Draft lesson.

Polynomial rings, finite resolutions and derived duality

Written by GPT-6.1 Sol (OpenAI), October 2026. Public domain (CC0).

A bounded cohomology condition is a condition on an object of the derived category, not on every complex representing it. Over a polynomial ring, finite generation of those cohomology modules gives a bounded representative made of finitely generated projectives. Dualizing that representative then gives both a bounded derived dual and an actual natural bidual isomorphism. We prove the algebraic finiteness statements as well as the signs and independence needed for this construction.

Let \(k\) be any field, \(n\geq0\), and \(R=k[x_1,\ldots,x_n]\). All modules and bases are small in the fixed universe. “Finite” for a module means finitely generated; a finite projective module is a finitely generated projective module. Write \(D^b_{\mathrm{fg}}(R)\) for the full subcategory of \(D(R\text{-}\mathrm{Mod})\) whose cohomology is finite and vanishes outside some finite interval. The terms of a chosen representative need not be finite or bounded.

Complexes have cochain differentials, \(P[r]^i=P^{i+r}\), with shifted differential \((-1)^r d_P\). Our Hom differential is \(d_Qf-(-1)^j f d_P\) in degree \(j\). Derived Hom and bounded resolution pairs, Sections 1–7, proves this normalization, its localization and its single projective computations. Cohomological bounds, denominators and truncation adjunctions, Sections 1–5, proves the full unbounded roof tests and the natural truncation triangles used below. Determinant signs and subset Koszul complexes, Sections 1–5, supplies the integral exterior signs and the complete ordered regularity criterion.

1. Polynomial Noetherianity and finite kernels

Theorem 1.1. The ring \(R\) is Noetherian. Every submodule of a finite \(R\)-module is finite.

Proof. A ring is Noetherian when every ideal is finite. This is equivalent to the ascending chain condition: a union of an increasing chain is an ideal, and its finite generating set belongs to one member, so the chain stabilizes. Conversely, if an ideal were not finite, choosing a new element outside the ideal generated by the preceding choices would produce a strictly increasing chain.

We prove the polynomial step for a commutative Noetherian ring \(A\). Let \(J\) be an ideal of \(A[t]\). For \(d\geq0\) put

Zero is included. Addition of polynomials and multiplication by elements of \(A\) make \(I_d\) an ideal. Multiplication by \(t\) shows \(I_d\subseteq I_{d+1}\). Choose \(N\) such that these ideals are constant for \(d\geq N\). For each \(0\leq d\leq N\), choose a finite list \(f_{d,s}\in J\) of degree at most \(d\) whose degree-\(d\) coefficients generate \(I_d\). A zero ideal requires no choices.

These finitely many polynomials generate \(J\). Indeed, if a nonzero \(f\in J\) has degree \(m\leq N\), its leading coefficient is a linear combination of the leading coefficients of the \(f_{m,s}\). Subtract that combination from \(f\); the result has smaller degree. If \(m>N\), use \(t^{m-N}f_{N,s}\) instead, since \(I_m=I_N\). Again the leading coefficient cancels. Induction on degree expresses \(f\) in the ideal generated by the chosen finite list; the zero remainder ends the procedure. This proves \(A[t]\) Noetherian. A field has only the ideals zero and itself. Iterating the polynomial step proves the first assertion, including \(n=0\).

For the module assertion, first prove that every submodule of \(R^r\) is finite by induction on \(r\). For \(r=1\) this is the ideal assertion; for \(r=0\) it is immediate. Project a submodule of \(R^r\) onto its last coordinate. Its image is a finite ideal, so choose finitely many lifts of ideal generators. Its kernel is a submodule of \(R^{r-1}\), finite by induction. Those kernel generators and the chosen lifts generate the original submodule: subtract a linear combination of the lifts to place any element in the kernel. Finally, if \(R^r\twoheadrightarrow M\) and \(L\subseteq M\), the inverse image of \(L\) is finite by what was just proved, and its images generate \(L\). \(\square\)

Consequently kernels and cohomology of finite complexes of finite modules are finite. A finite projective module is a direct summand of a finite free module: lift the identity through a finite free cover to split it. Its dual \(\operatorname{Hom}_R(P,R)\) is also a direct summand of a finite free module, by dualizing that split direct sum, and is therefore finite projective.

