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An irreducible complex analytic hypersurface is a closed analytic subset of pure complex codimension one which cannot be expressed as the union of two proper closed analytic subsets. Here a local defining function of a complex analytic hypersurface means a reduced local equation: a holomorphic germ generating its vanishing ideal, with each local branch occurring once. The word reduced matters. Merely prescribing the zero set would allow both $f$ and $f^2$, and the requested uniqueness would then be false.

The local facts used are these: in holomorphic coordinates the local ring ${\mathcal{O}}_{X,p}$ is $\mathbb{C}\{z_1,\dots ,z_n\}$, a regular local ring and a unique factorization domain; its units are exactly the germs nonzero at $p$; and the analytic Nullstellensatz identifies the zero-set ideal of a germ $f$ with $\sqrt{(f)}$. A reduced germ is a product of distinct irreducible factors. Unique factorization makes its principal ideal radical: if $f\mid h^r$, every irreducible factor of $f$ divides $h$, so $f\mid h$. Thus two reduced germs with the same hypersurface germ have $(f)=(g)$. Write $g=uf$, $f=vg$. The local ring is a domain, so $uv=1$. Hence $u$ is a holomorphic unit and $u(p)\ne 0$. This proves reduced local defining functions differ by a holomorphic unit. Global irreducibility does not imply local irreducibility at a singular point; the same argument uses the product of all local branches and therefore covers that case too.

A divisor on a complex manifold is a locally finite integer combination of irreducible analytic hypersurfaces. On a compact manifold only finitely many have nonzero coefficients. Choose local meromorphic equations $f_i$ for a divisor $D$, by products of powers of reduced local equations. Their ratios are holomorphic units on overlaps. For the holomorphic line bundle associated to a divisor, choose frames satisfyingThe ratios obey the cocycle identity and so define a holomorphic line bundle $[D]$. This convention makes the local meromorphic sections $f_i{e}_i$ agree, with divisor $D$. Changing the equations by units changes frames and gives an isomorphic bundle.

For hyperplanes $H_1,H_2$ in Complex projective space, choose nonzero linear equations ${\ell }_1,{\ell }_2$. The degree-zero quotient $r={\ell }_1/{\ell }_2$ is a global meromorphic function and has divisor $H_1-H_2$. A principal divisor on a complex manifold gives a trivial associated bundle: using the same global meromorphic equation on every chart makes all frame transitions equal to one. The allowed tensor-product rule now givesIf the hyperplanes coincide the same conclusion is immediate. Equivalently, their common bundle is the hyperplane bundle $\mathcal{O}(1)$.

Locally at the centre of a blowup of a complex manifold at a point, replace a coordinate neighbourhood in ${\mathbb{C}}^n$ byAway from zero the direction is uniquely $[z]$, so this map is an isomorphism there; glue it to $X\setminus \{p\}$. In the chart ${\ell }_j\ne 0$, write $z_j=u$ and $z_i=u{t}_i$ for $i\ne j$. These are smooth holomorphic coordinates on the blowup. The exceptional divisor is $u=0$, globally $E={\sigma }^{-1}(p)\cong \mathbb{P}(T_{p}X)$.

The canonical bundle formula for a point blowup isThe local chart also explains the exponent: pulling back a coordinate volume form gives $\pm u^{n-1}du\wedge {\bigwedge }_{i\ne j}d{t}_i$, so the Jacobian section has divisor $(n-1)E$.

Since $Y$ is smooth at the centre, choose coordinates with $Y=\{z_1=0\}$. Its strict transform is a genuine analytic hypersurface. In a blowup chart $j\ne 1$, it is the closed set $t_1=0$: off $E$, the original equation $u{t}_1=0$ reduces to $t_1=0$, and taking the closure restores the points with $u=0$. In chart $j=1$, there are no such off-exceptional points, and the strict transform is empty there. These descriptions agree on overlaps and show analyticity, with $\tilde{Y}\cap E\cong {\mathbb{CP}}^{n-2}$. The total transform has the extra factor $u$ exactly once, soUnder the given anticanonical assumption, this yieldsTo prove the only-if direction, one must rule out an accidental trivial power of $[E]$. The local blowup is the tautological line bundle over ${\mathbb{CP}}^{n-1}$, with $E$ its zero section. Its normal bundle is therefore ${\mathcal{O}}_E(-1)$. The normal bundle of a smooth analytic hypersurface identifies $[E]{|}_E$ with this bundle. If $n\ge 2$, restrict to a projective line inside $E$; the restriction of $[E{]}^{\otimes (2-n)}$ has degree $n-2$. A trivial line bundle has degree zero, so it is trivial only if $n=2$. Conversely the displayed tensor factor is literally trivial when $n=2$. This is the anticanonical strict transform under a point blowup criterion:For completeness, dimension one cannot satisfy the premise in this compact setting: an irreducible analytic hypersurface is one point, and $[-Y]$ has degree $-1$ on its curve component, whereas $K_X$ has degree $2g-2$. Thus no exceptional dimension-one case invalidates the equivalence.

