Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 89 (Arithmetic Geometry)

An integral scheme is nonempty, reduced and irreducible. For an affine scheme $X=\mathrm{Spec}R$, reducedness is equivalent to $\sqrt{(0)}=0$: nilpotent elements vanish in every stalk, and conversely a nonzero element has nonzero localization somewhere. Irreducibility is equivalent to the nilradical being prime. Indeed, if $ab$ lies in every prime, then $V(ab)=V(a)\cup V(b)=X$; irreducibility makes one of these closed subsets all of $X$, so $a$ or $b$ lies in the nilradical. Conversely, if the nilradical is prime, its point is a generic point of the whole spectrum. Thus reducedness and irreducibility together say that zero is a prime ideal, with $R\ne 0$. Therefore

For a global idempotent $e$, its germ in any local ring is either zero or one. To see this, $e(1-e)=0$, and at least one of $e,1-e$ is a unit. Multiplication by that unit forces the other to vanish. On each affine chart,These equalities also hold globally. Hence $V(1-e)$ is a clopen set. Conversely, on a clopen subset $U$ and its complement, glue the sections one and zero. They define a global idempotent $e_U$, and the two constructions are inverse. This proves the idempotent–clopen correspondence.

Assume each connected component is open. Every clopen subset is a union of connected components, since its intersection with a connected component is clopen in that component. Each individual component is also closed, so it defines a nonzero idempotent. A decomposition into two nonempty disjoint clopen subsets corresponds exactly to a sum of two nonzero orthogonal idempotents. A union containing at least two components admits such a decomposition by separating one component from the rest. A single component admits none. Consequently the precise conclusion is

The printed definition needs both nonzero and orthogonal. Literally forbidding every sum of two nonzero idempotents gives a false statement in characteristic two. For $X=\mathrm{Spec}({\mathbb{F}}_2\times {\mathbb{F}}_2)$, the component idempotent $(1,0)$ decomposes as $(1,1)+(0,1)$; both summands are nonzero idempotents but are not orthogonal. In fact each of the three nonzero idempotents in this ring decomposes into the other two, although $X$ has two open connected components. Excluding zero is also necessary: zero corresponds to the empty subset, not to a component. With the standard primitive idempotent definition the preceding proof establishes the intended bijection in every characteristic.

The Absolute Frobenius morphism is the identity on the underlying topological space and acts on the structure sheaf by $a\mapsto a^p$. On an affine scheme it is induced by the ring homomorphism $A\to A$, $a\mapsto a^p$. This is additive in characteristic $p$, and the inverse image of a prime ideal $\mathfrak{p}$ is again $\mathfrak{p}$, since $a^p\in \mathfrak{p}$ exactly when $a\in \mathfrak{p}$. These affine maps agree under localization and hence glue.

A perfect ring of characteristic p is one whose Frobenius endomorphism is bijective. Since the Absolute Frobenius morphism fixes points, its map on the stalk at $x$ is precisely the Frobenius endomorphism of ${\mathcal{O}}_{X,x}$. A morphism of sheaves is an isomorphism exactly when it is an isomorphism on all stalks. ThereforeIn the reverse direction the inverse sheaf map, together with the identity on points, gives the inverse morphism of locally ringed spaces. This is perfection in the Frobenius sense, not the unrelated homological use of the word perfect.

For the function field assertions below, interpret the normal schemes as integral, as is implicit in writing a single field $k(X)$ or $k(Y)$. General normal schemes need not be irreducible; without this interpretation those fields are not defined globally as single fields. Set $K=k(Y)\subseteq L=k(X)$. For an affine open $V=\mathrm{Spec}A$ in $Y$, finiteness gives $f^{-1}V=\mathrm{Spec}B$, with $A\subseteq B$ finite and both rings normal domains. In fact $B$ is the integral closure of $A$ in $L$: any element of $L$ integral over $A$ is also integral over $B$ and therefore belongs to $B$. This local description yields the two forms of Frobenius factorization through a finite normal cover.