2. Finite global dimension, with an explicit polynomial step

For a module, projective dimension at most \(r\) means the existence of a projective resolution of length at most \(r\). Its projectives are not required to be finite. The global dimension is the least such common bound for all modules.

Theorem 2.1. Every \(R\)-module has projective dimension at most \(n\), and \(R\) has global dimension exactly \(n\).

Proof of the upper bound. Every vector space over \(k\) has a basis, so every \(k\)-module is free; this is the case \(n=0\). We prove the induction step from \(A\) to \(S=A[t]\). Let \(M\) be any \(S\)-module and let \(T=S\otimes_A M\), with \(S\) acting on its first factor. There is a sequence of actual \(S\)-linear maps

The formulas are balanced over \(A\), since multiplication by \(t\) commutes with its action on \(M\); they commute with \(t\) on the first factor, proving \(S\)-linearity. The underlying \(A\)-module of \(T\) is the direct sum of copies of \(M\) indexed by \(j\geq0\). If a nonzero finite sum has last nonzero coefficient \(m_d\), its image under \(\Delta\) has coefficient \(m_d\) in degree \(d+1\), so \(\Delta\) is injective. The map \(e\) is onto, and \(e\Delta=0\). To identify its kernel, use the finite identity

For \(j=0\) both sides are zero. For larger \(j\) the intermediate terms cancel in pairs. Thus a finite sum differs from \(1\otimes e\) of that sum by an element of \(\operatorname{im}\Delta\). If its augmentation is zero, it is in that image. This proves exactness of (2.1) for every module.

Suppose all \(A\)-modules have projective dimension at most \(r\). Resolve the underlying \(A\)-module \(M\) by projectives of length \(r\), and tensor that resolution with \(S\). This remains exact because \(S\), as a right \(A\)-module, is a direct sum of copies of \(A\); tensoring is the corresponding direct sum of exact sequences. Its terms are projective over \(S\): a projective \(A\)-module is a direct summand of a free module, and extension of scalars retains that splitting. We obtain an augmented projective chain resolution \(Q_\bullet\to T\), supported in \(0,\ldots,r\).

Lift \(\Delta\) to a chain endomorphism \(f:Q_\bullet\to Q_\bullet\). In degree zero, projectivity lifts \(\Delta\) composed with the augmentation through the target augmentation. Recursively, \(f_{p-1}d_p\) has image in the target cycles by its already established chain equation; those cycles are the image of the next boundary, by exactness. Projectivity of \(Q_p\) lifts it through that surjection to define \(f_p\). This includes the last nonzero term and gives \(d f=f d\).

The target-first homological cone has

The chain equation for \(f\) proves \(d_C^2=0\). The short exact row of complexes \(0\to Q\to C\to Q[1]\to0\), with homological shift \(Q[1]_p=Q_{p-1}\) and negative boundary, gives the long exact homology sequence. Its connecting map on \(H_0(Q)=T\) is \(\Delta\): lift a cycle \(b\) by \((0,b)\), whose boundary is \((fb,0)\). Hence (2.1) gives \(H_0(C)=M\), \(H_1(C)=0\), and every other homology group zero. The actual augmentation is \(C_0=Q_0\to T\xrightarrow eM\). Its terms are projective and vanish above \(r+1\), so it is a resolution of length at most \(r+1\). Applying this step \(n\) times proves the upper bound.

Proof of sharpness. Put \(V=k^n\) with its ordered basis and \(\mathfrak m=(x_1,\ldots,x_n)\). The homological Koszul complex

resolves \(k=R/\mathfrak m\). Here is a direct exactness verification. In one variable the multiplication complex \(k[x]e\xrightarrow{x}k[x]\) contracts onto constants: evaluation \(p\), inclusion \(j\), and \(h(f)=(f-f(0))e/x\) give \(dh+hd=1-jp\). The quotient in this formula is a polynomial in every characteristic. Tensor these \(n\) complexes over \(k\), with the boundary signs from the preceding factor degrees. They are exactly (2.4). Writing \(q_i=j_ip_i\), the tensor homotopy is

The degree-one \(h_i\) has the sign \((-1)^{p_1+\cdots+p_{i-1}}\) on a homogeneous tensor. Boundary terms in other factors cancel by that sign, while the factor-\(i\) term is the product of the preceding \(q\)'s with \(1-q_i\). Their sum telescopes to \(1-\bigotimes_iq_i\). This proves the augmentation to \(k\) is exact; for \(n=0\) it is the identity.

