Formal definition of the pack Intermediate Hartshorne Chapter III builds the cohomology of sheaves on a scheme as the right-derived functors of the global-sections functor, proves Serre's vanishing on Noetherian affine schemes, computes the cohomology of all line bundles on projective space by an explicit Čech calculation, packages duality as Serre duality, and develops the relative theory: higher direct images, cohomology and base change, flatness, and smoothness. Many of its exercises do not anchor to a single Babel Bible unit — they braid the Čech machinery 04.03.03 , the projective-space computation 04.03.04 , the vanishing/finiteness pair 04.03.05 , the derived-functor and Ext formalism 04.03.06 , higher direct images 04.03.07 , Serre duality 04.08.03 , and the smooth/flat morphism layer 04.02.05 together.
This pack collects nine such exercises — two easy, four medium, three hard — each with a hint and a full solution. It is meant to be read alongside its prerequisite units rather than as a standalone development. The exercises group loosely by Hartshorne section: dimension vanishing and Čech warm-ups (easy), projective-space cohomology and Euler-characteristic counts (medium), and Serre-duality, base-change, and flatness arguments (hard).
The conventions throughout are Hartshorne's: X is a Noetherian scheme over a field k ; F a coherent sheaf; O X ( d ) the d -th Serre twist on P k n ; H i ( X , F ) the derived-functor cohomology, equal to Čech cohomology for an acyclic affine cover of a separated scheme. The canonical sheaf on a smooth projective variety of dimension n is ω X , and Serre duality reads H i ( X , F ) ≅ H n − i ( X , ω X ⊗ F ∨ ) ∨ for locally free F .
Key theorem with full solution Intermediate Before the pack proper, we work one exercise in full as an exemplar of the format. The remaining eight follow the same structure (problem, hint, full answer in <details> blocks).
Lead exercise. Compute H i ( P k n , O ( d )) for all i and all d ∈ Z , and read off the two duality-paired nonzero pieces.
Solution. Cover P n = Proj k [ x 0 , … , x n ] by the n + 1 standard affine opens U j = { x j = 0 } . This cover is acyclic for every line bundle (each U j ≅ A n is affine, intersections are affine, and quasi-coherent sheaves have no higher cohomology on affines by Serre vanishing 04.03.05 ), so Čech cohomology on U = { U j } computes derived-functor cohomology 04.03.03 .
Pass to the graded module S = k [ x 0 , … , x n ] and form ⨁ d H i ( P n , O ( d )) . The Čech complex of U with values in ⨁ d O ( d ) is the complex computing the local cohomology of S at the irrelevant ideal m = ( x 0 , … , x n ) , shifted by one degree:
d ⨁ H ˇ i ( U , O ( d )) = H m i + 1 ( S ) ( i ≥ 1 ) , d ⨁ H ˇ 0 = S .
The localization S x 0 ⋯ x n = k [ x 0 ± , … , x n ± ] sits at the top of the Čech complex; the top cohomology is the quotient by the images of the S x 0 ⋯ x j ⋯ x n , which is the span of monomials x 0 a 0 ⋯ x n a n with every a j ≤ − 1 .
The result, Hartshorne III.5.1 :
H 0 ( P n , O ( d )) = S d , the space of degree-d forms, of dimension ( n n + d ) for d ≥ 0 and 0 for d < 0 .
H n ( P n , O ( d )) is dual to S − d − n − 1 , of dimension ( n − d − 1 ) for d ≤ − n − 1 , and 0 otherwise.
H i ( P n , O ( d )) = 0 for 0 < i < n , all d .
The two nonzero pieces are Serre-dual 04.08.03 : H n ( P n , O ( d )) ≅ H 0 ( P n , O ( − d − n − 1 ) ) ∨ , matching ω P n = O ( − n − 1 ) , since O ( d ) ∨ ⊗ ω = O ( − d − n − 1 ) . □
This is the load-bearing computation of the chapter. Every Euler-characteristic count, every Riemann-Roch application, and the canonical-sheaf identification ω P n = O ( − n − 1 ) route through the explicit Čech generators produced here.
Exercises Intermediate
Exercise 1 (easy, proof). Cohomological dimension of an affine scheme.
