Let $H$ be a closed hypersurface in $\mathbb{A}^{n}$, $n$ big enough over $\mathbb{C}$. Let $U$ be the complementary open subset.
Let $x\in H$, Is it possible to find an curve $C\subset\mathbb{A}^{n}$ such that and $C\cap H=\{x\}$.
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Sign up to join this communityLet $H$ be a closed hypersurface in $\mathbb{A}^{n}$, $n$ big enough over $\mathbb{C}$. Let $U$ be the complementary open subset.
Let $x\in H$, Is it possible to find an curve $C\subset\mathbb{A}^{n}$ such that and $C\cap H=\{x\}$.
In general no, thanks to this example. Take the curve $E\subset \mathbb{A}^2$ of the example (given by $y^2=x^3+ax+b$), and take $H=E\times \mathbb{A}^{n-2}\subset \mathbb{A}^{2}\times \mathbb{A}^{n-2}=\mathbb{A}^n$. Let $p:\mathbb{A}^n\rightarrow \mathbb{A}^{n-2}$ be the projection on the second factor. Suppose $C\cap H={(x,v)}$ for $x\in E$, $v\in \mathbb{A}^{n-2}$. Then $p(C)$ is a curve in $\mathbb{A}^2$, and $p(C)\cap E=\{x\} $, which is not possible for general $x$.