Apply \(\operatorname{Hom}_R(-,R)\). Evaluation identifies its degree-\(q\) term with \(R\otimes_k\bigwedge^qV^*\). The standard Hom differential is \((-1)^{q+1}\) times wedge multiplication by \(\sum_i x_i e_i^*\). Multiplication in degree \(q\) by \((-1)^{q(q+1)/2}\) converts it to that ordinary cochain Koszul differential, since the successive multipliers satisfy \(\varepsilon_{q+1}(-1)^{q+1}=\varepsilon_q\). The coordinate sequence is ordered regular: after quotienting by its first \(i-1\) members, multiplication by \(x_i\) is injective in the remaining polynomial ring. The complete regular-prefix criterion therefore gives

Finite projective resolutions compute these Ext groups by the actual augmentation and standard Hom differential. A resolution of length less than \(n\) would have zero degree-\(n\) Hom term and hence zero \(\operatorname{Ext}^n\), contradicting (2.6) when \(n>0\). For \(n=0\), the already proved upper bound and the nonzero module \(k\) give global dimension zero. \(\square\)

3. A finite last syzygy is projective

The preceding induction resolved arbitrary modules, so its projectives may have infinitely many generators. Finite generation is obtained separately.

Lemma 3.1 (comparison of kernels). If \(P\twoheadrightarrow M\) and \(P'\twoheadrightarrow M\) are projective covers with kernels \(K,K'\), then

Here “cover” only means a surjection from a projective, with no minimality requirement. Proof. Form the pullback \(E=\{(p,p'):\text{their images in }M\text{ agree}\}\). Its projection to \(P'\) is onto with kernel \(K\), and its projection to \(P\) is onto with kernel \(K'\). Both exact sequences split by projectivity of their targets. Their two direct-sum descriptions of \(E\) give (3.1). \(\square\)

Theorem 3.2. Every finite \(R\)-module has a resolution of length at most \(n\) by finite projectives.

Proof. For \(n=0\) a finite-dimensional vector space is finite free. For \(n>0\), repeatedly take finite free covers. Noetherianity makes every new kernel finite. With \(K_0=M\), obtain

where every \(F_i\) is finite free and every \(K_i\) is finite. Independently take the length-\(n\) projective resolution guaranteed by Theorem 2.1. Write its successive kernels as \(L_i\), with \(L_0=M\); its last kernel \(L_n\) is projective. Pad a shorter resolution with zero terms if necessary.

We prove inductively that \(K_i\oplus U_i\simeq L_i\oplus V_i\) for some projective modules \(U_i,V_i\). Start with zero auxiliary modules at \(i=0\). Suppose the comparison is known. Cover its two sides by \(F_i\oplus U_i\) and \(Q_i\oplus V_i\), using the resolution covers on the first summands and identity maps on the auxiliary summands. Identify their targets by the known isomorphism. Their kernels are exactly \(K_{i+1}\) and \(L_{i+1}\). Lemma 3.1 now gives

This is the next comparison with projective auxiliary modules. At \(i=n\), its right side is projective since \(L_n\) is. Consequently \(K_n\), a direct summand of that projective module, is projective. It is already finite. The sequence ending in \(K_n\) and the finite free \(F_i\)'s is the required finite projective resolution. \(\square\)

4. From bounded finite cohomology to a bounded projective complex

Let \(\mathcal P\) denote finite projective modules. A bounded complex of these modules is K-projective: if \(Z\) is acyclic, the Hom complex \(\operatorname{Hom}_R(P,Z)\) has a finite filtration by source columns, each a signed shift of \(\operatorname{Hom}_R(P^i,Z)\). Projectivity makes these columns acyclic. The precomposition part of the differential goes from source column \(i+1\) to \(i\), so columns \(i\leq r\) form the required subcomplex filtration. A finite induction using its short exact rows proves the total acyclic, even for unbounded \(Z\).

It follows directly that

for every complex \(Y\). Indeed a quasi-isomorphism \(s:U\to P\) induces a quasi-isomorphism \(\operatorname{Hom}_R(P,U)\to\operatorname{Hom}_R(P,P)\): its cone is the Hom of an acyclic cone, with the actual cone comparison in the Hom prerequisite. Thus there is a chain map \(t:P\to U\) with \(st\) homotopic to the identity. In the derived category \(t=s^{-1}\), so the incoming roof \((s,f)\) is the image of the chain map \(ft\). This proves surjectivity. If a chain map becomes zero in the derived category, the incoming roof equality test supplies such an \(s\) making its composite null-homotopic. Composing with \(t\) proves the original map null-homotopic, proving injectivity. All maps in (4.1) are the actual localization maps, so the bijection is natural.

