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The Picard group $\mathrm{Pic}(X)$ is the set of isomorphism classes of line bundles on $X$, with tensor product as addition, the trivial bundle as zero and the dual bundle as inverse. On a smooth projective surface, line bundles can be represented by divisors. The intersection pairing on the Picard group of a surface is the symmetric bilinear mapobtained by moving the divisors into proper position and counting their intersections with multiplicity.

The surfaceis the first Hirzebruch surface ${\mathbb{F}}_1$. If $S$ is its negative section and $F$ a fiber of the ruling, its Picard lattice of a Hirzebruch surface is

Now let $\pi :X\to {\mathbb{P}}^2$ be the given nonisomorphic birational morphism. A birational morphism between smooth projective surfaces factors as a nonempty sequence of point blowups. Let $H={\pi }^{\ast }[\text{line}]$, and take the total transform $E$ on $X$ of the exceptional curve of the first blowup. The intersection formula for blowing up a surface givesTherefore the nonzero Picard group element $H+E$ satisfiesThis is the isotropic divisor from a nontrivial birational morphism to the projective plane.

For a morphism $\varphi :{\mathbb{P}}^2\to {\mathbb{P}}^n$, the pullback of the hyperplane bundle has the formfor an integer $d\ge 0$. If a line $\ell$ is contracted to a point, this bundle restricts trivially to $\ell$, whereasIts degree is therefore zero, so $d=0$. The homogeneous sections defining $\varphi$ are then constants, and $\varphi$ is constant. This proves the morphism from the projective plane contracting a line criterion.

Finally choose an integer $r>C^2$ and blow up $r$ distinct points of the smooth curve $C$. For the resulting morphism $\pi :X^{\prime }\to X$, let $E_1,\dots ,E_r$ be the exceptional curves. The strict transform isand the self-intersection after blowing up points on a smooth curve formula givesBlowing up a smooth point of a smooth curve does not change that curve itself, so $\pi {|}_{C^{\prime }}:C^{\prime }\to C$ is an isomorphism.

For an irreducible algebraic variety $X$ and $p\in X$, the local ring ${\mathcal{O}}_{X,p}$ consists of germs of rational functions regular on a neighbourhood of $p$. If $X\subseteq {\mathbb{A}}^n$ has ideal $I(X)$, its Zariski tangent space is

The given variety is the blowup of the affine plane at the origin. On the chart $W\ne 0$, put $W=1$; then $Y=XZ$, so $(X,Z)$ are free affine coordinates. On the chart $Z\ne 0$, put $Z=1$; then $X=WY$, so $(W,Y)$ are free affine coordinates. These two smooth affine-plane charts cover $V$, hence every point of $V$ is smooth.

If $(X,Y)\ne (0,0)$, the equation forcesConsequently $\pi$ restricts to an isomorphism away from the origin, and is therefore birational. Above the origin, however,so $\pi$ is not injective and cannot be an isomorphism of algebraic varieties. Birationality gives

For any morphism $\phi :V\to V^{\prime }$ with $V^{\prime }$ affine, its restriction to the exceptional curve $E\cong {\mathbb{P}}^1$ is constant: every regular function on a projective line is constant, and the affine coordinate functions of $V^{\prime }$ therefore have constant pullbacks. Since $E$ has more than one point, $\phi$ is not injective. If $V$ itself were affine, its identity morphism would contradict this conclusion, so $V$ is not affine.

Over $\mathbb{C}$, $V$ is not compact in the Euclidean topology: the sequence $(n,0,[1:0])$ has no convergent subsequence. Every complex projective variety is Euclidean compact. Since compactness is preserved by homeomorphisms, $V$ cannot be homeomorphic to a projective variety.

Two irreducible varieties $X$ and $Y$ are birational if they contain nonempty Zariski-open subsets $U\subseteq X$ and $V\subseteq Y$ that are isomorphic. Equivalently, their function fields are isomorphic:For an irreducible variety,An isomorphism of function fields preserves transcendence degree, so $\dim X=\dim Y$. This is the dimension from the function field.

Now let $K/\mathbb{C}$ be a finitely generated field extension. Choose field generators $a_1,\dots ,a_m$ and setThen $A$ is a finitely generated integral $\mathbb{C}$-algebra and $\mathrm{Frac}(A)=K$. The affine irreducible varietyhas function field $K$. Embed $X_0$ as a closed affine subvariety of some affine space ${\mathbb{A}}^N$, identify that affine space with the standard open chart of projective space ${\mathbb{P}}^N$, and take the projective closure $X$ of $X_0$. The closure of an irreducible topological space is irreducible, and $X_0$ is dense and open in $X$. Thus $X$ is a projective variety andThis constructs the projective model of a finitely generated field.

For the curveputIn its function field, the defining equation givesHence $\mathbb{C}(X)=\mathbb{C}(t)$, so $X$ is rational. Equivalently, this is the normalization of y squared equals x times x minus one squared, whose smooth projective normalization is ${\mathbb{P}}^1$ and has geometric genus zero.

A smooth projective curve in the projective plane of degree $d$ has, by the genus of a smooth plane curve formula,Its genus can be zero only for $d=1$ or $d=2$. Conversely, a projective line and every smooth conic over $\mathbb{C}$ are rational, so each is birational to $X$. Therefore the required degrees are exactly

Finally, in ${\mathbb{A}}_{x,y,z}^3$ takeThe polynomial $F$ is irreducible: as a quadratic in $y$, it could factor only if $x(x-1{)}^2$ were a square polynomial, which it is not. Thus $S$ is an irreducible affine hypersurface of dimension two. Its function field isso $S$ is birational to ${\mathbb{A}}^2$.

The partial derivatives areSolving $F=F_x=F_y=0$ gives $x=1$, $y=0$, with $z$ arbitrary. By the Jacobian criterion,This is an irreducible subvariety of dimension one, giving the requested singular cylinder over a nodal curve.