Let X = Spec A be a Noetherian affine scheme and F a quasi-coherent sheaf on X . Show H i ( X , F ) = 0 for all i > 0 .
Hint
This is Serre's vanishing theorem 04.03.05 . Recall that flasque sheaves are acyclic, and that on a Noetherian affine scheme a quasi-coherent sheaf embeds into a flasque quasi-coherent resolution built from injective A -modules.
Answer
Write F = M for an A -module M (quasi-coherent sheaves on Spec A are exactly the M ). Choose an injective resolution 0 → M → I 0 → I 1 → ⋯ of M in A -modules. Applying ( − ) , which is exact on A -modules, gives a resolution 0 → F → I 0 → I 1 → ⋯ by quasi-coherent sheaves.
The key fact (Hartshorne III.3.4) is that on a Noetherian scheme the sheaf I associated to an injective A -module I is flasque. Flasque sheaves are acyclic for the global-sections functor, so this resolution computes H i ( X , F ) . But Γ ( X , I ∙ ) = I ∙ as a complex of A -modules, which is exact in positive degrees because I ∙ resolves M . Hence H i ( X , F ) = 0 for i > 0 .
This is Serre's affine vanishing — the cohomological characterization of affineness (Serre's criterion, III.3.7, is its converse for Noetherian separated schemes).
Exercise 2 (easy, numeric). H 1 of the punctured affine plane.
Let U = A k 2 ∖ { 0 } . Compute H 1 ( U , O U ) and conclude that U is not affine.
Hint
Cover U by the two affines U 1 = { x = 0 } and U 2 = { y = 0 } and run the Čech complex 04.03.03 . The intersection is Spec k [ x , y ] x y .
Answer
Take U = { U 1 , U 2 } with U 1 = Spec k [ x , y ] x , U 2 = Spec k [ x , y ] y , U 1 ∩ U 2 = Spec k [ x , y ] x y . The Čech complex for O U is
0 → k [ x , y ] x ⊕ k [ x , y ] y ( f , g ) ↦ g − f k [ x , y ] x y → 0.
H 0 is the kernel: functions regular on both U 1 and U 2 , which is k [ x , y ] (the punctured plane has the same global functions as A 2 , by Hartog-type extension). H 1 is the cokernel. A basis of k [ x , y ] x y is the monomials x a y b with a , b ∈ Z ; the image of k [ x , y ] x hits those with b ≥ 0 , and k [ x , y ] y those with a ≥ 0 . The cokernel is spanned by monomials with both a < 0 and b < 0 :
H 1 ( U , O U ) = a , b ≤ − 1 ⨁ k ⋅ x a y b ,
which is infinite-dimensional. Since H 1 = 0 for the structure sheaf, U is not affine (Exercise 1's vanishing would force H 1 = 0 ). This is Hartshorne III.4.3.
Exercise 3 (medium, numeric). Euler characteristic of a twisted ideal sheaf.
Let C ⊂ P k 2 be a smooth plane curve of degree d , with ideal sheaf I C = O P 2 ( − d ) . Compute χ ( O C ) and the genus g of C .
Hint
Use the short exact sequence 0 → O ( − d ) → O → O C → 0 and additivity of Euler characteristic. Then χ ( O C ) = 1 − g .
Answer
The curve C is cut out by one degree-d form, so I C ≅ O P 2 ( − d ) and the structure sequence is
0 → O P 2 ( − d ) → O P 2 → O C → 0.
Euler characteristic is additive on short exact sequences, so χ ( O C ) = χ ( O P 2 ) − χ ( O P 2 ( − d )) .
From the projective-space computation (lead exercise): χ ( O P 2 ) = h 0 = 1 (only H 0 is nonzero). For O ( − d ) with d ≥ 1 : h 0 = 0 , h 1 = 0 , and h 2 ( O ( − d )) = h 0 ( O ( d − 3 )) = ( 2 d − 1 ) by Serre duality with ω P 2 = O ( − 3 ) . So χ ( O ( − d )) = ( 2 d − 1 ) .
Therefore
χ ( O C ) = 1 − ( 2 d − 1 ) = 1 − 2 ( d − 1 ) ( d − 2 ) .