Assume $L^p\subseteq K$. For $b\in B$, its power $b^p$ belongs to $K$ and is integral over $A$, because $B$ is finite over $A$. Because $A$ is an integrally closed domain, $b^p\in A$. Thusis a ring homomorphism, defining a morphism $g_V:V\to f^{-1}V$. Its composition with $f:f^{-1}V\to V$ is the Absolute Frobenius morphism of $f^{-1}V$, because the composite ring map sends $b$ to $b^p$ in $B$.

These local morphisms glue. Explicitly, over a distinguished open $D(a)\subseteq V$, the ring rule isso it is exactly the localization of ${\gamma }_V$. For the underlying topology, each fibre of $f$ consists of a single point: if primes ${\mathfrak{q}}_1,{\mathfrak{q}}_2\subset B$ contract to the same $\mathfrak{p}$, thenso ${\mathfrak{q}}_1={\mathfrak{q}}_2$. The map $g_V$ sends that point of $V$ back to its unique point above it. Thus localization compatibility on a basis of a topology also gives compatibility on overlaps of arbitrary affine charts. We obtain

For uniqueness, $f$ is surjective on points, so the equation determines the point map of $g$. On each affine chart it then requires $f^{\ast }(g^{\ast }(b))=b^p$. The inclusion $A\hookrightarrow B$ is injective, so this determines $g^{\ast }(b)$ uniquely as $b^p$. Hence the sheaf map and the morphism are unique.

Assume $K\subseteq L^p$. Every $a\in A$ has a unique $p$th root $r(a)$ in $L$: existence is the field inclusion and uniqueness follows from injectivity of the Frobenius endomorphism in a field. Since $r(a{)}^p=a$, this root satisfies the monic polynomial $T^p-a$ over $B$. Because $B$ is an integrally closed domain, this gives $r(a)\in B$. The identitiesand uniqueness show thatis a ring homomorphism.

On $D(a)$ the rule sends a fraction to the fraction of its unique roots. Since $r(a{)}^p=a$, inverting $a$ is equivalent to inverting $r(a)$, so the rule localizes. Also, for a prime $\mathfrak{q}\subset B$, $r(a)\in \mathfrak{q}$ exactly when $a\in \mathfrak{q}$; hence the induced point map agrees with that of $f$. The localized rules consequently glue to a morphism $h:X\to Y$. Its pullback followed by the Frobenius endomorphism sends $a$ to $r(a{)}^p=a$, which provesIf another morphism satisfies this equation, it has the same point map because the Absolute Frobenius morphism fixes points, and its pullback of $a$ must be a $p$th root of $a$ in the normal domain $B$. That root is unique. Thus $h$ is unique.

A morphism of schemes $f:X\to Y$ is a flat morphism when each local ring ${\mathcal{O}}_{X,x}$ is a flat module over ${\mathcal{O}}_{Y,f(x)}$: tensoring with it preserves exact sequences. An open immersion induces isomorphisms on the local rings at its points, so it is flat.

For a closed immersion, the printed equivalence is valid with the additional hypothesis of finite presentation, for example over a locally Noetherian target, and with nonempty source. Here is the proof with that hypothesis. On an affine chart the map is $A\to A/I$, with $I$ finitely generated. If $\mathfrak{p}\supseteq I$, the flat cyclic module $A_{\mathfrak{p}}/I_{\mathfrak{p}}$ is free over the local ring $A_{\mathfrak{p}}$, and its residue-field dimension is one. Therefore its quotient map from $A_{\mathfrak{p}}$ is an isomorphism and $I_{\mathfrak{p}}=0$. Finite generation gives a neighbourhood of $\mathfrak{p}$ on which $I$ vanishes. The closed image $V(I)$ is consequently open. A nonempty clopen image in a connected target is the whole target; then $I$ vanishes at every prime and is zero. The immersion is an isomorphism. The converse is immediate. This proves the flat closed immersion of finite presentation assertion.

