Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 113 (Algebraic Geometry)

An open immersion $j:U\to X$ is a morphism of schemes that identifies $U$, including its structure sheaf, with an open subscheme of $X$.

Let $X={\mathbb{A}}_k^2$ and $U=X\setminus \{(0,0)\}$. The inclusion is an open immersion and $X$ is an affine scheme, but $U$ is not affine. Indeed, regular functions extend across the missing codimension-two point, soWere $U$ affine, the inclusion would therefore correspond to an isomorphism of coordinate rings and would be an isomorphism $U\cong X$, a contradiction.

For an example with both schemes affine, the principal open subschemeis a nontrivial open immersion. The affine line is connected because $k[t]$ has no nontrivial idempotent elements.

The even-degree subring is the second Veronese subringwhere $u,v,w$ have degree one in the regraded ring. The canonical invariance of the Proj construction under passage to a Veronese subring givesOn homogeneous points this is the degree-two Veronese embeddingIt is an isomorphism onto the closed subschemeThus the quotient homomorphism $k[u,v,w]\twoheadrightarrow R^{(2)}$ supplies the requested closed immersion of $\mathrm{Proj}{R}^{(2)}$ into the projective plane.

For a point $p\in {\mathbb{A}}_k^1$, let $a$ be the image of $t$ in its residue field $\kappa (p)$. The scheme-theoretic fiber isIf $p\ne (t)$, then $a\ne 0$. The quadratic polynomial $xy-a{z}^2$ is irreducible. Indeed, after setting $z=0$, any hypothetical linear factors must restrict, up to nonzero scalars, to $x$ and $y$; comparing the $xz$ and $yz$ coefficients then forces both $z$ coefficients to vanish, contradicting the nonzero $z^2$ coefficient. Hence its homogeneous coordinate ring is an integral domain, so $Y_p$ is an integral scheme.

At the origin $p=(t)$, the fiber is $V_{+}(xy)$, the union of the two distinct projective lines $x=0$ and $y=0$, and is therefore not irreducible. It is nevertheless a reduced scheme because the ideal $(xy)=(x)\cap (y)$ equals its radical. Every other fiber is integral and hence reduced. Thus the fiber is integral exactly away from the origin, and it is reduced at every point of ${\mathbb{A}}_k^1$.

A morphism of schemes is proper when it is of finite type, separated, and universally closed.

Each property is local on the target. Finite type is local by its affine definition. The restrictions of the diagonal morphism of $\phi$ over the open sets $U_i$ are closed immersions; since being a closed subset is local on an open cover, the diagonal itself is a closed immersion, so $\phi$ is separated. Finally, after any base change $Y^{\prime }\to Y$, the inverse images $U_i^{\prime }=Y^{\prime }{\times }_Y{U}_i$ cover $Y^{\prime }$. For every closed $C\subseteq X{\times }_Y{Y}^{\prime }$, its image has closed intersection with every $U_i^{\prime }$ because the restricted base-changed morphism is closed. The image is therefore closed in $Y^{\prime }$. Thus $\phi$ is universally closed and hence proper. This proves that properness is local on the target.

Write $f:U={\mathbb{A}}_{\mathbb{C}}^1\setminus \{1\}\to {\mathbb{A}}_{\mathbb{C}}^1$ for $f(t)=t^2$. For any closed point $a\in \mathbb{C}$, the fiber is the zero-dimensional closed subscheme of $U$ cut out by $t^2-a$. It is finite over $\mathbb{C}$, and every finite morphism is proper. This includes $a=1$, whose fiber in $U$ consists only of $t=-1$.

The morphism itself fails the valuative criterion for properness. Take the discrete valuation ring $R=\mathbb{C}[[s]]$ with fraction field $K=\mathbb{C}((s))$. The $K$-point $t=1+s$ lies in $U(K)$ and lies over the $R$-point of the target defined byAny extension $\mathrm{Spec}R\to {\mathbb{A}}^1$ of that $K$-point must still be given by $t=1+s$, whose closed point maps to the deleted point $t=1$. It therefore cannot factor through $U$. The required lift does not exist, so $f$ is not proper. This is an instance of the fact that proper closed-point fibers do not imply properness.

A Cartier divisor on an integral scheme is an open cover $(U_i)$ together with nonzero rational functions $f_i$ such that every ratio $f_i/f_j$ is a regular unit on $U_i\cap U_j$.

