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If we consider $\mathbb{A}^{1} \setminus \{0\}$ then we see this is isomorphic to $Z(xy-1) \subseteq \mathbb{A}^{2}$ via the map $g(t)=(t,\frac{1}{t})$. So this raises a question: let $n \geq 2$, is every proper open subset of $\mathbb{A}^{1}$, i.e a set of the form:

$\mathbb{A}^{1} \setminus \{c_{1},..,c_{n}\}$ isomorphic to a closed subset of $\mathbb{A}^{n+1}$? or $A^{k}$ for some $k \geq 2$?

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up vote 3 down vote accepted

You're on the right track, even in the sense that you suspect that your attempt isn't surjective: if I take $c_1 = 0$ and $c_2 = 1$ then $(1, 1, 1)$ is not in the image.

A finite collection of points $\{c_1, \ldots, c_n\}$ in $\mathbf A^1$ is the zero set of a polynomial $f(x) = (x - c_1) \cdots (x - c_n)$. You want to know whether $U = \mathbf A^1 - Z(f)$ is affine. $P \in U$ $\Leftrightarrow$ $f(P)$ is invertible, so it's natural to look at \[ H = Z(yf - 1) \subset \mathbf A^2. \] As before, you would define $\varphi\colon U \to H$ by $\varphi(a) = (a, f(a)^{-1})$. This should generalize to other $\mathbf A^n$ and other $f$'s. (On the other hand, in general there are certainly open sets that are not affine, e.g. $\mathbf A^2 - \{(0, 0)\}$.)

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Let $f_i(x)=(x-c_i)y-1 \in k[x,y]$ for $1 \leq i \leq n.$ Then,

$$\mathbb{A}^1 \setminus \{c_1, c_2, \cdots, c_n\}=\bigcap_i (\mathbb{A}^1 \setminus \{c_i\}) \cong \bigcap_i Z(f_i)=Z(\bigcup _i (f_i)) \subset \mathbb{A}^2.$$

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