Two successive point blowups of a smooth algebraic surface suffice. Let $X_2\to X_1={\mathbb{A}}^2$ be the blowup of the affine plane at the origin. In its chart $x=uy$, with coordinates $(y,u)$, the total-transform equation isRemoving the exceptional factor gives the strict transformAbove the original origin it has just one point, $(y,u)=(0,0)$, which is still singular. The other chart is $y=vx$, where the strict transform has equation $1-v^5{x}^3=0$ and does not meet the exceptional divisor $x=0$. Thus there are no other points above the origin to resolve.

Blow up the remaining point to obtain $X_3\to X_2$. In the chart $u=vy$, with coordinates $(y,v)$, the total transform of $C_2$ isand hence its strict transform isThe derivative of $v^2-y$ with respect to $y$ is $-1$, so this is nonsingular, even in characteristics two or five. In the other chart $y=wu$, the strict transform has equation $1-w^{3}u=0$ and does not meet the exceptional divisor $u=0$. Therefore the only point of $C_3$ mapping to the original origin is the smooth point $(y,v)=(0,0)$.

The required sequence isThe local parameter $v$ there gives $y=v^2$ and $x=v^5$, also verifying the resolved branch directly. This is the resolution of the (2,5) cusp by two blowups. Its tangency to an exceptional divisor does not affect the requested nonsingularity of the strict transform; making the whole total transform have normal crossings is a stronger task.

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.

Before blowing up, the fiber over zero is the smooth curveand the blowup center $p$ lies on it. Its total transform is the union of the strict transform ${\tilde{F}}_0$ and the exceptional divisor $E$. Blowing up a smooth point of a smooth curve does not change the curve, so ${\tilde{F}}_0\cong {\mathbb{P}}^1$, while $E\cong {\mathbb{P}}^1$. They meet transversely in the single point recording the tangent direction of $F_0$ at $p$. Thereforeis two copies of the complex projective line meeting at one point.

The exceptional divisor is $E\cong {\mathbb{P}}^{r-1}$, so its Chow groups are generated by the linear subspaces $[{\Gamma }_k]$. The complement $X\setminus E$ is isomorphic through the blow-up map to ${\mathbb{P}}^r\setminus \{p\}$. The localization sequence for $\{p\}\subseteq {\mathbb{P}}^r$ shows that the latter's Chow groups are generated by the restrictions of linear spaces $H_j$ through $p$ for $1\le j\le r$; its zero-dimensional Chow group vanishes.

Under the complement isomorphism, $H_j\setminus \{p\}$ corresponds to the open part of the strict transform ${\tilde{H}}_j$. A second localization sequence, now for $E\subseteq X$, shows that $A_{\ast }(X)$ is generated by the lifts $[{\tilde{H}}_j]$ together with the images $[{\Gamma }_k]$ from $A_{\ast }(E)$, exactly as claimed.

This irreducible plane branch has parametrization $x=t^5$, $y=t^2$. Its only singular point is the origin. The resolution of the (2,5) cusp by two blowups makes its strict transform smooth in every characteristic.

Blow up the origin and use the chart $x=uy$, giving the strict transform $u^2-y^3=0$. Blow up its remaining singular point and use $u=vy$, giving $v^2-y=0$, which is smooth because its derivative in $y$ is $-1$. The complementary chart at each stage has no point of the strict transform on its exceptional divisor. These two blowups of a smooth algebraic surface resolve the branch, although the smooth transform is still tangent to an exceptional divisor.