Without finite presentation, connectedness alone does not suffice. Take $A=C([0,1],\mathbb{R})$ and let $\mathfrak{m}=\{a:a(0)=0\}$. Let $I$ consist of functions vanishing on some neighbourhood of zero. The map $A\to A_{\mathfrak{m}}$ has kernel $I$: multiplying such a function by a continuous cutoff equal to one at zero kills it, whereas $ga=0$ with $g(0)\ne 0$ forces $a$ to vanish near zero. This map is surjective, because the reciprocal of any denominator nonzero at zero can be extended continuously from a small neighbourhood of zero to the whole interval. ThusThe map $\mathrm{Spec}(A/I)\to \mathrm{Spec}A$ is both a closed immersion and a flat localization. Its source is connected because its ring is local. Its target is connected because continuous idempotent functions on the connected interval can only be the constants zero and one. It is not an isomorphism: $I$ contains nonzero functions supported away from zero. This supplies a concrete counterexample to the unrestricted assertion. The kernel is a pure ideal; this is the general mechanism behind such examples, as explained in the Stacks Project's treatment of pure ideals.

For the finite-flat fibre assertion, first note the local algebra fact that a finitely generated flat module $M$ over a local ring $(A,\mathfrak{m})$ is free. Choose generators $m_1,\dots ,m_r$ whose residues form a basis of $M/\mathfrak{m}M$, using Nakayama's lemma. If a row $a=(a_1,\dots ,a_r)$ gives a relation $\sum a_i{m}_i=0$, the equational criterion for flatness supplies elements $n_j$ and a matrix $B=(b_{ij})$ such that $m_i={\sum }_j{b}_{ij}{n}_j$ and $aB=0$. Write each $n_j$ as a linear combination of the $m_i$, with coefficient matrix $C$. Then $m=BCm$. Independence modulo $\mathfrak{m}$ makes $BC$ congruent to the identity. This square matrix is invertible over the local ring, and $aBC=0$ implies $a=0$. Thus the chosen generators are a basis. This also justifies the cyclic-module step above; see the finite-flat local lemma.

Now let $\mathcal{A}=f_{\ast }{\mathcal{O}}_X$ for the finite flat morphism. On each affine chart this is the finite flat algebra defining its inverse image. Hence ${\mathcal{A}}_y$ is free over ${\mathcal{O}}_{Y,y}$, of some finite rank $r_y$. Let $\eta$ be the generic point of the irreducible target. Localizing that free module at $\eta$ givesEvery point specializes from the same $\eta$, so every $r_y$ equals the fixed rank at $\eta$. This remains valid if the irreducible target is nonreduced: one compares free ranks over its nonzero generic local ring, not necessarily over a field. The fibre is affine, andConsequentlyThis is finite flat rank over an irreducible scheme. No unjustified replacement of finite flat by finite locally free on an arbitrary base is needed.

For an algebraic curve of finite type over a field, its normalization of an integral scheme is finite and birational. If flat, the preceding rank is one, since the generic fibre has the same function field. Locally its normalization algebra $B$ is therefore a free $A$-module of rank one. The element $1_B$ generates: its image cannot vanish in the one-dimensional fibre algebra, and Nakayama's lemma applies. Thus $A\to B$ is an isomorphism. Conversely the identity normalization is flat.

