Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 3 (Commutative Algebra)

We use commutative rings with identity and unital modules.

An Artinian module is an $A$-module satisfying the descending chain condition on submodules: every chain $M_0\supseteq M_1\supseteq \cdots$ is eventually constant. An Artinian ring is a ring that is Artinian as a module over itself, so its descending chains of ideals stabilize.

We first prove the closure properties needed for finite generation. A submodule of an Artinian module is Artinian, because any descending chain in it is also a chain in the original module. A quotient module is Artinian, because inverse images turn a descending chain in the quotient into one upstairs. Conversely, consider a short exact sequencewith $N$ and $Q$ Artinian. For a descending chain $(M_i)$, the chains $M_i\cap N$ and $\pi (M_i)$ both stabilize. Choose $r$ beyond both stabilization indices. If $i\ge r$ and $x\in M_i$, equality of the images gives $y\in M_{i+1}$ with $\pi (y)=\pi (x)$. Thenso $x\in M_{i+1}$. Hence $M_i=M_{i+1}$. This proves the extension property of Artinian modules in a short exact sequence.

Apply this property inductively to $0\to A^{n-1}\to A^n\to A\to 0$. Every finite direct sum $A^n$ is Artinian when $A$ is an Artinian ring. If $m_1,\dots ,m_n$ generate $M$, the map $(a_i)\mapsto \sum a_i{m}_i$ is a surjection $A^n\twoheadrightarrow M$. The quotient property therefore provesThis argument also covers the zero module; it does not require a structure theorem for Artinian rings.

For a counterexample without finite generation, take any field $k$ and the moduleThe ring $k$ is Artinian, since its only ideals are $0$ and $k$. But $M_i⊋M_{i+1}$ for every $i$, since $e_i\notin M_{i+1}$. Thus an infinite-dimensional vector space over a field provides the requested non-Artinian module.

Finally, let $f:M\to M$ be an injective endomorphism with $M$ Artinian. Its image chain stabilizes, so $f^{r}M=f^{r+1}M$ for some $r$. Given $x\in M$, choose $y$ with $f^{r}x=f^{r+1}y=f^r(fy)$. Since $f^r$ is injective, $x=fy$. ThereforeThis is precisely the assertion that Artinian modules are co-Hopfian.

The spectrum of a ring $A$ is the set of its proper prime ideals. For an ideal $I$, set $V(I)=\{\mathfrak{p}:I\subseteq \mathfrak{p}\}$. These are the closed sets of the Zariski topology. They satisfyThe union identity follows because a prime ideal containing $IJ$ must contain $I$ or $J$: otherwise choose $a\in I\setminus \mathfrak{p}$ and $b\in J\setminus \mathfrak{p}$, contradicting $ab\in \mathfrak{p}$. These identities verify the closed-set axioms. The maximal spectrum has the subspace topology, whose closed sets are $V(I)\cap \mathrm{MaxSpec}A$.

For a finitely generated $\mathbb{C}$-algebra, the needed form of the Hilbert Nullstellensatz isHere an empty intersection means $A$. We prove this formula rather than assume the desired correspondence.

First establish the residue-field fact. If a field $L$ is a finitely generated algebra over $\mathbb{C}$, choose a maximal algebraically independent subset $t_1,\dots ,t_r$ of a finite generating set, with remaining generators $u_1,\dots ,u_s$. Each $u_j$ is algebraic over $\mathbb{C}(t_1,\dots ,t_r)$. Clearing the finitely many denominators in their monic equations gives a nonzero polynomial $d$ such that each $u_j$ is integral overSince $L$ is a field, it contains $1/d$, and $L=R[u_1,\dots ,u_s]$ is an integral extension of $R$. If a field is integral over a domain, the domain is itself a field: for $0\ne a\in R$, an integral equation for $a^{-1}$, multiplied by $a^{n-1}$, expresses $a^{-1}$ in $R$.

