Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $k \geq 2$?

share|improve this question

2 Answers 2

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. $\mathbf A^2 - \{(0, 0)\}$.)

share|improve this answer

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

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.