Theorem 4.1. If \(H^i(X)\) is finite and zero outside \([a,b]\), with finite \(a\leq b\), then \(X\) has a representative \(P\in K^b(\mathcal P)\) with

Consequently localization induces an equivalence \(K^b(\mathcal P)\simeq D^b_{\mathrm{fg}}(R)\).

Proof. In the full unbounded derived category use the natural truncation triangles

Their actual maps, last arrows and quasi-isomorphism comparisons are those proved in the truncation prerequisite, Section 5. The first truncation \(\tau_{\leq a-1}X\) is acyclic and hence zero in \(D(R)\). By Theorem 3.2, \(H^q(X)[-q]\) has a finite projective representative \(Q_q\) supported in \([q-n,q]\).

Assume \(\tau_{\leq q-1}X\) already has such a representative \(P_{q-1}\) supported in \([a-n,q-1]\). By (4.1), its connecting map in (4.3), after these identifications, is represented by a chain map \(f:Q_q\to P_{q-1}[1]\). Form \(C_f[-1]\), where our source-first cochain cone has terms \(C_f^j=Q_q^{j+1}\oplus P_{q-1}^{j+1}\). After shifting back, its terms and differential are

The entries of \(f^i\) are typed \(Q_q^i\to P_{q-1}^{i+1}\); the negative arises from the inverse shift of the cone. The chain equation for \(f\) proves that this displayed differential squares to zero. The twice inverse-rotated cone triangle has first map minus the inclusion of \(P_{q-1}\), second map minus the projection to \(Q_q\), and last arrow \(f:Q_q\to P_{q-1}[1]\). To compare with (4.3), rotate both triangles so this last arrow is first. Their first two vertices and that arrow agree. The triangle morphism axiom supplies a comparison on the remaining vertices; its cohomology maps are isomorphisms by their long exact sequences. Rotating back therefore identifies \(P_q\) with \(\tau_{\leq q}X\). Its support is the union of the two stated supports, contained in \([a-n,q]\), and all its terms are finite projective.

Induct to \(q=b\). The map \(\tau_{\leq b}X\to X\) is a quasi-isomorphism, so (4.2) follows. This construction uses the cohomology bounds of \(X\), not a term-bound assumption on its original representative. If all its cohomology is zero, choose the zero representative.

For the equivalence, (4.1) proves full faithfulness on bounded projectives; the construction proves essential surjectivity. Conversely, a bounded finite projective complex has bounded finite cohomology by Theorem 1.1. It therefore lies in the stated full subcategory. \(\square\)

5. The dual stays bounded and finite

For a bounded finite projective complex define \(D(P)=\operatorname{Hom}_R(P,R)\), where \(R\) is in degree zero. Its terms and actual differential are

This is an \(R\)-module complex since the polynomial ring is commutative. Composing two such differentials gives a sign times \(f d_Pd_P=0\). Its terms are finite projective by the split finite-free argument after Theorem 1.1.

If \(u:P\to Q\) is a chain map, \(D(u)^j(f)=f u^{-j}\). The two chain equations are equal after using \(u d_P=d_Qu\). Precomposition gives \(D(vu)=D(u)D(v)\) and preserves identities and addition. It also respects homotopy. If \(u-v=d_Qh+h d_P\), then

has \(d_{D(P)}H+H d_{D(Q)}=D(u)-D(v)\). The first summand evaluates to \(f h d_P\), because its two factors \((-1)^j\) cancel; the second evaluates to \(f d_Qh\), because its two factors \((-1)^{j+1}\) cancel. Thus \(D\) is an actual contravariant functor on \(K^b(\mathcal P)\).

Through the equivalence in Theorem 4.1 it defines the derived dual

This agrees with the single projective computation in the Hom prerequisite. Indeed choose an isomorphism \(P\to X\) in \(D(R)\). By (4.1) it has a chain representative; its cohomology maps are isomorphisms, so that representative is a quasi-isomorphism. Precomposition gives the actual comparison \(\operatorname{Hom}_R(X,R)\to\operatorname{Hom}_R(P,R)\). The K-projective argument proves the single-replacement computation with its normalized localization map. Equation (5.3) is used on the stated bounded finite-cohomology domain; it requires no assertion about a derived Hom bifunctor on every pair of unbounded objects.

