Show that \(\AA^2\slice{\CC} \smz\) is not an affine variety.
Solution
Put \[ U=\AA^2_\CC\setminus\{(0,0)\}, \qquad R=\CC[x,y]. \] Then \[ U=D(x)\cup D(y). \]
1. Every global regular function on \(U\) is the restriction of a uniquepolynomial in \(\CC[x,y]\):\[ \boxed{\OO(U)=\CC[x,y].} \]
Since \[ D(x)=\operatorname{Specm}R_x, \qquad D(y)=\operatorname{Specm}R_y, \] a regular function \[ f\in\OO(U) \] is represented by \[ f|_{D(x)}=\frac{a}{x^m}, \qquad f|_{D(y)}=\frac{b}{y^n} \] for some \[ a,b\in R, \qquad m,n\geq0. \] On the overlap \(D(xy)\) these representatives agree, so in \[ \Frac R=\CC(x,y) \] one has \[ \frac{a}{x^m} = \frac{b}{y^n}. \] Hence \[ ay^n=bx^m. \] The ring \(R\) is a UFD and \(x,y\) are relatively prime prime elements. Thus \[ x^m\mid a \qquad\text{and}\qquad y^n\mid b. \] Write \[ a=x^m c, \qquad b=y^n c' \] with \(c,c'\in R\). The equality \(ay^n=bx^m\) then gives \(c=c'\). Therefore \[ f|_{D(x)}=c=f|_{D(y)}, \] so \(f\) is the restriction of the polynomial \(c\).
Uniqueness follows because a polynomial vanishing on the nonempty open subset \(U\) of the irreducible affine plane must be zero.
2. Under the identification in step 1, the canonical map\[ \alpha:U\longrightarrow \operatorname{Specm}\OO(U) \cong \operatorname{Specm}\CC[x,y] = \AA^2_\CC \]is the ordinary inclusion\[ U\hookrightarrow\AA^2_\CC. \]
For a point \[ p=(a,b)\in U, \] the canonical map to the maximal spectrum of global functions sends \(p\) to the maximal ideal of global regular functions vanishing at \(p\).
By step 1, the global functions are exactly the restrictions of polynomials in \(\CC[x,y]\). The vanishing ideal at \((a,b)\) is therefore \[ (x-a,y-b)\subseteq\CC[x,y]. \] This is exactly the point \((a,b)\) of \(\AA^2_\CC\). Hence \(\alpha\) agrees pointwise with the inclusion of the punctured plane into the affine plane.
3. If \(U\) were affine, the map \(\alpha\) in step 2 would be anisomorphism.
For any affine variety \(Y\), the canonical morphism \[ Y\longrightarrow\operatorname{Specm}\OO(Y) \] is an isomorphism: this is the affine variety–coordinate ring correspondence. Thus affineness of \(U\) would force \(\alpha\) to be an isomorphism.
4. The punctured affine plane is not affine:\[ \boxed{ \AA^2_\CC\setminus\{(0,0)\} \text{ is not an affine variety}. } \]
By step 2, \(\alpha\) is the inclusion \[ U\hookrightarrow\AA^2_\CC. \] It is not surjective, because \((0,0)\) is not in its image. Therefore it is not an isomorphism. Step 3 shows that this is impossible if \(U\) is affine. Hence \(U\) is not affine.
5. Q.E.D.
Step 4 is the required conclusion.