Every prime Weil divisor on ${\mathbb{P}}_k^n$ is cut out by an irreducible homogeneous polynomial because the polynomial ring $k[x_0,\dots ,x_n]$ is a unique factorization domain. Hence any Weil divisor can be represented by a homogeneous rational expressionof some total degree $d$. On the standard chart $U_i=D_{+}(x_i)$, put $f_i=F/x_i^d$. This is a degree-zero rational function, and on $U_i\cap U_j$,is a regular unit. These local equations form a Cartier divisor whose associated Weil divisor is the original one.

Now put $P=X\times {\mathbb{P}}^n$ and let $H=X\times {\mathbb{P}}^{n-1}$ be the hyperplane divisor at infinity. Its complement is $X\times {\mathbb{A}}^n$. Iterating the given invariance under multiplication by ${\mathbb{A}}^1$ givesThe localization sequence for the divisor class group shows that every class on $P$ is a pullback of a class on $X$ plus an integer multiple of $[H]$. Restriction to the generic fiber ${\mathbb{P}}_{K(X)}^n$ kills pullbacks from $X$ and sends $[H]$ to the generator of $\mathrm{Cl}({\mathbb{P}}_{K(X)}^n)\cong \mathbb{Z}$. Therefore the sum is direct, provingThis is the divisor class group of a projective-space bundle with trivial vector bundle.

An ${\mathcal{O}}_X$-module $\mathcal{F}$ is a quasi-coherent sheaf when every affine open $V=\mathrm{Spec}A\subseteq X$ has $\mathcal{F}{|}_V\cong \tilde{M}$ for some $A$-module $M$. For a morphism $f:X\to Y$, its direct image sheaf iswhile the pullback of a sheaf of modules is

If $i:Z\hookrightarrow X$ is a closed immersion of Noetherian schemes, then on an affine open $V=\mathrm{Spec}A\subseteq X$ one has $Z\cap V=\mathrm{Spec}(A/I)$. The restriction of $i_{\ast }{\mathcal{O}}_Z$ corresponds to the cyclic $A$-module $A/I$, so it is finitely generated. Hence $i_{\ast }{\mathcal{O}}_Z$ is a coherent sheaf; more generally this is the direct image of a coherent sheaf under a closed immersion.

Coherence need not survive an arbitrary pushforward. Letbe the open immersion. The sheaf ${\mathcal{O}}_{D(x)}$ is coherent, butis not a finitely generated $k[x]$-module. Therefore $j_{\ast }{\mathcal{O}}_{D(x)}$ is not coherent.

Choose a finite affine cover $\mathcal{U}=(U_j)$ of the Noetherian scheme $X$. Because $X$ is separated, every finite intersection $U_J$ is affine. Its inverse image $V_J=Z\cap U_J$ under the closed immersion is also affine. By the definition of the direct image sheaf,The Čech complexes for $i_{\ast }\mathcal{F}$ on $\mathcal{U}$ and for $\mathcal{F}$ on the induced cover $(Z\cap U_j)$ are therefore identical, including their restriction maps. Both affine covers are acyclic for the relevant quasi-coherent sheaves, so the acyclic cover theorem givesfor every $q$. This is cohomology under a closed immersion.

For $X={\mathbb{P}}_k^n$, use its $n+1$ standard affine opens. The induced cover of $Z$ is still acyclic, and its Čech complex has no cochains in degrees greater than $n$. The cohomological dimension bound from an affine cover therefore yields

Let the degree-$d$ projective hypersurface $X_d$ be defined by the homogeneous polynomial $F$. Multiplication by $F$ identifies its ideal sheaf with ${\mathcal{O}}_{{\mathbb{P}}^n}(-d)$, giving the ideal-sheaf sequence of a projective hypersurfaceIn particular, the kernel in the question is ${\mathcal{O}}_{{\mathbb{P}}^n}(-d)$.

For $n=2$, additivity of the Euler characteristic in this short exact sequence and the line-bundle formulagiveHere $H^2(X_d,{\mathcal{O}}_{X_d})=0$ by the preceding cohomological-dimension argument. Moreover $H^0({\mathbb{P}}^2,\mathcal{O}(-d))=H^1({\mathbb{P}}^2,\mathcal{O}(-d))=0$ for $d>0$, so the long exact sequence gives $H^0(X_d,{\mathcal{O}}_{X_d})\cong k$. Consequentlywhich is the genus-degree formula for a projective plane curve.