The parenthetical definition in the PDF does not explicitly impose finite type. One can prove the conclusion even in that broader setting without assuming a finite normalization. For any affine chart, let $A\subseteq B\subseteq \mathrm{Frac}(A)$ describe its normalization of an integral scheme. Integrality makes the map on spectra surjective; if $B$ is flat it is faithfully flat. Faithful flatness gives contraction of ideals, $vB\cap A=vA$. For $b=u/v\in B$, with $u,v\in A$ and $v\ne 0$, one has $u=vb\in vB\cap A=vA$. Thus $u=va$ for some $a\in A$, so $b=a$. Therefore $B=A$ on every chart. We have proved the flat normalization is an isomorphism result and, in particular,

Use the polynomial graded ring $R[x_0,x_1,x_2,x_3]$, as in the PDF; the doubled brackets in the converted TeX are an OCR error. Let $I$ denote the homogeneous defining ideal. The four cubic coordinates in the proposed map never vanish simultaneously on ${\mathbb{P}}_F^1$, since $y_0^3$ and $y_1^3$ already have that property. Substitution makes each generator of $I$ zero, so they define a morphism into the generic fibre.

The two standard opens $D_{+}(x_0)$ and $D_{+}(x_3)$ cover $X$. Indeed, a homogeneous prime containing $x_0,x_3$ contains $x_2$ by $x_0{x}_3^2=x_2^3$, and then contains $x_1$ by $x_1^2={\pi }^2{x}_0{x}_2$. Such a prime contains the irrelevant ideal and gives no point of Proj.

On $D_{+}(x_0)$ put $a=x_1/x_0$, $b=x_2/x_0$, $c=x_3/x_0$. Over $F$, where $\pi$ is invertible, the equations giveThus the generic-fibre chart is exactly $\mathrm{Spec}F[t]$. The proposed map restricts to this isomorphism with $t=y_1/y_0$. On $D_{+}(x_3)$ put $s=x_2/x_3$. The equations giveso this chart is $\mathrm{Spec}F[s]$, and the map has $s=y_0/y_1$. On the overlap $s=t^{-1}$. These are the standard two charts of the projective line, proving the scheme-level isomorphismEquivalently, rescaling $x_1$ by ${\pi }^{-1}$ gives the usual twisted cubic. The chart proof also shows that the extra cubic relation introduces no additional generic-fibre scheme structure.

On $D_{+}(x_3)$, write $r=x_0/x_3$, $q=x_1/x_3$, and $s=x_2/x_3$. The equations include $q=\pi s^2$ and $r=s^3$; after making these substitutions, the remaining equations are identities. ThereforeThe other chart isThere is a homomorphism $B\to R[t]$ defined by $a\mapsto \pi t$, $b\mapsto t^2$, $c\mapsto t^3$. To prove injectivity rather than merely exhibit a parametrization, reduce monomials using the four relations. Powers of $a$ reduce to at most one $a$; any remaining product of $a$ with $b$ or $c$ eliminates $a$; powers of $c$ reduce to at most one $c$. Thus every element is an $R$-linear combination ofTheir images are $1,\pi t,t^{2n},t^{2n+3}$, with pairwise distinct degrees. Since $R$ is a domain and $\pi \ne 0$, these images are $R$-linearly independent. Hence they are a basis, the map is injective, andIn particular $B$ is torsion-free, indeed free, as an $R$-module.

The two charts cover $X$ by the prime-ideal argument in part i. Both coordinate rings are torsion-free over the discrete valuation ring $R$, hence flat: over a principal ideal domain a torsion-free module is a filtered union of finitely generated free submodules, and filtered colimits preserve flatness. Thus the morphism $X\to \mathrm{Spec}R$ is flat on an open cover and therefore flat everywhere. It is proper because $X$ is a closed subscheme of the proper projective space ${\mathbb{P}}_R^3$. Consequently $X$ is both proper and flat over $R$.

The overlap of the two charts is obtained by inverting $c=t^3$. Since $c^2=b^3$, this also inverts $b=t^2$; then $t=c/b$ and $t^{-1}=b/c$. Thus its coordinate ring is $R[t,t^{-1}]$. The other chart has coordinate $s=t^{-1}$. A global section is therefore an element in the intersectioninside $R[t,t^{-1}]$: a Laurent polynomial with both only nonnegative and only nonpositive exponents is constant.

