Find an open subset of $\mathbb{A}^2$ which cannot be isomorphic to any affine variety Find an open subset of $\mathbb{A}^2$ which (with its given Zariski topology) cannot be isomorphic to any affine variety. [Delete the point (0,0).]
My problem is I don't know how to prove it is impossible for a subset of $\mathbb{A}^2$ to be isomorphic to any affine variety. 
 A: Let me just mention a slightly more elementary (local-cohomology free) version of David Speyer's answer in the thread mentioned in the comments. Let $X$ be an affine normal scheme, and let $Z$ be a subset of codimension at least 2 (but nonempty). Then $X-Z$ is not affine. Indeed, $X-Z$ has the same ring of global sections as $X$, by the so-called "algebraic Hartogs lemma" (namely, that singularities of a rational function occur in codimension one for a normal scheme). So if $X-Z$ were affine, the natural map $X-Z \to X$ would be an isomorphism (as both have the same regular functions), which it is clearly not.
A: And let me mention an even more elementary answer, expanding on Amitesh's hint.
Let $U$ be the punctured affine plane. Assuming that we have shown that there are no regular functions on $\mathbb{A}^2$ which vanish only at one point (I believe this follows from Krull's Hauptidealsatz), then it is clear that a regular function on $U$ is just the same thing as a regular function on $\mathbb{A}^2$. So its ring of regular functions is isomorphic to the ring of regular functions of $\mathbb{A}^2$.
Now consider the embedding $\Phi : U \hookrightarrow \mathbb{A}^2$. If $U$ were affine, then, by the equivalence of categories between $k\textbf{-Var}$ and (a certain full subcategory of) $k\textbf{-Alg}$, $\Phi$ is an isomorphism of varieties if and only if it induces an isomorphism of coordinate rings, and $\Phi$ does induce an isomorphism on each ring of regular functions. Yet $\Phi$ clearly cannot be an isomorphism as it is not surjective on points—a contradiction. So $U$ isn't affine.