But if $r>0$, choose $c\in \mathbb{C}$ for which $t_1-c$ does not divide $d$. Such a choice exists because a nonzero polynomial has only finitely many distinct factors $t_1-c$. The element $t_1-c$ is a nonunit even after inverting $d$: an inverse would imply that $t_1-c$ divides a power of $d$. Thus $R$ would not be a field, a contradiction. Hence $r=0$, and $L$ is algebraic over the algebraically closed field $\mathbb{C}$, forcing $L=\mathbb{C}$. This proves the Zariski lemma in the form used here.

Now if $f\notin \sqrt{I}$, the localization of a ring $B=(A/I)[1/f]$ is nonzero: vanishing would mean $f^n\in I$ for some $n$. It is still a finitely generated algebra over $\mathbb{C}$, since one need only adjoin an inverse of $f$. Choose a maximal ideal $\mathfrak{n}$ of $B$. By the preceding argument, $B/\mathfrak{n}=\mathbb{C}$. The composite $A\to B\to \mathbb{C}$ is surjective because it contains the constant scalars. Its kernel is consequently a maximal ideal $\mathfrak{m}$ of $A$, contains $I$, and does not contain $f$, whose image is invertible. This proves that the displayed intersection is contained in $\sqrt{I}$. The reverse inclusion follows because every maximal ideal is prime and contains the radical of an ideal whenever it contains that ideal.

For a closed subset $Z\subseteq \mathrm{MaxSpec}A$, put $I(Z)={\bigcap }_{\mathfrak{m}\in Z}\mathfrak{m}$. It is a radical ideal. The formulas now showare inverse maps on radical ideals and closed sets. Indeed the first followed by the second gives $\sqrt{I}=I$, while if $Z=V(J)\cap \mathrm{MaxSpec}A$, the second gives $\sqrt{J}$ and $V(\sqrt{J})=V(J)$. Thus the correspondence is a bijection reversing inclusion. This is the relation between radical ideals and closed subsets of the maximal spectrum.

For $A=\mathbb{C}[x]$, the maximal ideals are ${\mathfrak{m}}_c=(x-c)$, for $c\in \mathbb{C}$. Every nonzero proper radical ideal is generated by a squarefree polynomial, and hence has the formfor a finite nonempty subset $S\subset \mathbb{C}$. The associated closed set is $\{{\mathfrak{m}}_c:c\in S\}$. A repeated factor of a polynomial does not change its zero set, but passing to its radical ideal removes that repeated factor. The two endpoint cases areThe empty-product convention makes $I_{\varnothing }=(1)$. Under $c\mapsto {\mathfrak{m}}_c$, the closed sets are exactly the finite subsets and the whole affine line.

An extension $A\subseteq B$ is an integral extension when every $b\in B$ is an integral element over $A$, meaning thatfor some monic polynomial. The Krull dimension of a ring is the supremum of the lengths $r$ of strict chains ${\mathfrak{p}}_0⊊\cdots ⊊{\mathfrak{p}}_r$ of prime ideals; the number counted is the number of inclusions.

We prove the prime-chain facts needed for integral extensions preserve Krull dimension. Taking quotients preserves an integral equation. Localizing both rings by the same multiplicative subset of $A$ also preserves integrality: for $b/s$, divide the equation for $b$ by $s^n$, so its new coefficients are $a_i/s^{n-i}$. Localizing just $B$ at all elements outside one of its primes would not be the same operation.

Two elementary field observations are useful. If a field $C$ is integral over a subdomain $R$, then $R$ is a field, by the inverse argument given in Question 2. Conversely, if an integral domain $C$ is integral over a field $R$, every nonzero $c\in C$ has a monic equation with nonzero constant term: cancel a factor of $c$ from an equation if necessary. Dividing by $c$ and by that constant term expresses $c^{-1}$ as a polynomial in $c$. Thus $C$ is a field. In particular, the contraction of a maximal ideal under an integral extension is maximal.

