Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 20 (Topics in Algebraic Geometry)

A morphism of schemes is a morphism of locally ringed spaces. Thus it consists of a continuous map $f:|X|\to |Y|$ and a homomorphism of sheaves of ringssuch that, for every $x\in X$, the induced map on stalksis a local homomorphism: it carries the maximal ideal of the source into the maximal ideal of the target. Equivalently, the inverse image of the target maximal ideal is the source maximal ideal. The locality condition on every stalk distinguishes a morphism of schemes from a general morphism of ringed spaces.

The affine-target adjunction for schemes gives the natural bijectionA morphism of schemes $f$ gives the homomorphism on global sections induced by $f^{\#}$, using $\Gamma (\mathrm{Spec}A,\mathcal{O})=A$.

Conversely, let $\alpha :A\to \Gamma (X,{\mathcal{O}}_X)$ be a ring homomorphism. Choose an affine open subscheme cover $U_i=\mathrm{Spec}{B}_i$ of $X$. Restriction of global sections gives homomorphisms $A\to B_i$, hence morphisms of schemes $f_i:U_i\to \mathrm{Spec}A$. On any affine open subscheme $W\subseteq U_i\cap U_j$, both restrictions correspond to the same homomorphism $A\to \Gamma (W,{\mathcal{O}}_W)$. They therefore agree on $W$. Such affine opens cover the overlap, so the $f_i$ glue uniquely to a morphism of schemes $f:X\to \mathrm{Spec}A$.

The two constructions are inverse: the first recovers $\alpha$ on each $U_i$, hence on all of $X$, and the second recovers every restriction $f{|}_{U_i}$ of a given $f$. This proof needs neither affineness nor quasi-compactness of $X$.

Let $\mathfrak{m}$ be the maximal ideal of the local ring $A$. Every open subset of $\mathrm{Spec}A$ containing the closed point $\mathfrak{m}$ is the whole spectrum of a commutative ring: it contains a principal open subscheme $D(a)$ with $a\notin \mathfrak{m}$, and that $a$ is a unit, so $D(a)=\mathrm{Spec}A$.

Given $f:\mathrm{Spec}A\to {\mathbb{P}}_{\mathbb{Z}}^n$, choose a standard affine open subscheme $D_{+}(x_i)$ containing $f(\mathfrak{m})$. Its preimage is consequently all of $\mathrm{Spec}A$. The affine-target adjunction for schemes expresses $f$ in this chart by elements $b_j\in A$ for $j\ne i$, the images of $x_j/x_i$. It is represented by homogeneous coordinates with $a_i=1$ and $a_j=b_j$.

Conversely, a tuple $(a_0,\dots ,a_n)$ with some $a_i\in A^{\times }$ defines a morphism of schemes into $D_{+}(x_i)$ by $x_j/x_i\mapsto a_j/a_i$. Choosing another unit entry gives the same morphism of schemes, since the usual projective space transition functions identify the ratios. Multiplying all entries by one unit does not change any ratio. If two such tuples define the same morphism of schemes, choose a unit entry $a_i$ in the first and a unit entry $a_j^{\prime }$ in the second. In the second chart, the function $x_i/x_j$ pulls back to $a_i^{\prime }/a_j^{\prime }$. Because the whole map lies in $D_{+}(x_i)$, this ratio is a unit, so $a_i^{\prime }$ is a unit too. Equality in this chart gives $a_j/a_i=a_j^{\prime }/a_i^{\prime }$ for every $j$, hence $a_j^{\prime }=(a_i^{\prime }/a_i)a_j$.

Thus the correspondence is exactlyThis is the projective coordinates over a local ring description.

For a general ring, the key open-neighbourhood argument fails. Even a tuple generating the unit ideal need not have any unit entry. For example, over $A=k\times k$ the pair $((1,0),(0,1))$ defines a map to ${\mathbb{P}}^1$ whose two points have images $[1:0]$ and $[0:1]$. Neither coordinate is a unit, and no common unit multiple changes that fact. The map lies in no single standard chart. More generally, maps into projective space correspond to invertible sheaf quotients of $A^{n+1}$; the quotient need not be a free rank-one module outside the local ring case.

