Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 113 (Algebraic Geometry)

A morphism $f:X\to S$ is a separated morphism when its diagonal morphismis a closed immersion.

For the requested example, take the affine plane with doubled origin: glue two copies $U,V\cong {\mathbb{A}}_k^2$ by the identity onThe opens $U$ and $V$ are affine, while their intersection is the punctured affine plane, which is not affine. Indeed, its regular functions are still $k[x,y]$; if it were affine, the canonical map to $\mathrm{Spec}k[x,y]={\mathbb{A}}_k^2$ would be an isomorphism, contrary to the missing origin. The resulting scheme is not separated: in a separated scheme, the intersection of two affine opens is the inverse image of the closed diagonal inside their affine product and is therefore affine.

On the standard charts $D_{+}(x)$ and $D_{+}(y)$ of ${\mathbb{P}}_k^1$, a section of the twisting sheaf on projective space $\mathcal{O}(d)$ is represented by a degree-zero element of the corresponding localization of $k[x,y](d)$. A global section is therefore a homogeneous polynomial of degree $d$ when $d\ge 0$. There are no nonzero global sections for $d<0$. Hence

Local isomorphism on every member of a fixed affine cover does not imply a global isomorphism: the local identifications may have different transition functions. For example, $\mathcal{O}$ and $\mathcal{O}(1)$ are both trivial on the two standard affine charts of ${\mathbb{P}}^1$, but they are not isomorphic because their spaces of global sections have dimensions one and two.

An injective map between line bundles need not be an isomorphism. Multiplication by a nonzero section giveson ${\mathbb{P}}^1$; its cokernel is a nonzero skyscraper sheaf supported at the zero of the section.

For $n\ge 2$, the omitted point has codimension at least two in the normal integral scheme ${\mathbb{P}}_k^n$. Regular functions extend across such a subset, soThuson $\mathrm{Spec}k$, and it is a coherent sheaf because it corresponds to the one-dimensional $k$-vector space $k$.

The complement of a rational point in ${\mathbb{P}}_k^1$ is ${\mathbb{A}}_k^1$. Its regular functions are $k[t]$, which is infinite-dimensional over $k$. Its pushforward to $\mathrm{Spec}k$ is therefore quasi-coherent but not coherent.

For an example on $X$, choose a projective line $L\subset {\mathbb{P}}_k^n$ through $p$. Thenis closed in $X$. For the closed immersion $i:C\hookrightarrow X$, the sheaf $\mathcal{F}=i_{\ast }{\mathcal{O}}_C$ is coherent, butConsequently ${\pi }_{\ast }\mathcal{F}$ is not coherent on $\mathrm{Spec}k$.

The fibre product of schemes $X{\times }_{S}Y$ comes with projections to $X$ and $Y$ having equal composites to $S$, and is universal with this property. A morphism is universally closed when every base change is closed on underlying topological spaces.

The structural morphism ${\mathbb{A}}_k^1\to \mathrm{Spec}k$ is closed because its target has one point, but it is not universally closed. After base change by ${\mathbb{A}}_k^1$, it becomes the projection ${\mathbb{A}}_k^2\to {\mathbb{A}}_k^1$; the closed hyperbola $V(xy-1)$ has image $D(x)$, which is not closed.

For a finite-type universally closed nonseparated example, glue two copies of ${\mathbb{P}}_k^1$ along the complement of one rational point, producing a projective line with a doubled point. It is of finite type. After every base change, each of its two projective-line charts maps closedly to the base, so the image of any closed subset, being the union of the two closed images, is closed. Thus the structural morphism is universally closed. It is not separated because the two doubled points violate uniqueness in the valuative criterion for separatedness, or equivalently because its diagonal is not closed.

Fix $x\in X$ and choose affine opens $x\in U\subseteq X$ and $f(x)\in W\subseteq S$ with $f(U)\subseteq W$. The open subset $U{\times }_{W}U$ of $X{\times }_{S}X$ contains $\Delta (x)$. If $U=\mathrm{Spec}A$ and $W=\mathrm{Spec}R$, its diagonal is induced by the surjectionso it is a closed immersion. Hence every diagonal morphism is locally a closed immersion into an open subset, and therefore is a locally closed immersion.