Theorem 5.1. If \(H^i(X)\) is finite and zero outside \([a,b]\), then \(D_R(X)\) has finite cohomology and

In particular it is an object of \(D^b_{\mathrm{fg}}(R)\).

Proof. Choose the representative (4.2). Formula (5.1) makes its dual terms zero outside \([-b,n-a]\), and all remaining terms are finite projective. Their kernels are finite by Noetherianity, and quotienting those kernels by the finite images retains finite generation. These are exactly the cohomology modules of the computing dual. Isomorphism in the derived category preserves them, proving (5.4) independently of the original terms of \(X\). \(\square\)

6. Signed evaluation is a natural chain isomorphism

For a finite projective module \(F\), ordinary evaluation \(F\to F^{**}\) is an isomorphism. For a finite free module this follows by its coordinate basis and dual basis. For a direct summand of a finite free module, use its split direct sum; evaluation is the direct sum of the two evaluations by naturality. Hence its restriction to either summand is an isomorphism.

For complexes the degree sign is part of the map. Define

Theorem 6.1. The maps (6.1) give a natural chain isomorphism \(\beta_P:P\to D^2(P)\) on \(K^b(\mathcal P)\), with coherence

Proof. Evaluate its chain equation on \(g\in D(P)^{-i-1}=\operatorname{Hom}_R(P^{i+1},R)\). The inner dual differential is

Using (5.1) for the outer dual and then (6.1) gives

Equality on every functional proves the chain equation. Each component is an ordinary finite-projective evaluation isomorphism multiplied by a unit, so it is a chain isomorphism.

For a chain map \(u:P\to Q\), evaluation on \(f\in\operatorname{Hom}_R(Q^i,R)\) gives

Thus \(D^2(u)\beta_P=\beta_Qu\). It descends through homotopies by (5.2).

Finally let \(f\in D(P)^j\), and evaluate the left side of (6.2) on \(x\in P^{-j}\). The map \(D(\beta_P)^j\) is precomposition with \(\beta_P^{-j}\), whereas \(\beta_{D(P)}^j(f)\) sends \(z\) to \((-1)^j z(f)\). Their composite has value

This proves the actual coherence equation in every degree. It includes characteristic two, where the signs become one. The inverse of \(\beta_P\) is the termwise inverse evaluation with the same sign, and (6.2) says that the inverse comparison on the dual is exactly the dual of the comparison on \(P\). \(\square\)

7. Representative independence and the contravariant equivalence

Theorem 7.1. Derived duality is a contravariant equivalence

and \(D_R(\beta_X)\beta_{D_R(X)}=1\). The equivalence has the opposite-category triangle and translation conventions of the derived Hom prerequisite.

Proof. On \(K^b(\mathcal P)\), (5.1)–(5.2) define the contravariant functor, (6.1) is its natural bidual isomorphism, and (6.2) supplies its inverse coherence. These equations already exhibit \(D\), with the reversed variance, as its own quasi-inverse: one composite is \(D^2\), identified with the identity by \(\beta\), and the other is the same comparison on the opposite category. Thus it is an equivalence there.

Here is the representative comparison explicitly. Choose quasi-isomorphisms \(u:P\to X\) and \(u':P'\to X\) with \(P,P'\) bounded finite projective, whose existence was proved in Sections 4–5. Full faithfulness (4.1) gives a unique homotopy class \(a:P\to P'\) whose derived image is \((u')^{-1}u\). This is an isomorphism in \(K^b(\mathcal P)\): lift the inverse derived arrow by the same bijection, and use its injectivity to obtain the two inverse equations in the homotopy category. Applying \(D\) gives the actual inverse comparison

For three representatives, their transition classes satisfy the composition law by uniqueness in (4.1), and the dual comparisons satisfy the reversed law because \(D\) is a functor. For a derived morphism \(X\to Y\), lift its transported arrow between the chosen projective representatives using the same bijection, and apply \(D\). Changing either representative conjugates this dual arrow by (7.2), again by functoriality. Identities and composition therefore agree under all the comparisons, proving independence as a functor, not just existence of isomorphic values.

