_1^2-z_2^3+z_1$, how do you take roots here???

homosapien
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  • No, I mean $\mathbb{C}^2,0$, that is, functions holomorphic on $\mathbb{C}^2$ centered at the origin. The text has it written how I have. Using Huybrechts. – homosapien Jan 27 '22 at 18:09
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    replace $z_2^3$ by $z_1-z_1^2$ in the second polynomial giving a polynomial in $z_1$ from which you'll find $\le 12$ prime (and maximal) ideals containing $I$ and $\sqrt{I}$ is their intersection, the ideal of holomorphic functions vanishing at $12$ given points. – reuns Jan 27 '22 at 18:10
  • @reuns how did you know to make that substitution? – homosapien Jan 27 '22 at 18:16

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