Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 4 (Advanced Topics in Commutative Algebra)

Let $\mathcal{U}=(U_0,\dots ,U_m)$ be a finite affine open cover of the projective scheme $X$. For a coherent sheaf $\mathcal{F}$, form the Čech cochain complexThe two ways to omit any pair of indices have opposite signs, giving ${\delta }^2=0$. Its cohomology groups areA projective scheme over a field is a separated scheme. Finite intersections of its affine opens are affine, and a coherent sheaf is quasi-coherent. Higher sheaf cohomology of a quasi-coherent sheaf on an affine scheme vanishes. Thus this is an acyclic cover and the acyclic cover theorem identifies the displayed Čech cohomology with $H^p(X,\mathcal{F})$. In particular its degree-zero kernel glues compatible local sections to $\Gamma (X,\mathcal{F})$. A cover by $m+1$ opens also gives vanishing for $p>m$.

For the cohomology of twists on projective space, take $S=k[X_0,\dots ,X_r]$ with the usual grading and $U_i=D_{+}(X_i)$. On an intersection indexed by the nonempty set $J$,This follows directly from the construction of the twisting sheaf on projective space: sections of the twist are the degree-$n$ homogeneous fractions in the corresponding localization. The Čech differential preserves each Laurent monomial $X^a=X_0^{a_0}\cdots X_r^{a_r}$, where $a_i\in \mathbb{Z}$ and ${\sum }_i{a}_i=n$. Such a monomial appears precisely in summands withThe complex therefore decomposes as the direct sum of finite-dimensional cochain complexes indexed by these exponent vectors. Their differentials have only the alternating signs of the simplex incidence maps.

If $N(a)=\varnothing$, this is the ordinary unaugmented simplex cochain complex: its degree-zero kernel consists of the common value on every vertex and has dimension one, while all higher cohomology vanishes. If $\varnothing \ne N(a)\ne \{0,\dots ,r\}$, choose $v\notin N(a)$. A cochain homotopy inserting $v$ into the ordered index list, with the sign of that insertion, contracts this subcomplex. Insertion or deletion of $v$ preserves the condition $N(a)\subseteq J$; in $\delta h+h\delta$, terms inserting and deleting different vertices cancel in pairs, while the term inserting then deleting $v$ is the identity. Thus this entire monomial subcomplex is acyclic. Finally, if all $a_i<0$, the monomial occurs only in the full intersection, in degree $r$, and contributes one copy of $k$ there.

For $r\ge 1$, these three cases give the complete answer:In the first case counting nonnegative exponent vectors gives $h^0=\left(\genfrac{}{}{0ex}{}{n+r}{r}\right)$. In the top case write $a_i=-1-b_i$ with $b_i\ge 0$; then $\sum b_i=-n-r-1$, givingThis is the Laurent-monomial description of top cohomology on projective space. All intermediate degrees vanish for every twist, and the top degree vanishes when $n\ge -r$. If $r=0$, then ${\mathbf{P}}_k^0=\mathrm{Spec}k$ and every twist is trivial, so $H^0({\mathbf{P}}_k^0,\mathcal{O}(n))=k$ for every integer $n$, with all higher groups zero.

Use homogeneous coordinates $X_0,\dots ,X_n$. Multiplication by these degree-one sections defines the sheaf morphismOn $U_i=D_{+}(X_i)$, write $t_j^{(i)}=X_j/X_i$, with $t_i^{(i)}=1$, and let $E_j^{(i)}$ be the standard vector in component $j$ using the local line bundle frame $X_i^{-1}$ of $\mathcal{O}(-1)$. Then $ϵ(E_j^{(i)})=t_j^{(i)}$, so $ϵ$ is surjective since $ϵ(E_i^{(i)})=1$. Its kernel is a locally free sheaf with basisIndeed any vector in the kernel is uniquely a sum of these vectors, by solving for its $i$-th component.