For the Lying-over theorem, fix $\mathfrak{p}\in \mathrm{Spec}A$ and localize by $S=A\setminus \mathfrak{p}$. The nonzero domain $S^{-1}B$ is integral over the local ring $A_{\mathfrak{p}}$. Choose a maximal ideal $\mathfrak{n}$ of $S^{-1}B$. Its contraction is maximal in $A_{\mathfrak{p}}$, so it is the unique maximal ideal $\mathfrak{p}{A}_{\mathfrak{p}}$. The prime ideal correspondence for localization now pulls $\mathfrak{n}$ back to a prime ideal $\mathfrak{q}$ of $B$ with $\mathfrak{q}\cap A=\mathfrak{p}$.

For the going-up theorem, suppose $\mathfrak{p}\subseteq {\mathfrak{p}}^{\prime }$ and $\mathfrak{q}\cap A=\mathfrak{p}$. The injection $A/\mathfrak{p}\hookrightarrow B/\mathfrak{q}$ is still integral. Apply the just-proved Lying-over theorem to ${\mathfrak{p}}^{\prime }/\mathfrak{p}$. Pulling the resulting prime back gives ${\mathfrak{q}}^{\prime }\supseteq \mathfrak{q}$ with ${\mathfrak{q}}^{\prime }\cap A={\mathfrak{p}}^{\prime }$.

For the incomparability theorem for integral extensions, suppose $\mathfrak{q}\subseteq {\mathfrak{q}}^{\prime }$ have the same contraction $\mathfrak{p}$. Write $R=A/\mathfrak{p}$ and $C=B/\mathfrak{q}$. The latter is a domain integral over $R$. Localizing both by $R\setminus \{0\}$ makes $S^{-1}C$ integral over the fraction field of $R$, so $S^{-1}C$ is a field. The prime ${\mathfrak{q}}^{\prime }/\mathfrak{q}$ is disjoint from $S$ and therefore localizes to a proper prime, which must be zero in this field. Since $C$ is a domain and all denominators are nonzero, an element of $C$ whose localization is zero was already zero. Hence ${\mathfrak{q}}^{\prime }/\mathfrak{q}=0$, proving ${\mathfrak{q}}^{\prime }=\mathfrak{q}$.

A strict prime chain in $B$ therefore contracts to a strict chain in $A$, giving $\dim B\le \dim A$. Conversely, start over the bottom of any finite prime chain in $A$ using Lying-over theorem, and successively use going-up theorem to lift the entire chain. Distinct contractions ensure that the lifted inclusions are strict, giving the reverse inequality. Taking suprema, including when arbitrarily long chains exist, proves

For an algebraic extension $K/\mathbb{Q}$, its ring $\mathcal{O}$ of algebraic integers is an integral extension of $\mathbb{Z}$. The primes of $\mathbb{Z}$ are $(0)$ and the maximal ideals $(p)$, so the longest strict chains have length one; for example $(0)⊊(2)$. The preceding argument yieldsThis Krull dimension of rings of algebraic integers argument also covers infinite algebraic extensions. It uses integrality, not an assumption that $K$ is a number field or that $\mathcal{O}$ is Noetherian.

Let $K=Q(A)$ be the fraction field of the integral domain $A$. The Picard group of a ring consists of isomorphism classes of finitely generated projective modules that have rank one after localization at every prime ideal. Its product is induced by the tensor product of modules, its identity is $[A]$, and the inverse of $[L]$ is $[L^{\vee }]$, where $L^{\vee }={\mathrm{Hom}}_A(L,A)$. Evaluation $L{\otimes }_A{L}^{\vee }\to A$ is an isomorphism: after localization it is evaluation on a free rank-one module, and a map whose kernel and cokernel vanish at every prime is an isomorphism.

An invertible fractional ideal is a nonzero finitely generated $A$-submodule $I\subseteq K$ admitting another fractional ideal $J$ with $IJ=A$. The Cartier divisors of $\mathrm{Spec}A$ can be represented by these ideals. Concretely a Cartier divisor has local nonzero rational equations $f_i$ on a cover by distinguished open sets, with $f_i/f_j$ a regular unit on each overlap. We use the convention assigning to it $\mathcal{O}(-D)$, whose local fractional ideals are $f_i{A}_{s_i}$. Their unit ratios make these modules agree on overlaps. Thus the Cartier divisor group $\mathrm{Cart}(A)$ is realized by invertible fractional ideals with multiplication. The identity is $A$ and a principal Cartier divisor of $f\in K^{\times }$ corresponds to $fA$.