For ${\mathbb{A}}_k^1$, the product is ${\mathbb{A}}_k^2=\mathrm{Spec}k[x,y]$ and the diagonal is $V(x-y)$. Its complement is the principal affine open

Take instead the separated scheme $X={\mathbb{A}}_k^2$. The complement of its diagonal in $X{\times }_{k}X\cong {\mathbb{A}}_k^4$ is ${\mathbb{A}}_k^4\setminus {\mathbb{A}}_k^2$, where the removed diagonal has codimension two. Its global functions still form the polynomial ring of ${\mathbb{A}}_k^4$. Were the complement affine, its canonical morphism to $\mathrm{Spec}$ of this ring would identify it with all of ${\mathbb{A}}_k^4$, which is impossible. Thus this diagonal complement is not affine.

For a Noetherian separated integral scheme regular in codimension one, a Weil divisor is a finite integer combination of integral codimension-one closed subschemes. Principal divisors are the valuation divisors of nonzero rational functions, and the divisor class group is

For an open immersion $U\subseteq X$, restriction induces a surjectionevery prime divisor of $U$ closes to a prime divisor of $X$, while divisors supported in $X\setminus U$ form the kernel. Therefore $\mathrm{Cl}(X)=0$ implies $\mathrm{Cl}(U)=0$.

Affine schemes need not have trivial class group. For example,is a normal affine surface with $\mathrm{Cl}(X)\cong \mathbb{Z}/2\mathbb{Z}$. The height-one prime $(x,z)$ represents the nonzero class: twice this divisor is the principal divisor of $x$, but the divisor itself is not principal.

A sheaf $\mathcal{F}$ of ${\mathcal{O}}_X$-modules is quasi-coherent when every affine open $U=\mathrm{Spec}A$ has $\mathcal{F}{|}_U\cong \tilde{M}$ for some $A$-module $M$.

Cover the target ${\mathbb{P}}_k^1$ by $D_{+}(y_0)$ and $D_{+}(y_1)$. On the first chart put $s=y_1/y_0$; its inverse image is $D_{+}(x_0)$ with coordinate $t=x_1/x_0$, and the morphism on rings isAs a $k[s]$-module,is free of rank two. The same calculation on the other standard chart uses $s^{-1}\mapsto t^{-2}$. Hence $f_{\ast }{\mathcal{O}}_X$ is locally free of rank two on the target.

The complement $Y$ is covered by the $m$ principal affine opens $D(f_1),\dots ,D(f_m)$. Every finite intersection is again a principal affine open, so this is an acyclic cover for ${\mathcal{O}}_Y$. Its Čech cochain complex has no degree-$m$ term because there is no intersection of $m+1$ distinct cover members. Therefore

Translate $p$ to the origin. The complement ${\mathbb{A}}_k^3\setminus \{0\}$ has the affine cover $D(x),D(y),D(z)$, and its second Čech cohomology isThe class of $x^{-1}{y}^{-1}{z}^{-1}$ is nonzero, so $H^2({\mathbb{A}}_k^3\setminus \{p\},\mathcal{O})\ne 0$. On the other hand, ${\mathbb{A}}_k^3\setminus \ell =D(x)\cup D(y)$, and the first part with $m=2$ gives $H^2({\mathbb{A}}_k^3\setminus \ell ,\mathcal{O})=0$. Since sheaf cohomology is invariant under scheme isomorphism, the two complements are not isomorphic.

Two nonisomorphic punctual schemes areTheir underlying spaces each have one point, but the second has a nonzero nilpotent element and the first is reduced.

After translating their common support to the origin, punctual closed subschemes of ${\mathbb{A}}_k^1$ correspond to ideals of $k[t]$ whose radical is $(t)$. Since $k[t]$ is a principal ideal domain, each such ideal is $(t^r)$ for a unique $r\ge 1$. Its coordinate ring has basis $1,t,\dots ,t^{r-1}$ and hence dimension $r$. Equal dimensions force equal exponents, so $Z$ and $Z^{\prime }$ are in fact the same closed subscheme after the common coordinate choice, and in particular are isomorphic.

In ${\mathbb{A}}_k^2$, the idealsdefine distinct punctual closed subschemes supported at the origin. Both quotient rings have dimension two over $k$, with bases $1,y$ and $1,x$ respectively. Thus they have equal-dimensional global-section spaces despite being distinct embedded closed subschemes.