The Kähler differential sheaf ${\Omega }_{{\mathbf{P}}^n/k}^1$ has basis $d{t}_j^{(i)}$ on $U_i$, since this chart is a polynomial affine scheme. Define the local map $d{t}_j^{(i)}\mapsto F_j^{(i)}$. To check that these isomorphisms glue, on $U_i\cap U_{\ell }$ put $u=t_i^{(\ell )}$ and $v=t_j^{(\ell )}$. ThenThe twisting sheaf on projective space frame changes by $E_j^{(i)}=u^{-1}{E}_j^{(\ell )}$, henceFor $j=\ell$, use $t_{\ell }^{(\ell )}=1$, $d{t}_{\ell }^{(\ell )}=0$ and $F_{\ell }^{(\ell )}=0$; the same formula applies. Thus the differential and kernel frames have identical transition matrices. We obtain the cotangent form of the Euler sequence:This construction works over any field, including positive characteristic.

The Kähler differential sheaf has rank $n$. Taking determinant line bundles in this locally split short exact sequence of sheaves givesThe left side is the tensor product of $n+1$ copies of $\mathcal{O}(-1)$, and $\det (\mathcal{O})=\mathcal{O}$. ConsequentlyIt is the canonical bundle of projective space. The determinant identity can also be seen directly by taking the wedge of the kernel basis followed by a lift of the quotient basis; changing that lift by a kernel vector leaves the wedge unchanged.

The identification with ${\mathbf{P}}^1$ means that $C$ is a smooth plane conic. An arbitrary singular or nonreduced conic over an algebraically closed field cannot be identified with the projective line, so this is a necessary qualification implicit in the requested description.

Choose an equation $\ell =0$ for $L\subset {\mathbf{P}}^3$ and a degree-two equation $q=0$ whose restriction to $L$ defines $C$. Its homogeneous ideal in ${\mathbf{P}}^3$ is $(\ell ,q)$. These equations form a regular sequence: $\ell$ is a non-zero-divisor in the polynomial ring, and modulo $\ell$ the ring is a polynomial ring in three variables, in which the nonzero quadratic $q$ is again a non-zero-divisor. The generator classes therefore form a basis of the conormal sheaf, with their grading shifts:One can verify the absence of relations from the two-generator Koszul complex: a relation between $\ell ,q$ is a multiple of $(q,-\ell )$, whose coefficients become zero modulo ${\mathcal{I}}_C$. This is the normal sheaf of a projective complete intersection calculation. Dualizing the conormal sheaf gives

These twists are restrictions from ${\mathbf{P}}^3$, not intrinsic degree-one twists on ${\mathbf{P}}^1$. A line in the plane cuts the degree-two curve in two points counted with multiplicity, so the degree of a line bundle ${\mathcal{O}}_C(1)$ is two. The Picard group of the projective line then givesEquivalently, the degree-two Veronese embedding $[s:t]\mapsto [s^2:st:t^2]$ pulls each linear coordinate back to a quadratic. Thus the normal bundle of a smooth plane conic isThe same splitting follows from the normal sequence for the two embeddings:Its outer terms are ${\mathcal{O}}_{{\mathbf{P}}^1}(4)$ and ${\mathcal{O}}_{{\mathbf{P}}^1}(2)$. The extension group class lies in ${\mathrm{Ext}}^1(\mathcal{O}(2),\mathcal{O}(4))=H^1({\mathbf{P}}^1,\mathcal{O}(2))=0$, by the preceding Čech cohomology calculation, so the sequence splits. Both methods distinguish the ambient twist from the intrinsic degree on the curve.

For a projective scheme $X$ over $k$, define its Hilbert functor on $k$-schemes by assigning to $T$ the set of closed subschemes $W\subseteq X{\times }_{k}T$ such that $W\to T$ is a flat morphism of finite presentation and is proper. For projective $X$, the properness is automatic for these closed families. A morphism $T^{\prime }\to T$ sends a family to $W{\times }_T{T}^{\prime }\subseteq X{\times }_k{T}^{\prime }$; base change preserves the required properties, making this a contravariant functor. The embedding in $X\times T$ is part of the data, so these are embedded families rather than families modulo automorphisms of $X$.

Choose a very ample line bundle on $X$. In a flat projective family of finite presentation the fibre Hilbert polynomial is locally constant. Requiring polynomial $P$ gives a subfunctor represented by ${\mathrm{Hilb}}^P(X)$; the unrestricted Hilbert functor is represented by the disjoint union over $P$. For the specified $Z$, use its polynomial $P_Z$ and let $h=[Z]$ be the resulting $k$-point of the Hilbert scheme $H$.