To check explicitly that local divisor data give an invertible fractional ideal, set $I={\bigcap }_i{f}_i{A}_{s_i}\subseteq K$. Take a finite distinguished-open cover; this is possible because an affine spectrum is quasi-compact. On an overlap, $f_i/f_j\in A_{s_i{s}_j}$. For each fixed $i$, clear the powers of $s_i$ in these finitely many expressions: there is $N_i$ with $s_i^{N_i}{f}_i\in f_j{A}_{s_j}$ for every $j$. Thus $t_i=s_i^{N_i}{f}_i$ belongs to $I$, and $I_{s_i}=f_i{A}_{s_i}$. The finite module $J={\sum }_{i}A{t}_i$ has the same localizations, so $I/J$ vanishes on the cover and $I=J$. Repeating with $f_i^{-1}$ gives a finitely generated fractional ideal $I^{\prime }$ with $I_{s_i}^{\prime }=f_i^{-1}{A}_{s_i}$. Therefore $I{I}^{\prime }$ is contained in $A$ and equals $A$ after localization on the cover, whence $I{I}^{\prime }=A$. These conclusions use the elementary localization test proved below. Conversely, the next argument constructs precisely such a finite principal cover from an inverse ideal. Thus the local-divisor and ideal descriptions agree, without assuming that $A$ is Noetherian.

Here is the algebra underlying that realization and the map to the Picard group. If $IJ=A$, write $1={\sum }_{i=1}^r{u}_i{v}_i$, with $u_i\in I$ and $v_i\in J$. On the distinguished open where $s_i=u_i{v}_i$ is invertible, $I_{s_i}=u_i{A}_{s_i}$: for any $x\in I$, the expressionis in $A_{s_i}$. These open sets cover the spectrum because $\sum s_i=1$. The generators on overlaps differ by units. Moreover the maps ${\varphi }_i:I\to A$, ${\varphi }_i(x)=v_{i}x$, satisfy $x=\sum u_i{\varphi }_i(x)$. Consequently the compositeis the identity. This explicitly makes $I$ a direct summand of a finite free module, and hence a projective module, with rank one locally.

Its inverse isThe inclusion $J\subseteq I^{-1}$ follows from $IJ=A$. For the other inclusion, $v\in I^{-1}$ gives $v=\sum (v{u}_i)v_i\in J$. Local multiplication is an isomorphism between free rank-one modules, so multiplication $I{\otimes }_{A}J\to IJ$ is an isomorphism. In particular $I\mapsto [I]$ respects the group laws.

Conversely, suppose $L$ is a finitely generated projective module of rank one. It is a direct summand of a finite free module, so it is torsion-free and injects into $L{\otimes }_{A}K$. The latter is a one-dimensional $K$-space. Choose a basis to identify it with $K$, and let $I$ be the image of $L$. Its finitely many generators have a common nonzero denominator, making $I$ a fractional ideal. Every $A$-linear map $I\to A$ extends to a $K$-linear map $K\to K$, which is multiplication by some $v\in K$. Thus $I^{\vee }$ identifies with $I^{-1}$. Evaluation is an isomorphism locally, so its image ideal $I{I}^{-1}$ localizes to $A_{\mathfrak{p}}$ at every prime. A proper ideal would be contained in a maximal ideal, contradicting that localization; hence $I{I}^{-1}=A$. This proves that every Picard group class has an invertible fractional ideal representative, and also that locally principal fractional ideals on a finite open cover are invertible.

For completeness, these local tests do detect a zero module without a finite-generation assumption. If $0\ne x\in N$, its annihilator is proper; choose a maximal ideal containing it. Then $x/1$ is nonzero in that localization, since any denominator killing $x$ would belong to its annihilator. Applying this observation to a kernel and cokernel justifies the local isomorphism tests above.

