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By definition (in the book that I am following), an affine variety is zero set of a collection of finitely many polynomials.

Question is to prove that $\mathbb{A}^1\setminus\{0\}$ is an affine variety.

For that I need to prove that $\mathbb{A}^1\setminus\{0\}$ is zero set of (finitely many) polynomial(s).

I was having this misconception that I have to prove that $\mathbb{A}^1\setminus\{0\}$ is zero set of polynomial in one variable as we are in $\mathbb{A}^1$.

Then I read the definition again and it says:

$X\subset k^n$ is an affine variety if it is zero set of polynomials in $n$ variables.

As $\mathbb{A}^1\setminus\{0\}$ and $\{(x,y)\in k^2:xy=1\}$ are homeomorphic and $\{(x,y)\in k^2:xy=1\}$ being zero set of polynomial $f(x,y)=xy-1$, is an affine variety and so $\mathbb{A}^1\setminus \{0\}$ should also be an affine variety though not as subvariety of $\mathbb{A}^1$.

Is this justification sufficient enough to say that $\mathbb{A}^1\setminus\{0\}$ is an affine variety or am I missing something?

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    Yes, $\mathbb A^1 -\{0\}$ is an affine variety, since it is isomorphic to the affine variety $V(xy-1) \subset \mathbb A^2$. – 2017-01-10
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    @MooS : I am atill confused little bit.. Can you say some thing more – 2017-01-10
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    What precisely are you confused about? The isomorphism? The conclusion? – 2017-01-11
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    @MooS The definition given is "an affine variety is a zero set of a collection of finitely many polynomials." It doesn't say anything about closing under isomorphism (in what category?). So just asserting that something is an affine variety because it's isomorphic to an affine variety doesn't make sense, and cello's confusion is justified! Of course, the problem here is that the given definition is deficient... – 2017-01-12

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Let $X=\mathbb{A}^1\setminus\{0\}$, it is an open subset of $\mathbb{A}^1$ (w.r.t. Zariski topology); by defintion, only the finite subsets (a.k.a. Zariski closed subsets) of $\mathbb{A}^1$ are affine (sub)varieties of $\mathbb{A}^1$: $X$ is not an affine subvariety of $\mathbb{A}^1$!

As the OP-user affirms: $X$ is homeomorphic to hyperbola $V(xy-1)=\{(x,y)\in\mathbb{A}^2\mid xy-1=0\}$ (w.r.t. Zariski topologies); therefore one can pull-back from $V(xy-1)$, via a fixed homeomorphism, an affine algebraic variety structure on $X$, and $X$ with a such structure is an abstract affine algebraic variety.

The same trick works, for example, with the graph $G$ of the absolute function $|\cdot|$: as subset of $\mathbb{R}^2$, it is homeomorphic to $\mathbb{R}$ (w.r.t. natural topologies); therefore $G$ can be structured as an analityc manifold.

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    I fail to understand How does this answer Alex Kruckman's question. – 2017-01-19
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    @cello I answered your question! And I had add an example *extra* of the same trick. – 2017-01-20