I'm trying to prove, via the Cech cohomology, that $S=\mathbb A^2_k\setminus\{0\}$ with the induced Zariski topology is not an affine variety. Consider the structure sheaf $\mathcal O_{\mathbb A^2_k}\big|_S:=\mathcal O_S$ (which is quasi coherent), i must show that $\exists n$ such that $\check H^n(S,\mathcal O_S)\neq 0$. It is enough to prove that $\check H^n(\mathcal U,\mathcal O_S)\neq0$ for a certain affine cover of $S$ (and a certain $n$); so let's choose $\mathcal U=\{D(X), D(Y)\}$ where $D(X)=\{(x,y)\in S\,:\, x\neq 0\}$ and $D(Y)=\{(x,y)\in S\,:\, y\neq 0\}$. Clearly for $n\ge 2$ we have that $\check H^n(S,\mathcal O_S)=0$, so i must show that $\check H^1(\mathcal U,\mathcal O_S)\neq0$. The Cech complex is: $$\mathcal O_S(D(X))\times\mathcal O_S(D(Y))=\Gamma(S)_X\times\Gamma(S)_Y\longrightarrow \mathcal O_S(D(X)\cap D(Y))=\Gamma(S)_{XY}\longrightarrow 0\cdots$$

with the homomorphism: $d^0: (f,g)\mapsto g|_{{D(X)\cap D(Y)}}-f|_{{D(X)\cap D(Y)}}$. To complete the proof i should conclude that $d^0$ is not surjective, but why is this true?



First notice that the restriction morphism $\Gamma(\mathbb A^2_k,\mathcal O_{ \mathbb A^2_k})\to \Gamma(S, \mathcal O_S)$ is bijective because the affine plane $A^2_k$ is normal ("Hartogs phenomenon").
Hence we may identify $\Gamma(S, \mathcal O_S)$ with the polynomial ring $k[X,Y]$

a) The open set $D(X)$ is isomorphic to $\mathbb G_m\times \mathbb A^1_k$ where $\mathbb G_m=\operatorname {Spec} k[T,T^{-1}]$, the affine line with origin deleted.
Hence $\Gamma(D(X),\mathcal O_{ A^2_k})=k[X,X^{-1},Y]$.

b) Similarly $D(Y)$ is isomorphic to $\mathbb A^1_k \times \mathbb G_m$.
Hence $\Gamma(D(Y),\mathcal O_{ A^2_k})=k[X,Y, Y^{-1}]$.

c) Finally the open set $D(X)\cap D(Y)$ is isomorphic to the product $\mathbb G_m\times_k \mathbb G_m$ .
Hence $\Gamma(D(X)\cap D(Y),\mathcal O_{ A^2_k})=k[X,X^{-1}]\otimes _k k[Y,Y^{-1}]= k[X,X^{-1},Y,Y^{-1}]$.

d) With these identifications established, the first cohomology group $\check H^1(\mathcal U,\mathcal O_S)$ of the structural sheaf is the cohomology of the complex $$ k[X,X^{-1},Y]\times k[X,Y,Y^{-1}] \to k[X,X^{-1},Y,Y^{-1}] \to 0 $$ where the non trivial map is $$(f(X,X^{-1},Y),g(X,Y,Y^{-1}))\mapsto g(X,Y,Y^{-1})-f(X,X^{-1},Y)$$ e) Hence we see that the required cohomology is the following infinite dimensional $k$-vector space , spectacularly violating vanishing of cohomology for affine schemes, which $S$ is thus not.

Final result $$ \check H^1(\mathcal U,\mathcal O_S)=\check H^1(S,\mathcal O_S)=\oplus _{i,j\gt 0} \; k\cdot X^{-i} Y^{-j} $$


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