The natural maps in the Cartier divisor exact sequence for an integral domain are nowThe first map is injective. The ideal $fA$ is the identity divisor $A$ precisely when $f$ and $f^{-1}$ lie in $A$, that is, precisely when $f\in A^{\times }$. Next $[I]=[A]$ precisely when some isomorphism $A\to I$ sends $1$ to an element $f\in K^{\times }$ generating $I$, so $I=fA$. Finally the construction from $L$ proves surjectivity onto the Picard group. We have therefore checked exactness at every term:No Noetherian, normality or factoriality hypothesis is required.

Put $D=\mathbb{Q}/\mathbb{Z}$. The character module $M^{\ast }={\mathrm{Hom}}_{\mathbb{Z}}(M,D)$ has the natural actionAdditivity is immediate, and $a(b\varphi )(m)=\varphi (bam)=\varphi (abm)$, using commutativity of $A$. The identity of $A$ acts as the identity. We establish the extension and duality tools used in the four implications, so no flatness or injectivity criterion is assumed without proof.

An injective module $E$ is one for which every homomorphism from a submodule $U\subseteq V$ extends to $V$. The Baer criterion says it suffices to test the inclusions of ideals $I\subseteq A$. Necessity follows directly from the definition. For sufficiency, suppose every $I\to E$ extends to $A$. Given $h:U\to E$, use Zorn's lemma to choose a maximal extension $(W,h_W)$ inside $V$; extensions along a chain have a well-defined union. If $v\in V\setminus W$, letThis is an ideal and an $A$-linear map. Its extension to $A$ has the form $a\mapsto ae$ for some $e\in E$. Thenis well-defined: if two expressions represent the same element, their coefficient difference lies in $I$, where the two prescribed values agree. It extends $h_W$ to the strictly larger module $W+Av$, a contradiction. Thus $W=V$, proving the Baer criterion for arbitrary modules and ideals.

The group $D$ is divisible. A homomorphism $n\mathbb{Z}\to D$, for $n>0$, extends to $\mathbb{Z}$ by choosing $d\in D$ with $nd$ equal to its value at $n$; the zero ideal causes no restriction. The proved Baer criterion over $\mathbb{Z}$ therefore makes $D$ injective. It is also a cogenerator: a nonzero element $x$ of an abelian group generates either a finite cyclic group of order $n>1$, which maps $x$ to $1/n+\mathbb{Z}$, or an infinite cyclic group, which maps $x$ to $1/2+\mathbb{Z}$. Injectivity extends this character to the whole group. Thus every nonzero element is detected by an additive character.

It follows that a group homomorphism $u:T\to U$ is injective exactly whenis surjective. For an injection, extension into the injective group $D$ proves surjectivity. Conversely, a character nonzero on an element of $\ker u$ cannot be the pullback of a character on $U$. This is the detection property of the injective cogenerator of abelian groups.

Finally, for any $A$-module $N$ there is a natural character form of the Tensor-hom adjunction:An $A$-linear map $h$ gives the pairing $(n,m)\mapsto h(n)(m)$. It is balanced because $h(an)(m)=(ah(n))(m)=h(n)(am)$. Conversely a balanced additive pairing determines $h$ by this formula. The universal property of the tensor product of modules makes these constructions mutually inverse, and compatible with restriction maps.

A flat module is one for which tensoring preserves injections. This is equivalent to exactness of the tensor functor, since the tensor product of modules is always right exact: its universal balanced maps identify tensoring a cokernel with the cokernel of the tensored map.

Suppose $M$ is flat. For every inclusion $N^{\prime }\hookrightarrow N$, the map $N^{\prime }{\otimes }_{A}M\to N{\otimes }_{A}M$ is injective. The detection property proved above therefore makes its character-dual map surjective. Under the character Tensor-hom adjunction, this is exactlyEvery map from a submodule into $M^{\ast }$ thus extends, proving

Conversely, if the character module $M^{\ast }$ is injective, every restriction map in the preceding display is surjective. The character Tensor-hom adjunction then makes the dual of $N^{\prime }{\otimes }_{A}M\to N{\otimes }_{A}M$ surjective. Since $D$ detects injections, that tensor map is injective for every inclusion $N^{\prime }\hookrightarrow N$. Hence $M$ is flat, andTogether these two implications prove the Lambek theorem, with both the duality and the injective-cogenerator facts justified.