Since C is a smooth connected curve, χ ( O C ) = h 0 − h 1 = 1 − g . Hence
g = ( 2 d − 1 ) = 2 ( d − 1 ) ( d − 2 ) ,
the genus-degree formula for smooth plane curves (Hartshorne I.7.2 via III.5). For d = 3 this gives g = 1 , recovering the elliptic curve 04.04.03 .
Exercise 4 (medium, symbolic). The dualizing sheaf of projective space.
Identify ω P k n explicitly as a twist O ( m ) , using only the cohomology computation of the lead exercise and the defining property of a dualizing sheaf.
Hint
A dualizing sheaf ω ∘ for an n -dimensional projective variety satisfies H n ( X , ω ∘ ) ≅ k and the pairing H i ( F ) × H n − i ( F ∨ ⊗ ω ∘ ) → k is perfect. Test against F = O ( d ) .
Answer
Suppose ω ∘ = O ( m ) is dualizing on P n . The duality pairing applied to F = O ( d ) requires
H n ( P n , O ( d )) ≅ H 0 ( P n , O ( − d ) ⊗ O ( m ) ) ∨ = H 0 ( P n , O ( m − d ) ) ∨ .
From the lead exercise, H n ( O ( d )) is nonzero exactly when d ≤ − n − 1 , with dimension ( n − d − 1 ) . The right side H 0 ( O ( m − d )) is nonzero exactly when m − d ≥ 0 , i.e. d ≤ m , with dimension ( n m − d + n ) .
Matching the thresholds: the nonvanishing ranges d ≤ − n − 1 and d ≤ m must coincide, forcing m = − n − 1 . Matching dimensions: ( n − d − 1 ) = ( n ( − n − 1 − d ) + n ) = ( n − d − 1 ) , which holds identically. The normalization H n ( P n , ω ∘ ) = H n ( O ( − n − 1 )) ≅ k checks (dim = ( n n ) = 1 ).
Therefore
ω P n = O ( − n − 1 ) ,
agreeing with the differential-forms computation ω P n = ∧ n Ω P n 1 = O ( − n − 1 ) from the Euler sequence 04.08.02 . Hartshorne III.7.1.
Exercise 5 (medium, proof). Cohomology and the dimension bound.
Let X be a Noetherian topological space of dimension n . Show that H i ( X , F ) = 0 for all sheaves of abelian groups F and all i > n (Grothendieck vanishing).
Hint
Induct on dim X . Reduce to F generated by a single section over a closed irreducible subset, and use that on an irreducible space the constant sheaf is flasque.
Answer
This is Grothendieck's vanishing theorem (Hartshorne III.2.7). Proceed by induction on n = dim X .
For the base n = 0 : a Noetherian space of dimension 0 is a finite discrete-on-components space, and every sheaf is flasque, hence acyclic; H i = 0 for i > 0 .
Inductive step. By a standard reduction (taking direct limits, which commute with cohomology on a Noetherian space, and filtering by subsheaves), it suffices to treat F = Z U , the extension by zero of the constant sheaf on an open U with irreducible closed complement, or sheaves supported on a proper closed subset Z . For the latter, H i ( X , F ) = H i ( Z , F ∣ Z ) and dim Z < n , so induction gives vanishing for i > n − 1 , a fortiori for i > n .
For F = Z U , use 0 → Z U → Z X → Z X ∖ U → 0 . The constant sheaf Z X on an irreducible space is flasque (every open is connected, so restriction maps are surjective), hence acyclic. The long exact sequence then gives H i ( X , Z U ) ≅ H i − 1 ( X , Z X ∖ U ) for i ≥ 2 , and dim ( X ∖ U ) < n when U is dense, so by induction the right side vanishes for i − 1 > n − 1 , i.e. i > n .
Assembling these cases through the Noetherian induction on the support gives H i ( X , F ) = 0 for all i > n .
Exercise 6 (medium, numeric). Hilbert polynomial is constant in a flat family.
Let f : X → T be a projective morphism with T integral Noetherian, and let F be a coherent sheaf on X , flat over T , with O X ( 1 ) relatively very ample. Show that the Hilbert polynomial P t ( m ) = χ ( X t , F t ( m )) is independent of t ∈ T .
