Past exam of the introduction to schemes course of the University of Oxford / 2017 (A15253W1)

Source: released examination PDF, bundle pages 114–115.

Apply the surjectivity of $f(Y)$ to the identity morphism $1_Y$. It gives a morphism of schemes $g:Y\to X$ satisfying $f\circ g=1_Y$. To obtain the other identity, compare $g\circ f$ and $1_X$ in $X(X)$. Their images under $f(X)$ agree, sinceThe injectivity of $f(X)$ therefore gives $g\circ f=1_X$. ThusOnly surjectivity for the test scheme $Y$ and injectivity for the test scheme $X$ were needed. This is the isomorphism-detection argument behind the Yoneda lemma for the functor of points.

No. Let $k={\mathbb{F}}_2$ be the field with two elements, put $A=k[\epsilon ]/({\epsilon }^2)$, and take the closed immersioninduced by the ring homomorphism $q:A\twoheadrightarrow k$ with $q(\epsilon )=0$.

For every field $K$, a ring homomorphism $\alpha :A\to K$ must send $\epsilon$ to zero: $\alpha (\epsilon {)}^2=0$, and a field has no nonzero nilpotent elements. Consequently $\alpha$ factors uniquely through $q$. Passing from ring homomorphisms to morphisms of schemes between affine schemes gives a bijectionThis concerns all morphisms of schemes, without requiring them to be over $k$: if $K$ does not have characteristic two, both sets are empty; otherwise each contains exactly one element.

Nevertheless, $f$ is not an isomorphism of schemes. Its map on global sections is $q$, whose kernel of a ring homomorphism contains the nonzero element $\epsilon$, so $q$ is not a ring isomorphism. The target is the nonreduced scheme defined by the dual numbers, whereas the source is a reduced scheme. This illustrates why field-valued points do not detect nilpotents.

Yes. First, the hypothesis implies that $f(T)$ is injective for every scheme $T$, even when $T$ is not an affine scheme. Indeed, suppose $a,b:T\to X$ satisfy $f\circ a=f\circ b$. Choose an affine open cover $T={\bigcup }_{\lambda }{V}_{\lambda }$. On each $V_{\lambda }$, injectivity of $f(V_{\lambda })$ gives $a{|}_{V_{\lambda }}=b{|}_{V_{\lambda }}$. By gluing morphisms of schemes, $a=b$.

Now choose an affine open cover $Y={\bigcup }_i{U}_i$, and denote its inclusion maps by $j_i:U_i\hookrightarrow Y$. Surjectivity of $f(U_i)$ gives morphisms of schemes $g_i:U_i\to X$ with $f\circ g_i=j_i$. On $U_i\cap U_j$, the two restrictions have the same composite with $f$. By the injectivity just established for arbitrary test schemes,In particular, the overlaps need not be affine schemes. Gluing morphisms of schemes supplies $g:Y\to X$ with $g{|}_{U_i}=g_i$, and $f\circ g=1_Y$ because this holds on every $U_i$.

Finally,Injectivity of $f(X)$, already proved without an affineness assumption, gives $g\circ f=1_X$. ThereforeThis proves the relevant instance of affine test schemes determine scheme morphisms: the restricted functor of points still retains the information needed to construct the inverse.

The localization of a ring $R_r$ is generated over $R$ by the single element $1/r$, since every fraction can be written asThe presentation of a localization at one element isTo verify the presentation even when $R$ has zero divisors, observe that $r$ is invertible in $R[t]/(rt-1)$ with inverse $t$. Hence the map from $R$ to this quotient ring extends uniquely to $R_r$. This extension and the displayed map are inverse ring homomorphisms: they agree on $R$ and send the designated inverse of $r$ to the designated inverse of $r$.

Thus $R_r$ is a finitely generated algebra over a ring, and the criterion for a morphism of finite type between affine schemes applies to $u:\mathrm{Spec}{R}_r\to \mathrm{Spec}R$. There is just one affine scheme in the required cover of the source, with one algebra generator. Therefore $u$ is of finite type. If $r$ is nilpotent, the same presentation is the zero ring, and the source is the empty scheme; the conclusion still holds.

Put $C=\overline{\{z\}}$, where the closure is taken in the underlying topological space of $X$. It is nonempty because it contains $z$.

Suppose $C=A\cup B$ for two closed subsets of $C$ in the subspace topology. Since $C$ is a closed subset of $X$, both $A$ and $B$ are also closed subsets of $X$. The point $z$ lies in one of them, say $A$. By the defining minimality of the closure, every closed subset of $X$ containing $z$ contains $C$. Hence $C\subseteq A$, so $A=C$.