For an ideal $I$, the canonical multiplication maphas image $IM$. Consequently the asserted isomorphism is the natural surjection $I{\otimes }_{A}M\twoheadrightarrow IM$ being injective. This is the flatness criterion using ideals, whose proof follows from the extension criterion above.

Under the character Tensor-hom adjunction, the dual map ${\mu }_I^{\ast }$ identifies withIndeed, for $\varphi \in M^{\ast }$, its restriction corresponds to $i\otimes m\mapsto \varphi (im)$. If $M^{\ast }$ is injective, this restriction is surjective, so the cogenerator criterion makes ${\mu }_I$ injective. Conversely, if every ${\mu }_I$ is injective, every restriction map in this display is surjective. The proved Baer criterion then makes $M^{\ast }$ injective. ThusIn particular, (1) also implies (3) directly by tensoring $I\hookrightarrow A$. No finite-generation condition on $I$, and no Noetherian condition on $A$, has been inserted.

To compute the first Tor of a cyclic quotient, start a free resolution of $A/I$ with $P_0=A$, take a free module $P_1$ surjecting onto $I$, and take $P_2$ surjecting onto the kernel of $P_1\to I$. Tensor right exactness identifiesBy definition, the degree-one Tor functor is the kernel of $P_1{\otimes }_{A}M\to A{\otimes }_{A}M$ modulo the image from $P_2{\otimes }_{A}M$. Quotienting first by that image consequently gives the natural identificationEquivalently the calculation gives the exact sequenceThus its first term vanishes for every ideal exactly when each multiplication map is injective, provingTogether with the preceding implications, all four conditions are equivalent. The proof used explicit extension arguments, character duality and the definition of the Tor functor, rather than an unproved flatness or injectivity criterion.

Write $R^{i}F$ for the cohomological indexing of the right derived functor. For the contravariant construction, assume $F$ is additive and left exact, so an exact sequence $M^{\prime }\to M\to M^{\prime \prime }\to 0$ gives $0\to F(M^{\prime \prime })\to F(M)\to F(M^{\prime })$. Module categories have enough projective modules: a free module on a generating set surjects onto any module, and repeating the construction on its kernel produces a projective resolutionContravariance turns it into the cochain complexThe composite of consecutive differentials is zero because $F$ is additive and reverses composition. Applying left exactness to $P_1\to P_0\to M\to 0$ gives $F(M)=\ker (F(d_1))$, so $R^{0}F\simeq F$.

Here are the comparison and exactness arguments behind this construction. A map $M\to N$ lifts to a chain map between projective resolutions. In degree zero, lift through the surjection $Q_0\twoheadrightarrow N$ by projectivity of $P_0$. Inductively, the required composite from $P_i$ lands in the cycle module of the target resolution, which is the image of $Q_i$ by exactness, so projectivity supplies the next lift. Two such lifts are chain homotopic. Their difference in degree zero lands in $\ker (Q_0\to N)=\mathrm{im}{d}_1$, so it lifts to a homotopy map $P_0\to Q_1$. Subtracting the corresponding homotopy term in degree one gives another map into the next cycle module; continue inductively. Applying additive contravariant $F$ turns this chain homotopy into a cochain homotopy, so the maps on cohomology agree. Applying this to lifts of identity maps between two resolutions gives mutually inverse maps on cohomology. Thus the contravariant right derived functor is well-defined, contravariant and natural in $M$.

For $0\to M^{\prime }\to M\to M^{\prime \prime }\to 0$, construct compatible projective resolutions by the horseshoe lemma. Its construction is elementary: lift the projective cover of $M^{\prime \prime }$ into $M$ and add the projective cover of $M^{\prime }$; their sum surjects onto $M$. The kernels fit into a new short exact sequence of the corresponding kernels, so repeat at each degree. This gives a degreewise split sequenceAn additive contravariant functor takes split sequences to split sequences with their arrows reversed. The resulting short exact sequence of cochain complexes givesThe connecting map is obtained by lifting a cocycle from the quotient complex, taking its differential, and viewing the result in the subcomplex; changing a lift changes the result by a coboundary. This construction proves both exactness and naturality of the long exact sequence.