Equation (6.4) makes \(\beta\) commute with every transition and every lifted morphism. It consequently transports to the natural isomorphism \(\beta_X\) in (7.1), independently of representatives. Equation (6.2) transports as well, giving exactly the stated inverse coherence. Theorem 5.1 ensures that all objects in these comparisons remain in \(D^b_{\mathrm{fg}}(R)\).

The standard derived Hom construction is separately triangulated in the opposite source variable with its checked signed shifts and whole cone comparisons. Its single finite-projective computation agrees with this functor by (5.3). Hence its triangle and translation comparisons restrict to (7.1); no new choice of an unsigned source translation is being imposed. For \(n=0\) this is the usual derived duality of bounded complexes with finite-dimensional cohomology over \(k\), with the same degree signs. The zero object uses the zero representative and every comparison is its identity. \(\square\)

8. Four graded exercises with complete solutions

Exercise 1 (24 points). Let \(R=k[x,y]\) and \(M=R/(x^2,xy,y^2)\), over any field. Prove exactness of the resolution whose nonzero maps are

Compute every \(\operatorname{Ext}^q_R(M,R)\), including representatives and the dimension of the nonzero group over \(k\).

Solution. The displayed composite is zero. To describe the middle kernel, suppose \(a x^2+bxy+c y^2=0\). Reduction modulo \(x\) gives \(\bar c y^2=0\) in the domain \(k[y]\), so \(c=xc_1\). Divide the original equation by \(x\) in the domain \(R\), obtaining \(a x+by+c_1y^2=0\). Modulo \(y\) this forces \(a=ya_1\); substitution gives \(b=-xa_1-yc_1\). Thus \((a,b,c)=-a_1(-y,x,0)+c_1(0,-y,x)\), in the image of \(B\). Its injectivity follows from its first and last coordinates: \(-yu=0\) and \(xv=0\) imply \(u=v=0\). The image of \(A\) is exactly the defining ideal, proving the full augmented exactness.

The standard dual complex has terms \(R\to R^3\to R^2\) in degrees \(0,1,2\), with maps \(-A^t\) and \(B^t\). The first has zero kernel since \(x^2\) is a nonzero divisor. If \(-ya+xb=0\) and \(-yb+xc=0\), first reduce modulo \(x\), then cancel \(x\), to get \(a=xu\), \(b=yu\). The second equation gives \(y^2u=xc\), so \(u=xv\) and \(c=y^2v\). Therefore \((a,b,c)=(x^2v,xyv,y^2v)\), which is in the image of \(-A^t\) after replacing \(v\) by \(-v\). Hence \(\operatorname{Ext}^0\) and \(\operatorname{Ext}^1\) vanish.

The degree-two cokernel has generators \(e_1,e_2\), with relations

It has basis \(e_1,e_2,t\), where \(t=xe_1=ye_2\). Both \(x\) and \(y\) kill \(t\), so all higher-degree multiples vanish. The constant generators are independent because all relations have positive degree. In degree one, the four possible multiples have the three independent relations displayed above, leaving the nonzero class \(t\). Thus its dimension is three. All other integer Ext degrees vanish since the computing complex has only degrees \(0,1,2\). In particular the derived dual is this three-dimensional module shifted by \([-2]\), whereas ordinary \(\operatorname{Hom}_R(M,R)\) is zero; Theorem 7.1 recovers \(M\) from its derived bidual.

Exercise 2 (12 points). Let \(R=k[t]\), and let \(P^{-1}=P^0=R\), with \(d_P^{-1}=t\). Compute \(D(P)\), \(D^2(P)\), and every component of \(\beta_P\). Verify the chain equation and inverse coherence. When does the unsigned degreewise evaluation fail to be a chain map?

Solution. The first dual is \(R\xrightarrow{-t}R\) in degrees \(0,1\). The double dual is \(R\xrightarrow{-t}R\) in degrees \(-1,0\). Formula (6.1) is minus the identity in degree \(-1\) and the identity in degree zero. Its chain equation is \((-t)(-1)=1\cdot t\). For the coherence on the first dual, degree zero has two factors \(+1\); degree one has two factors \(-1\). Each product is \(+1\), exactly (6.2). Unsigned evaluation would require \(-t=t\); it fails when \(2t\ne0\), which occurs here when the field has characteristic other than two. In characteristic two all the same formulas remain valid, with the signs equal. The original complex resolves \(R/(t)\), its derived dual has that module in degree one, and its derived bidual has it in degree zero.