Hint
Flatness over an integral base makes the fiberwise Euler characteristic locally constant; on an integral base "locally constant" means constant. Use the cohomology-and-base-change / semicontinuity package 04.03.07 .
Answer
Fix m . By Serre's finiteness 04.03.05 each H i ( X t , F t ( m )) is finite-dimensional, so χ ( X t , F t ( m )) = ∑ i ( − 1 ) i h i ( X t , F t ( m )) is well defined.
The cohomology-and-base-change theorem (Hartshorne III.12.8, packaged in 04.03.07 ): for F flat over T , the function t ↦ χ ( X t , F t ( m )) is locally constant on T . The proof represents R f ∗ F ( m ) locally by a bounded complex K ∙ of finite free O T -modules whose fiberwise cohomology computes that of F t ( m ) ; the alternating sum of the ranks of K ∙ is both ∑ ( − 1 ) i rk K i (visibly constant on connected components) and χ ( X t , F t ( m )) (by the base-change spectral sequence).
Since T is integral, it is connected, so "locally constant" forces constant. Thus P t ( m ) = χ ( X t , F t ( m )) is the same for every t , and as a function of m it is the common Hilbert polynomial P . This is the foundational fact behind the Hilbert scheme: flat families are exactly those with locally constant Hilbert polynomial (Hartshorne III.9.9).
Exercise 7 (hard, proof). Serre duality for a smooth projective curve.
Let X be a smooth projective curve of genus g over an algebraically closed field k , and L a line bundle. Using Serre duality, prove the Riemann-Roch relation h 0 ( L ) − h 1 ( L ) = deg L + 1 − g implies h 1 ( L ) = h 0 ( ω X ⊗ L − 1 ) , and deduce deg ω X = 2 g − 2 .
Hint
Serre duality on a curve (n = 1 ): H 1 ( X , L ) ≅ H 0 ( X , ω X ⊗ L − 1 ) ∨ 04.08.03 . Apply Riemann-Roch to L = ω X and to L = O X and compare.
Answer
Serre duality on the smooth projective curve X (dimension n = 1 ) gives, for any line bundle L ,
H 1 ( X , L ) ≅ H 0 ( X , ω X ⊗ L − 1 ) ∨ ,
so h 1 ( L ) = h 0 ( ω X ⊗ L − 1 ) . Substituting into Riemann-Roch,
h 0 ( L ) − h 0 ( ω X ⊗ L − 1 ) = deg L + 1 − g . ( ∗ )
Now extract deg ω X from two evaluations.
Take L = O X : then h 0 ( O X ) = 1 (X connected, k algebraically closed), h 0 ( ω X ) = g by definition of the genus (g = dim H 0 ( X , ω X ) , the space of global differentials), and deg O X = 0 . Equation ( ∗ ) reads 1 − g = 0 + 1 − g — a consistency check, no new information.
Take L = ω X : then h 0 ( ω X ) = g , and h 0 ( ω X ⊗ ω X − 1 ) = h 0 ( O X ) = 1 . Equation ( ∗ ) becomes
g − 1 = deg ω X + 1 − g ,
hence deg ω X = 2 g − 2 .
This is the canonical-degree identity (Hartshorne IV.1.3.3), the engine behind Hurwitz 04.04.02 and the genus-degree formula. The argument uses only Serre duality plus the numerical Riemann-Roch; both inputs are theorems of Chapter III–IV, not assumptions.
Exercise 8 (hard, proof). Vanishing forces local freeness — base change.
Let f : X → T be projective, F coherent and flat over T , T reduced and Noetherian. Suppose H 1 ( X t , F t ) = 0 for every t ∈ T . Show that f ∗ F is locally free and that the natural map ( f ∗ F ) ⊗ k ( t ) → H 0 ( X t , F t ) is an isomorphism for every t .
Hint
This is the Grauert / cohomology-and-base-change corollary 04.03.07 . Represent R f ∗ F by a finite complex K ∙ of locally free sheaves; vanishing of H 1 on fibers controls the cokernel maps.