Thus $C$ cannot be expressed as the union of two proper closed subsets, proving that $C$ is an irreducible closed subset. This proves closure of a point is irreducible in every topological space; the scheme assumption is unnecessary. Equivalently, $z$ is a generic point of its own closure.

Yes: every scheme is a Kolmogorov space. Choose an affine open subscheme $V=\mathrm{Spec}A$ containing $x$. If $y\notin V$, take $U=V$.

Otherwise, the distinct points $x,y\in V$ correspond to distinct prime ideals ${\mathfrak{p}}_x,{\mathfrak{p}}_y\subseteq A$. There is an element in one of these prime ideals and not the other. After interchanging the names of the two points if necessary, chooseThe principal open subschemecontains $y$ and does not contain $x$. It is an open subset of $V$, hence an open subset of $X$ because $V$ is open in $X$. This establishes $(\ast )$ in every case, proving schemes are Kolmogorov spaces.

Write ${\mathbb{P}}^r={\mathbb{P}}_k^r$ over its ground field; the argument also works over any Noetherian affine base. Since $\iota$ is a closed immersion, $X$ is a noetherian scheme and a quasi-compact scheme. LetHere the pullback of a sheaf of modules makes $L$ an invertible sheaf, the restriction of the twisting sheaf on projective space. It is an ample line bundle.

By global generation by sufficiently positive twists, choose $n\ge 0$ such that $F(n)$ is a globally generated sheaf. Finitely many global sections suffice: at each point, finite generation of the stalk of the coherent sheaf $F(n)$ lets us select finitely many generating germs of global sections. Those sections generate on a neighbourhood, because their cokernel sheaf is a coherent sheaf with zero stalk at that point. A finite subcover, using that $X$ is a quasi-compact scheme, supplies finitely many global sections generating everywhere. This is finite global generation on a quasi-compact scheme, and gives a surjective sheaf morphism

Tensor with $(L^{\vee }{)}^{\otimes n}$, where $L^{\vee }$ is the dual of a sheaf. Here tensoring with an invertible sheaf is exact: locally the tensor product of sheaves is tensoring with a free module of rank one. Using the permitted identity $L\otimes L^{\vee }\cong {\mathcal{O}}_X$, we obtainA finite direct sum of sheaves of invertible sheaves is a locally free sheaf of finite rank, and it is a coherent sheaf on the noetherian scheme $X$. This constructs the required $G$. No regularity or smoothness assumption on $X$ is needed.

Use homogeneous coordinates $x_0,\dots ,x_r$ on projective space and letThese $r+1$ open subschemes cover $X$. Each is a closed subscheme of the affine scheme $D_{+}(x_i)$, hence is an affine scheme.

Every nonempty finite intersection is affine as well. Indeed, inside $D_{+}(x_{i_0})$ the intersection with $D_{+}(x_{i_1}),\dots ,D_{+}(x_{i_p})$ is the principal open subscheme obtained by invertingIntersecting it with the closed subscheme $X$ gives $U_{i_0}\cap \cdots \cap U_{i_p}$, again an affine scheme.

For any quasi-coherent sheaf $F$, vanishing of quasi-coherent cohomology on an affine scheme makes every such intersection acyclic for $F$. The acyclic cover theorem therefore identifies the sheaf cohomology of $F$ with the Čech cohomology of this cover. Its alternating Čech cochain complex has termswhich vanish for $p>r$: one cannot choose $p+1$ distinct indices from $r+1$ indices. ConsequentlyThe choice $N=\max \{1,r\}$ meets the strict requirement $N>0$ even when $r=0$. It is independent of $F$, and applies to all quasi-coherent sheaves, including those which are not coherent sheaves. This is the cohomological dimension bound from an affine cover.

Fix a coherent sheaf $F$ and an integer $k>M$. Set $K_0=F$. Repeatedly apply a to construct short exact sequences of sheaveswhere each $G_j$ is a coherent sheaf and a locally free sheaf. Each kernel sheaf $K_{j+1}$ is a coherent sheaf because $X$ is a noetherian scheme. Thus the construction can be repeated as many times as needed; it does not require the kernel sheaves themselves to be locally free sheaves.

For every integer $q>M$, the long exact sequence in sheaf cohomology containsBoth outer groups vanish by the hypothesis, since $q$ and $q+1$ exceed $M$. Hence ${\delta }_j$ is an isomorphism. Iterating this dimension shifting in sheaf cohomology gives

Choose $s\ge 0$ with $k+s>N$, where $N$ is the uniform bound from b. The coherent sheaf $K_s$ is a quasi-coherent sheaf, so b makes the last group zero. ThereforeThe argument uses a finite initial segment of a locally free resolution. There is no need to assume that every coherent sheaf has a finite locally free resolution: the uniform bound from b is what terminates the argument.