Let $k=R/(\pi )$. Reducing the presentation of $B$ givesAs a vector space and as a ring extension, this iswhere $a^2=0$ and $a$ is annihilated by $b,c$. In particular $a\ne 0$: the free basis of $B$ in part ii remains a basis after tensoring with $k$. Although $a$ mapped to $\pi t$ before reduction, reducing the subring is not the same operation as taking its image inside $k[t]$; that distinction is exactly where the extra section arises.

On the overlap, $c$ is invertible and $ac=0$, so $a$ restricts to zero. Therefore the section $a$ on the first chart glues to zero on the second chart and defines a nonzero global nilpotent section $\eta$ on $X_s$. More precisely the reduced first-chart ring is $k[t^2,t^3]$, andin $k[t,t^{-1}]$. The only extra compatible sections are multiples of $a$. HenceThe map printed in the question reduces $R$ modulo $\pi$ and includes its image as constants. Its kernel is $\pi R$ and its cokernel is $k\eta$. More strongly, even the usual base-change map after tensoring,is not surjective. This is global-section base-change failure in a flat projective family, despite the properness and flatness proved above.

Choose an ordered affine open cover $\mathcal{U}=(U_i)$ of the separated scheme. For a quasi-coherent sheaf $\mathcal{F}$, the Čech cochain groups and differential areTerms corresponding to empty intersections are zero. Restriction commutes with restriction, so the terms in $d^2$ cancel in pairs. DefineHere an affine cover computes the Čech groups independently of the chosen affine cover. The reason is that finite intersections of affine opens in a separated scheme are affine: the intersection of two is the inverse image of the closed diagonal inside their affine product. Repeating gives the finite-intersection assertion. Quasi-coherent sheaves have no higher cohomology on these intersections. Apply an injective resolution of the sheaf and form the double complex of its Čech cochains. In one direction the higher cohomology vanishes on every affine intersection, leaving precisely the displayed Čech complex. In the other direction the augmented Čech complex of the flasque resolution terms is exact, leaving the global-section resolution computing sheaf cohomology. Thus the two cohomologies agree, and refinement maps between affine-cover complexes induce the same groups. No such affine-intersection claim is being made for arbitrary nonseparated schemes.

For an exact sequence of quasi-coherent sheaves, sections on an affine intersection give an exact sequence of modules, including surjectivity: this follows from the equivalence between quasi-coherent sheaves and modules on an affine scheme. Taking products yields a short exact sequence of Čech cochain complexes. To describe the connecting map, lift a degree-$p$ cocycle in $C^p({\mathcal{F}}^{\prime \prime })$ to a cochain in $C^p(\mathcal{F})$. Its differential maps to zero in $C^{p+1}({\mathcal{F}}^{\prime \prime })$ and therefore is a cocycle in $C^{p+1}({\mathcal{F}}^{\prime })$. Changing the lift adds a coboundary there; changing the original representative does the same. This defines the connecting homomorphism. The same lift-and-differentiate argument shows that a class is in its kernel exactly when it lifts to a cocycle in the middle complex, and verifies exactness at the other terms. Thus the resulting sequence isThe sequence must be in quasi-coherent sheaves for this affine-sections argument; that is the intended category here.

For the punctured affine plane, use $U=D(x)$ and $V=D(y)$. Their rings are $k[x^{\pm 1},y]$ and $k[x,y^{\pm 1}]$, and the intersection ring is $k[x^{\pm 1},y^{\pm 1}]$. The two-open Čech complex isConsequentlyThe denominator contains precisely the span of Laurent monomials for which at least one exponent is nonnegative. Every Laurent polynomial has a unique monomial expansion, so the remaining monomials are linearly independent in the quotient and span it. Thereforea countably infinite-dimensional vector space. The brackets here denote cohomology classes. This calculates the Čech cohomology of the punctured affine plane, rather than merely observing that some nonzero cohomology exists.