For a point $y\in Y$ with residue field $\kappa (y)$, the scheme-theoretic fibre is the base change of a morphism of schemesThe map $\mathrm{Spec}\kappa (y)\to Y$ is the canonical point map. This fibre product of schemes retains the structure sheaf, including any nilpotents, rather than just the set $f^{-1}(y)$. For an affine scheme map induced by $A\to B$, at $\mathfrak{p}\in \mathrm{Spec}A$ it is

Compute each scheme-theoretic fibre by tensoring with the residue field of the chosen base point.

For the first map, put $L=\kappa (\mathfrak{p})$ for a point $\mathfrak{p}$ of $\mathrm{Spec}k[T]$ and let $t$ be the image of $T$ in $L$. ThenIf the characteristic of a field is not two and $t\ne 0$, the two factors $U-t$ and $U+t$ are coprime, so the Chinese remainder theorem gives $L[U]/(U^2-t^2)\simeq L\times L$: the scheme-theoretic fibre is two distinct $L$-points. If $t=0$, its ring is $L[U]/(U^2)$, a dual number ring, so it is a nonreduced double point. In characteristic two, $U^2-t^2=(U-t{)}^2$ at every point, giving a nonreduced double point in every scheme-theoretic fibre. This covers the generic point, where $L=k(T)$, as well as closed points defined by irreducible polynomials.

For the arithmetic map, the generic scheme-theoretic fibre isOver a closed point $(p)$ it is $\mathrm{Spec}{\mathbb{F}}_p[T]/(T^2+1)$. At $p=2$ this is $\mathrm{Spec}{\mathbb{F}}_2[ϵ]/({ϵ}^2)$, since $T^2+1=(T+1{)}^2$. For odd $p$, the finite field multiplicative group is cyclic, and $-1$ is a square exactly when $p\equiv 1(\mathrm{mod}4)$. ThusIn the second case there is one degree-two closed point over ${\mathbb{F}}_p$, which becomes two points after extending the residue field to an algebraic closure. The case $p=2$ remains nonreduced after such extension.

For $\mathrm{Spec}\mathbb{C}\to \mathrm{Spec}\mathbb{Z}$, the unique source point maps to the generic point $(0)$. Since all nonzero integers are invertible in $\mathbb{C}$,Indeed $\mathbb{C}{\otimes }_{\mathbb{Z}}\mathbb{Q}=\mathbb{C}$, whereas $\mathbb{C}{\otimes }_{\mathbb{Z}}{\mathbb{F}}_p=\mathbb{C}/p\mathbb{C}=0$. The distinction between a reduced split scheme-theoretic fibre and a nonreduced double point is essential in the first two examples.

Necessity follows from commutativity of the fibre product of schemes square: if $p(z)=x$ and $q(z)=y$, then $f(x)=g(y)$.

For sufficiency, put $s=f(x)=g(y)$ and $K=\kappa (s)$, $L=\kappa (x)$, $L^{\prime }=\kappa (y)$. The maps on local rings induce field embeddings $K\hookrightarrow L,L^{\prime }$. The canonical point maps into $X$ and $Y$ therefore give a morphism of schemesThe tensor product of commutative algebras $L{\otimes }_K{L}^{\prime }$ is nonzero. To see the hinted fact directly, choose a $K$-basis of $L$ containing $1$; tensoring that basis with $L^{\prime }$ makes $1\otimes 1$ nonzero. A nonzero unital commutative ring has a prime ideal, so its spectrum of a commutative ring has a point $w$. Its projections to the two field spectra are their unique points. The image $z$ of $w$ consequently projects to $x$ and $y$.

The desired point exists precisely when the base images coincide. This is the point-lifting property of a scheme fibre product. It does not claim that such a point is unique: different prime ideals of the tensor product may give different points over the same pair.

