Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 20 / 5

A line bundle $\mathcal{L}$ is a globally generated line bundle when the evaluation mapis surjective. Equivalently, its global sections span each fibre of the line bundle; locally at every point, some section is a generator.

Choose a finite generating family $s_0,\dots ,s_n$. On the open set where $s_i$ generates $\mathcal{L}$, the ratios $s_j/s_i$ are regular functions, giving a morphism of schemes into the standard chart $D_{+}(x_i)$ of projective space. The ratios obey the usual transition rules on overlaps, so these chart maps glue toThe tuple is computed using any local trivialization of $\mathcal{L}$; changing that trivialization multiplies all entries by the same invertible function. The construction has ${\phi }^{\ast }\mathcal{O}(1)\simeq \mathcal{L}$ and pulls back the coordinate sections to the chosen $s_i$.

There is a finiteness qualification for an arbitrary $X$: global generation alone need not provide such a finite family. It does if $X$ is quasi-compact, since the open sets on which individual sections generate have a finite subcover. Without that hypothesis, take $X={\coprod }_{m\ge 1}{\mathbb{P}}_k^m$ and let $\mathcal{L}$ restrict to $\mathcal{O}(1)$ on each component. This line bundle is globally generated, but any $N$ global sections have a common zero on a component ${\mathbb{P}}^m$ with $m\ge N$. Thus this is a globally generated line bundle without finite generators. This is finite global generation on a quasi-compact scheme. The finite-family construction is automatic in the projective case asked next.

For projective nonsingular $X$, put $V=⟨s_0,\dots ,s_n⟩$. The length-two criterion for a very ample linear system says that $\phi$ is a closed immersion precisely when, after extending to an algebraic closure, $V$ separates distinct points and tangent directions. Equivalently, the evaluationis surjective for every length-two geometric closed subscheme $E\subseteq X$. Two distinct points give point separation; a nonreduced length-two subscheme supported at one point gives separation of a direction in the Zariski tangent space. For the complete space $V=H^0(X,\mathcal{L})$, this says exactly that $\mathcal{L}$ is a very ample line bundle. For a chosen smaller $V$, it is the chosen linear system of divisors which must be a very ample linear system; mere global generation is insufficient.

Let $U={\mathbb{P}}^2\setminus C$. The localization sequence for the divisor class group givesEvery prime Weil divisor of $U$ extends by closure to one of ${\mathbb{P}}^2$, and the only removed prime Weil divisor is $C$. Rational functions have the same function field on the two spaces, so the kernel on divisor class groups consists exactly of multiples of $[C]$.

The divisor class group of ${\mathbb{P}}^2$ is $\mathbb{Z}$, generated by the class $[H]$ of a line. To see the degree identification, if a plane curve $D$ has degree $e$ and homogeneous equation $F_D$, then $F_D/{\ell }^e$, for a line equation $\ell$, is a rational function with principal Weil divisor $D-eH$. Degrees of principal Weil divisors are zero, so $[H]$ has infinite order. In particular, $[C]=d[H]$. The localization sequence therefore yieldsThis is the divisor class group of a plane-curve complement. It includes $d=1$, when the group is zero, and does not require the removed curve to be nonsingular.

The hyperplane sections through $P$ are spanned by the three linear forms $x_1,x_2,x_0-x_3$. Their common zero on $X$ is precisely $P$, so away from $P$ the linear system of divisors is base-point-free and defines the projection from a point on a smooth quadricIn particular, the chart-ratio construction in part (a) makes this a morphism of schemes on $X\setminus \{P\}$, not merely a rational map there.

Write a target point as $[u:v:w]$. The corresponding line through $P$ consists of pointsSubstituting in the quadric equation givesRemoving $P$ means $\mu \ne 0$, so set $\mu =1$. Over the residue field $K$ of the target point, its scheme-theoretic fibre is consequentlyIf $u\ne v$, this is one reduced point, with $\lambda =-uw/(u-v)$. If $u=v$ and $uw\ne 0$, it is empty: the corresponding line meets the quadric only at the removed point, with intersection multiplicity two. If $u=v$ and $uw=0$, the line lies entirely on the quadric, and the scheme-theoretic fibre is ${\mathbb{A}}_K^1$, the line with $P$ removed.

The last case occurs at exactly $[0:0:1]$ and $[1:1:0]$. Their lines are respectivelythe two rulings of a smooth quadric surface through $P$. Thus the image consists of the complement of the line $u=v$, together with those two points. The line $u=v$ parametrizes directions in the tangent plane $x_1=x_2$ at $P$. The calculation describes the scheme-theoretic fibres over arbitrary target points and works in every characteristic.

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.