Positive derived functors vanish on projective modules, since such a module has a resolution concentrated in degree zero. For $0\to K\to P\to M\to 0$ with $P$ projective, the preceding sequence gives $R^{i+1}F(M)\simeq R^{i}F(K)$ for $i\ge 1$. This is dimension shifting. If $F$ is exact, all positive derived functors vanish because applying $F$ preserves exactness of the resolution. A natural transformation of additive left-exact contravariant functors induces natural transformations of their derived functors by applying it degreewise. These derived functors form the universal contravariant cohomological delta functor: successive projective surjections, on which the positive functors vanish, and the connecting maps determine any extension of a degree-zero natural transformation uniquely.

The displayed example in the PDF has a genuine variance mismatch. The functor $G(M)={\mathrm{Hom}}_A(A/(a),M)$ is covariant in $M$, not contravariant. For a map $u:M\to N$, its induced map sends $h$ to $u\circ h$, in the same direction. We compute that displayed example with its correct variance, and then explain the contravariant variant.

For an additive covariant left-exact functor, use an injective resolution $0\to M\to I^0\to I^1\to \cdots$ instead of a projective resolution, and set $R^{i}G(M)=H^i(G(I^{\bullet }))$. Such resolutions exist for arbitrary modules. In fact $J={\mathrm{Hom}}_{\mathbb{Z}}(A,D)$, with $(a\varphi )(b)=\varphi (ab)$, is injective: the adjunctionsends a map to its value at $1$, and has inverse $\lambda \mapsto [n\mapsto (b\mapsto \lambda (bn))]$. Extension on the right was proved in Question 5, so it gives extension on the left. Products of injective modules are injective by extending every component separately. The mapis $A$-linear and injective, because a nonzero $m$ is detected by some character at $b=1$. Applying the same embedding to successive cokernels produces the stated injective resolution. Its comparison and homotopy arguments are dual to the projective ones, and covariant derived functors have the long exact sequence in the original order of $M^{\prime },M,M^{\prime \prime }$.

Let $B=A/(a)$. Evaluation at $1+(a)$ identifiesa naturally defined $B$-module. It is left exact: a compatible element in the kernel of $G(M)\to G(M^{\prime \prime })$ is exactly an element of $M^{\prime }$ killed by $a$.

Because $a$ is a non-zero-divisor on $A$, multiplication by $a$ is surjective on every injective module $I$. Indeed, for $m\in I$, the map $aA\to I$ given by $ax\mapsto xm$ is well-defined and $A$-linear. It extends to $A\to I$, and if the extension takes $1$ to $v$, then $av=m$. Thus for an injective resolution we have a short exact sequence of cochain complexesThe cohomology of $I^{\bullet }$ is $M$ in degree zero and zero in higher degrees. Its long exact sequence therefore givesConsequently the Ext from a principal non-zero-divisor quotient computation isAll these groups are naturally $B$-modules. Equivalently they are ${\mathrm{Ext}}_A^i(B,M)$; the two-term free resolution $0\to A\stackrel{a}{\to }A\to B\to 0$ gives the same kernel and cokernel after applying ${\mathrm{Hom}}_A(-,M)$.

If one instead wants a genuinely contravariant example, take $H(N)={\mathrm{Hom}}_A(N,B)$. Then $R^{i}H(N)={\mathrm{Ext}}_A^i(N,B)$, obtained by resolving the variable $N$. There is no claim of vanishing above degree one for arbitrary $N$. At $N=B$, the same two-term resolution has zero differential after applying ${\mathrm{Hom}}_A(-,B)$, and gives ${\mathrm{Ext}}_A^0(B,B)=B$, ${\mathrm{Ext}}_A^1(B,B)=B$, and zero higher groups. This keeps the printed covariant calculation distinct from the general contravariant construction.