Let $f:X\to Y$ be a surjective morphism of schemes and let $Y^{\prime }\to Y$ be any morphism of schemes. Take any point $y^{\prime }\in Y^{\prime }$ and let $y$ be its image in $Y$. Surjectivity supplies $x\in X$ with $f(x)=y$. By the point-lifting property of a scheme fibre product proved in part (c), there is a point of $X{\times }_Y{Y}^{\prime }$ projecting to both $x$ and $y^{\prime }$. Thus every point of $Y^{\prime }$ is hit by the projection.

This proves that surjectivity is preserved by base change. No restriction is imposed on the base change of a morphism of schemes. No flatness or finite-type assumption is required.

For a smooth projective curve $C$ of geometric genus $g$ and a divisor on an algebraic curve $D$, write $\ell (D)=h^0(C,{\mathcal{O}}_C(D))$ and let $K_C$ be a canonical divisor. The curve form of the Riemann-Roch theorem isHere $\mathrm{deg}{K}_C=2g-2$ and $\ell (K_C)=g$. In line bundle form,The usual curve convention is a smooth geometrically integral projective curve; geometric embedding tests may be checked after extending the ground field to an algebraic closure.

First use the canonical map to obtain the embedding. Its canonical divisor has degree six and $h^0(K_C)=4$. Over an algebraically closed ground field, every effective degree-two divisor on an algebraic curve $E$ satisfies $h^0(E)=1$: a second section would give a nonconstant rational function with poles bounded by $E$, hence a map $C\to {\mathbb{P}}^1$ of degree at most two. Degree one would make $C$ rational, and degree two would make it hyperelliptic, both excluded here.

The Riemann-Roch theorem now givesThus the complete linear system of divisors $|K_C|$ separates every pair of points and every tangent direction, including the tests $E=2p$. By the length-two criterion for a very ample linear system, it is a very ample linear system. Consequently the canonical map embeds $C$ in ${\mathbb{P}}^3$, with degree of a projective curve six. Its image is a nondegenerate projective variety, since the four canonical sections are linearly independent.

The space of homogeneous quadrics in four variables has dimension ten, while the Riemann-Roch theorem givesRestriction therefore has a nonzero kernel, giving a quadric $Q$ containing $C$. This quadric is irreducible: a reducible quadric is a union of two planes, or a double plane, and an integral curve lying in it would lie in a plane, contradicting nondegeneracy.

Similarly, homogeneous cubics form a twenty-dimensional space, whereasThere are at least five independent cubics vanishing on $C$. The multiples of $Q$ by linear forms give only four dimensions, so choose a cubic $F$ vanishing on $C$ which is not divisible by $Q$.

The polynomials $Q,F$ are a regular sequence, hence their intersection $D$ is a projective complete intersection of pure dimension one and degree $2\cdot 3=6$. It contains $C$, which already has degree six. The unmixedness of a complete intersection excludes embedded components. Equal degrees force $C$ to be the only irreducible component and force multiplicity one at its generic point. By generic reducedness with no embedded components, this makes $D$ reduced, so $D=C$ as schemes. We have provedThis is a canonical genus-four curve as a quadric-cubic intersection. The quadric may be smooth or a cone; the argument does not require it to be smooth. The same construction descends over the ground field when the curve is geometrically nonhyperelliptic, since the restriction maps and canonical embedding are defined there.

Let $\mathcal{L}={\mathcal{O}}_C(1)$. Its degree is five. Because the linear span of $C$ has dimension at least three, restrictions of linear forms supply at least four independent global sections, soIf $h^1(C,\mathcal{L})=0$, the Riemann-Roch theorem gives $h^0(\mathcal{L})=6-g$, and therefore $g\le 2$ immediately.