Exercise 3 (18 points). For \(0\leq c\leq n\), put \(I=(x_1,\ldots,x_c)\) and \(M=R/I\). Compute every \(\operatorname{Ext}^q_R(M,R)\) and \(\operatorname{Tor}_p^R(M,M)\). Specify the orientation lines, derived shifts and both empty cases \(c=0\) and \(n=0\).

Solution. Let \(V_c=k^c\) with its ordered coordinate basis. The Koszul chain complex for those \(c\) coordinates resolves \(M\): tensor the one-variable contractions in (2.5), and then tensor with the remaining polynomial ring \(k[x_{c+1},\ldots,x_n]\). The projection to constants in the first coordinates is the actual quotient to \(M\). Dualizing and using the degree multiplier in Section 2 gives the cochain Koszul complex, whose ordered regular-prefix proof has only top cohomology. Its top quotient is \(M\otimes_k\det(V_c)^*\). Consequently

For tensor, every boundary coefficient \(x_i\), \(i\leq c\), acts as zero on \(M\). The entire computing tensor complex therefore has zero differential and terms \(M\otimes_k\bigwedge^pV_c\) in chain degrees \(0,\ldots,c\). Thus

and it is zero at every other integer index. Its derived object is the direct sum of these terms shifted by \([p]\), which places them in cochain degree \(-p\). When \(c=0\), the module is \(R\), the empty determinant and exterior degree zero are \(k\), and only Ext and Tor degree zero remain. When \(n=0\), necessarily \(c=0\) and these reduce to \(k\). The order specifies each top-line identification; the natural bidual uses (6.1) and is coherent independently of any chosen basis of that line.

Exercise 4 (16 points). Over the field \(k\), let \(V=\bigoplus_{i\geq0} k\), viewed in degree zero. Prove that evaluation \(V\to V^{**}\) is injective but not surjective. Explain which hypothesis of Theorem 7.1 is absent despite global dimension zero.

Solution. A functional on \(V\) is uniquely a sequence of its values on the coordinate vectors, so \(V^*=\prod_{i\geq0}k\). A nonzero vector of \(V\) has a nonzero coordinate, and its coordinate functional detects it, proving injectivity of evaluation.

Inside \(V^*\), let \(W=\bigoplus_{i\geq0}k\) denote the finite-support sequences. The constant sequence \(\mathbf1\) is not in \(W\), so its class is nonzero in \(V^*/W\). Choose a linear functional on this quotient taking that class to one: define it on its one-dimensional span and extend a basis. Pull it back to a functional \(\Phi\in V^{**}\). It vanishes on every coordinate sequence, yet \(\Phi(\mathbf1)=1\). If \(\Phi\) were evaluation at a finite-support vector \(v\), its value on each coordinate sequence would equal the corresponding coordinate of \(v\). All those coordinates would be zero, forcing \(v=0\), contradicting its value on \(\mathbf1\). Thus evaluation is not onto.

Every vector space is projective, so global dimension zero still holds. Finite generation is absent: this object is not in \(D^b_{\mathrm{fg}}(k)\), and its degree-zero term is not finite projective. The finite-projective bidual argument therefore has no claim about this object. The proof uses algebraic duals, with no topology or restricted continuous dual.

Freely accessible references

• The Stacks Project, Noetherian rings, Lemmas 10.31.1 and 10.31.4, for polynomial Noetherianity and finiteness of submodules. Section 1 gives the required complete algebraic proofs directly.
• The Stacks Project, Proposition 10.114.2, for the global dimension of a polynomial ring. Sections 2–3 give an independent induction using the polynomial-step exact sequence, cone resolution and kernel comparison, including arbitrary modules and finite generators.
• The Stacks Project, Perfect complexes, especially Lemmas 15.76.7 and 15.76.14; and Lemma 15.76.15, for the perfect representative and derived bidual statements. Sections 4–7 provide the complete representative construction, differential, signed evaluation, naturality and inverse coherence in the conventions used here.
• Determinant signs and subset Koszul complexes, Sections 1–5; Derived Hom and bounded resolution pairs, Sections 1–7; and Cohomological bounds, denominators and truncation adjunctions, Sections 1–5, for the complete programme proof interfaces used above.

These are classical algebraic and categorical results.