Answer
Locally on T = Spec A , choose a bounded complex K ∙ : 0 → K 0 → K 1 → ⋯ of finite free A -modules such that for every A -algebra B ,
H i ( X × T Spec B , F B ) = H i ( K ∙ ⊗ A B )
(Hartshorne III.12.2 — the existence of such a "computing complex" for a flat sheaf). We may take K 0 in degree 0 and arrange H i ( K ∙ ) = 0 for i < 0 .
By hypothesis H 1 ( X t , F t ) = H 1 ( K ∙ ⊗ k ( t )) = 0 for all t . Consider the map d 0 : K 0 → K 1 . The cokernel of d 1 : K 1 → K 2 controls H 1 ; more directly, apply the standard "base-change criterion": H 1 commutes with base change at every t iff a certain map is surjective there, and vanishing of H 1 ( X t , F t ) for all t makes coker ( d 0 ⊗ k ( t )) have locally constant rank.
Concretely: H 0 ( K ∙ ⊗ k ( t )) = ker ( d 0 ⊗ k ( t )) . Because H 1 vanishes on all fibers and T is reduced, the function t ↦ dim k ( t ) H 0 ( X t , F t ) is locally constant (semicontinuity is an equality when the next cohomology vanishes everywhere, Hartshorne III.12.9). On a reduced Noetherian scheme a coherent sheaf with locally constant fiber dimension is locally free; here that sheaf is f ∗ F = ker d 0 (the degree-0 cohomology of K ∙ ). Hence f ∗ F is locally free, and the base-change map
( f ∗ F ) ⊗ k ( t ) → H 0 ( X t , F t )
is an isomorphism for every t (Grauert's theorem, III.12.9). This is the workhorse that builds vector bundles on T out of fiberwise global sections — the construction underlying relative Proj embeddings and moduli of sheaves.
Exercise 9 (hard, proof). Smoothness and the relative cotangent sequence.
Let f : X → S be a smooth morphism of relative dimension n and g : S → T smooth of relative dimension m . Show g ∘ f is smooth of relative dimension n + m , and that the relative canonical sheaves satisfy ω X / T ≅ ω X / S ⊗ f ∗ ω S / T .
Hint
Use the cotangent sequence f ∗ Ω S / T → Ω X / T → Ω X / S → 0 04.02.05 ; smoothness of f makes it left-exact with locally free terms. Take top exterior powers.
Answer
For a smooth morphism, Ω X / S is locally free of rank equal to the relative dimension 04.02.05 . The fundamental cotangent sequence for the tower X → S → T is
f ∗ Ω S / T α Ω X / T β Ω X / S → 0.
When f is smooth, this sequence is also left-exact and α is a subbundle inclusion (Hartshorne II.8.11 / III.10.0): each term is locally free, of ranks m , and n respectively, and the sequence splits locally. Counting ranks, Ω X / T is locally free of rank m + n . Smoothness of g ∘ f then follows from the criterion "flat with locally free relative differentials of the expected rank, and geometrically regular fibers" — flatness is preserved under composition, the fibers of g ∘ f are smooth over the fibers of g hence regular, so g ∘ f is smooth of relative dimension n + m .
Now take the top exterior power of the short exact sequence
0 → f ∗ Ω S / T → Ω X / T → Ω X / S → 0.
For a short exact sequence of locally free sheaves 0 → A → B → C → 0 with ranks a , a + c , c , one has ∧ top B ≅ ∧ top A ⊗ ∧ top C (the determinant is multiplicative on extensions). Applying this with B = Ω X / T , A = f ∗ Ω S / T , C = Ω X / S and writing ω X / T = ∧ m + n Ω X / T , etc.:
ω X / T ≅ f ∗ ω S / T ⊗ ω X / S .
This relative-canonical multiplicativity is the engine behind adjunction 04.05.07 and the behaviour of canonical sheaves in families (Hartshorne III.10, II.8.13). Specializing T = Spec k recovers ω X ≅ ω X / S ⊗ f ∗ ω S for a smooth fibration over a smooth base.
Exercise pack EP1 for 04-algebraic-geometry/03-cohomology. Hartshorne Chapter III supplement: sheaf and Čech cohomology, cohomology of projective space, Serre duality, higher direct images, flat and smooth morphisms (§III.2, §III.4, §III.5, §III.7, §III.8, §III.9, §III.10).