If instead $\mathcal{L}$ is special, the Clifford inequality for curves givesso its integer dimension is at most three, a contradiction. Here is a short proof of the required Clifford inequality for curves, so the bound need not be assumed. Put $r=h^0(\mathcal{L})>0$ and $s=h^0({\omega }_C\otimes {\mathcal{L}}^{-1})>0$. At a smooth geometric point, choose bases of these two section spaces with strictly increasing orders of vanishing $a_1<\cdots <a_r$ and $b_1<\cdots <b_s$. Such bases exist because each successive order-of-vanishing quotient has dimension at most one. The products of the first basis with the first element of the second basis, followed by products of the last element of the first basis with the remaining elements of the second, have ordersThey are linearly independent global sections of ${\omega }_C$, proving the product dimension bound for sections on a curve. Thus $r+s-1\le h^0({\omega }_C)=g$. Combining this with Riemann-Roch theorem, $r-s=\mathrm{deg}\mathcal{L}+1-g$, gives $2r\le \mathrm{deg}\mathcal{L}+2$, which is exactly the displayed Clifford inequality for curves.

The special case is impossible, and henceThis is the genus bound for a nonplanar degree-five curve. The geometric argument can be carried out after extension to an algebraic closure, which preserves these section dimensions.

A closed immersion $i:X\to Y$ is a morphism of schemes whose underlying map is a homeomorphism onto a closed subset and whose map of structure sheavesis surjective. Equivalently, $X$ is the closed subscheme defined by a quasi-coherent ideal sheaf of a closed subscheme on $Y$. The affine-local characterization is particularly useful: for every affine open subscheme $V=\mathrm{Spec}A\subseteq Y$,for an ideal $I\subseteq A$, with the map induced by the quotient homomorphism.

Work over an affine open subscheme $V=\mathrm{Spec}A$ of $Z$. Since $g$ is a closed immersion, its inverse image is $\mathrm{Spec}(A/I)$ for some ideal $I$. Since $f$ is a closed immersion, the inverse image of this affine scheme in $X$ is $\mathrm{Spec}((A/I)/J)$ for an ideal $J\subseteq A/I$. Let $\tilde{J}$ be the inverse image of $J$ in $A$. The composite on this open is induced by the surjectionThe affine-local characterization from part (a) therefore shows that $g\circ f$ is a closed immersion. In particular, closed-subscheme ideals compose by taking the inverse image of the second quotient ideal.

Over an affine open subscheme $\mathrm{Spec}A$ of the target, a closed immersion has the form $\mathrm{Spec}(A/I)\to \mathrm{Spec}A$. Pulling it back along an affine scheme map $\mathrm{Spec}B\to \mathrm{Spec}A$ givesThe identification follows directly by tensoring the quotient presentation; it does not require $B$ to be flat over $A$. Since $B\to B/IB$ is surjective, this is again a closed immersion. These affine calculations cover every base change of a morphism of schemes.

A morphism of schemes $f:X\to Y$ is a separated morphism if its diagonal morphismis a closed immersion. The diagonal morphism is induced by the pair of identity maps on $X$. Thus separatedness is a condition relative to the given target $Y$, not a requirement that $X$ be absolutely separated.

Since $g$ is a separated morphism, the diagonal morphism ${\Delta }_g:Y\to Y{\times }_{Z}Y$ is a closed immersion. The squareis cartesian. Indeed, on test schemes its pullback condition is $(f(x),y)=(t,t)$, forcing $y=t=f(x)$, and leaving exactly the choice of $x$. The graph morphism of schemes ${\Gamma }_f$ is therefore a base change of a morphism of schemes of ${\Delta }_g$. Part (c) shows that ${\Gamma }_f$ is a closed immersion.

Now assume additionally that $h=g\circ f:X\to Z$ is a closed immersion. The projection $p_Y:X{\times }_{Z}Y\to Y$ is its base change of a morphism of schemes along $g$, so is a closed immersion by part (c). Sincepart (b) shows that $f$ is a closed immersion. This covers the further assumption in the final unheaded paragraph.

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.