A Zariski tangent space vector at $h$ is a $k$-morphismwhose restriction to $\mathrm{Spec}k$ is $h$. Here $D$ is the ring of dual numbers. Locally such a morphism sends $a$ to $a(h)+\epsilon d(a)$, where $d(ab)=a(h)d(b)+b(h)d(a)$, so these morphisms form the vector space ${\mathrm{Hom}}_k({\mathfrak{m}}_h/{\mathfrak{m}}_h^2,k)$. By the representing property of the Hilbert scheme, the same tangent vector is exactly a first-order embedded deformationwith the last equality an equality of embedded closed subschemes.

Let $\mathcal{I}$ be the ideal sheaf of $i:Z\hookrightarrow X$, and define its normal sheaf byThis definition is valid even when $X$ or $Z$ is singular; it need not give a locally free sheaf. We now derive the tangent-space identification by constructing inverse maps, rather than merely invoking the normal sheaf's name.

Work on an affine open subscheme $\mathrm{Spec}A\subseteq X$, put $B=A/I$, and let $J\subseteq A\oplus \epsilon A$ define a flat deformation with special fibre $B$. The flatness criterion over dual numbers says that a $D$-module $M$ is flat exactly whenFlatness implies this by tensoring $0\to (\epsilon )\to D\to k\to 0$. Conversely lift a $k$-basis of $M/\epsilon M$ to $M$. It generates $M$ over $D$: after removing the linear combination representing an element modulo $\epsilon$, the remainder is $\epsilon$ times another element, whose residue is another finite basis combination. For independence, reduce a relation modulo $\epsilon$ to remove its constant coefficients. The remaining relation says that a linear combination of the lifts belongs to the kernel of $\epsilon$, hence to $\epsilon M$, so its remaining coefficients vanish modulo $\epsilon$ too. The lifts are a free $D$-basis, establishing flatness.

Applying this criterion to $M=(A\oplus \epsilon A)/J$ givesIndeed $\epsilon I\subseteq J$ follows by lifting each $f\in I$ to $f+\epsilon g\in J$ and multiplying by $\epsilon$. Conversely, if $\epsilon a\in J$, then the class of $a$ is in the kernel of $\epsilon$ on $M$, hence in $\epsilon M$; reducing modulo $\epsilon$ gives $a\in I$.

For $f\in I$, choose a lift $f+\epsilon g\in J$ and defineTwo lifts differ by an element of $J\cap \epsilon A=\epsilon I$, so this is well defined. Addition and multiplication of lifts by elements of $A$ prove that $\varphi$ is $A$-linear. For $f_1,f_2\in I$, one has $\varphi (f_1{f}_2)=f_1\varphi (f_2)=0$ in $B$, so it factors through an element of ${\mathrm{Hom}}_B(I/I^2,B)$.

Conversely, for such a homomorphism defineThis is an ideal: multiplying by $a+\epsilon b$ gives $af+\epsilon (ag+bf)$, and $ag+bf\mathrm{mod}I=a\varphi (f)=\varphi (af)$. Its reduction is $I$. To prove that its quotient is flat, suppose $\epsilon [a+\epsilon b]=0$. The condition $\epsilon a\in J_{\varphi }$ says $a\in I$. Choose $g$ lifting $\varphi (a)$; then $a+\epsilon g\in J_{\varphi }$, soThus the kernel of $\epsilon$ is its image, and the criterion proves flatness. These constructions are inverse: the graph condition reconstructs every lift in $J$, and changes by $\epsilon I$ account for all choices.

Localization respects both constructions. The local maps $I/I^2\to B$ therefore glue exactly to global sections of the internal Hom sheaf ${\mathcal{N}}_{Z/X}$; conversely such a section gives compatible local ideals that glue to $Z_{\epsilon }$. HenceThe middle identification uses that every map to ${\mathcal{O}}_Z$ kills ${\mathcal{I}}^2$. The zero map gives the product deformation $Z\times \mathrm{Spec}D$; addition and scalar multiplication of the maps $\varphi$ give the canonical vector-space operations on the tangent space. This proves the Zariski tangent space of a Hilbert scheme formula without a smoothness or regular-embedding hypothesis.