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The question is

For any quasi-affine variety $X$, there exists $n, m$ such that $X$ is isomorphic to a closed subvariety of the quasi-affine variety $\mathbb{A}^n\setminus \mathbb{A}^m$.

Intuitively I know that given an quasi-affine variety $X$, I need to take $n,m$ "big enough" so that I can embed $X$ linearly (maybe not?!) into $\mathbb{A}^n\setminus \mathbb{A}^m$. But how do you do that? I tried some examples and came up empty.

I think if somebody can give me an example of how to show this for a specific quasi-affine variety (which is not completely trivial). I think a good example which is general enough can be:

$X=Z\cap U$ where $Z=\{(x,y)\in \mathbb{A}^2|y^3-x^2=0\}$ and $U$ is complement of twisted cubic curve $U=\{(x,y)\in \mathbb{A}^2|y^2-x^2(x+1)\neq 0\}$.

I specifically tried working this out but after spending hours playing around, drawing stuff and staring at it, I realized I don't really know what I'm doing and don't know really where to begin.

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Suppose your quasi-affine variety is $Z \setminus Y$, where $Y \subseteq Z \subseteq \mathbb{A}^m$ are closed subsets. By definition, that means there are regular functions $f_1, \ldots, f_q, g_1, \ldots, g_r : \mathbb{A}^m \to \mathbb{A}^1$ such that $Y = V (f_1, \ldots, f_q) \subseteq \mathbb{A}^m$ and $Z = V (g_1, \ldots, g_r)$. Now, $$Z' = \{ (x_1, \ldots, x_m, t_1, \ldots, t_q) : g_j (x_1, \ldots, x_m) = 0, t_i - f_i (x_1, \ldots, x_m) = 0 \} \subseteq \mathbb{A}^{m + q}$$ is a closed subset isomorphic to $Z$ by construction, and moreover, $$Y' = Z' \cap (\mathbb{A}^m \times \{ (0, \ldots, 0) \})$$ hence, $$Z \setminus Y \cong Z' \setminus Y' = Z' \cap \mathbb{A}^{m+q} \setminus (\mathbb{A}^m \times \{ (0, \ldots, 0) \})$$ as required.

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