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An extension group is a value of an Ext functor. In degree one it classifies short exact sequences $0\to N\to E\to M\to 0$ up to isomorphisms fixing the endpoints. The zero class is the split extension; its addition is the Baer sum. Higher degrees can be represented by longer exact extensions, or computed from a projective resolution.

For a non-zero-divisor $a\in A$, the free resolution $0\to A\stackrel{a}{\to }A\to A/(a)\to 0$ gives degree-zero kernel and degree-one cokernel, and all higher Ext functors vanish. Directly in an injective resolution of the second variable, multiplication by $a$ is surjective in every term: the map $aA\to I$ taking $ax$ to $xm$ extends to $A\to I$. Thus $0\to {\mathrm{Hom}}_A(A/(a),I^{\bullet })\to I^{\bullet }\stackrel{a}{\to }{I}^{\bullet }\to 0$ is a short exact sequence of complexes and gives the same result. The displayed Hom functor is covariant in $M$; changing it to ${\mathrm{Hom}}_A(M,A/(a))$ changes the variable that is resolved.

Partition the isomorphism classes of simple modules into two sets ${\mathcal{C}}_1,{\mathcal{C}}_2$, with ${\mathrm{Ext}}^1(S,T)={\mathrm{Ext}}^1(T,S)=0$ whenever $S,T$ lie in different sets. Every module $M$ of finite composition length then has a unique decomposition $M=U_1\oplus U_2$ into its largest submodules whose Jordan–Hölder factors belong to the respective sets.

For a simple $S\in {\mathcal{C}}_1$ and a finite-length $N$ with factors in ${\mathcal{C}}_2$, induction through the long exact sequence of the Ext functor gives ${\mathrm{Ext}}^1(S,N)=0$. Induct on the length of $M$: take $0\to N\to M\to S\to 0$, split $N=N_1\oplus N_2$, and suppose $S\in {\mathcal{C}}_1$. The quotient extension $0\to N_2\to M/N_1\to S\to 0$ splits. The inverse image of its $S$-summand complements $N_2$ in $M$ and has only factors in ${\mathcal{C}}_1$. Finally, Hom between modules with factors in different sets is zero, since any nonzero image would have a factor in both sets. Projecting any proposed submodule to the opposite summand therefore proves maximality and uniqueness.

On ${\mathbb{P}}_k^n$ with $n>0$, let $\mathcal{F}=i_{\ast }k$ at a rational point. It has one-dimensional global sections, but its dual of a sheaf is zero: a local map from the residue field into the local polynomial ring must have image annihilated by a nonzero parameter, which is impossible in an integral domain. Thus $H^n({\mathcal{F}}^{\vee }(-n-1))=0$. Serre duality for arbitrary coherent sheaves instead uses an Ext functor; replacing this by the ordinary dual can lose torsion information.

For an ideal generated by a finite regular sequence in a commutative ring, the self-Ext functor algebra is exterior on the degree-one dual generators. Contraction chain maps on the Koszul resolution represent those generators, square to zero, and anticommute. The exterior square-zero relations also apply in characteristic two.

A short exact sequence with specified kernel module $A$ and quotient module $C$. Its equivalence class is determined by the degree-one Ext functor; equivalence must preserve both end modules.

A free resolution of $\mathbb{Q}$ has generators $a_n=1/n!$ and independent relations $a_n-(n+1)a_{n+1}$. Thus the Ext functor gives the cokernel of the map on integer sequence productsThe constant sequence one is not in its image. A preimage would require $u_1\equiv {\sum }_{j=1}^{N}j!(\mathrm{mod}(N+1)!)$ for every $N$. For $N\ge 3$, the sum is below $2N!$, tends to infinity, and its distance from $(N+1)!$ also tends to infinity. No fixed integer satisfies all these congruences. Hence ${\mathrm{Ext}}_{\mathbb{Z}}^1(\mathbb{Q},\mathbb{Z})\ne 0$.

Put $\Gamma ={\pi }_1(X,x)$ and $R=\mathbb{Z}[\Gamma ]$. Give $\mathbb{Z}$ its trivial left $R$-module structure, through the augmentation $\epsilon :R\to \mathbb{Z}$, $\sum n_{g}g\mapsto \sum n_g$. The universal cover $\tilde{X}$ has the lifted cell structure, with $\Gamma$ acting by deck transformations.

For each cell of $X$, choose one lifted cell and orient every translate compatibly with its projection. The translates form one copy of the regular $R$-module. Thusis a free left $R$-module, and its boundary is $R$-linear. The augmentation $C_0(\tilde{X})\to \mathbb{Z}$ sends every vertex to one. The contractibility of $\tilde{X}$ says that its reduced homology vanishes, so the augmented complexis exact. It is therefore a free resolution of the trivial module. By the definition of the Ext functor,

An $R$-linear cochain with values in the trivial module has the same value on every translate of a lifted cell. It is therefore exactly an ordinary integral cellular cochain on $X$. The correspondence does not depend on which lift was initially chosen. It also respects the coboundary: applying a constant-on-orbits cochain to the lifted cellular boundary sums precisely the incidence coefficients of the boundary in the base. Henceas cochain complexes, not just as graded groups. Cellular cohomology computes the cohomology of a cell complex, givingAt $i=0$ both sides are $\mathbb{Z}$, by connectedness and ${\mathrm{Hom}}_R(\mathbb{Z},\mathbb{Z})=\mathbb{Z}$. This is the cellular free resolution from a contractible universal cover description of group cohomology.

We prove the Birkhoff–Grothendieck theorem by induction on the rank of the vector bundle $E$. Rank zero is immediate; rank one is the classification $\mathrm{Pic}({\mathbb{P}}_k^1)\cong \mathbb{Z}$ established above.

First there is a largest integer $a$ with $H^0(E(-a))\ne 0$. To see existence and boundedness directly, trivialize $E$ on the two standard affine charts: the associated finite projective modules over the principal ideal domains $k[t]$ and $k[t^{-1}]$ are free. Let $A(t)\in G{L}_r(k[t,t^{-1}])$ be its transition matrix, so a section of $E(-a)$ is a pair satisfyingIf $a$ exceeds the largest exponent in any entry of $A$, the right side has only strictly negative powers; the left side has nonnegative powers. Hence both vanish. For $a$ sufficiently negative, take a nonzero constant vector $p$ and put $q=t^{a}A(t{)}^{-1}p$; all its exponents are nonpositive, so it supplies a nonzero section. Thus a largest $a$ exists.

Such a section gives a nonzero map $\mathcal{O}(a)\to E$. It is injective, since it is nonzero generically and the source is torsion-free. Saturate its image. The resulting line subbundle is $\mathcal{O}(a+d)$ for the degree $d\ge 0$ of the effective divisor of zeros of the original map. If $d>0$, its inclusion in $E$ would give a nonzero section of $E(-a-d)$, contradicting maximality. Therefore $d=0$ andhas a locally free quotient. By induction, $Q\cong {⨁}_j\mathcal{O}(b_j)$.

Twist this exact sequence by $\mathcal{O}(-a-1)$. The outer groups $H^0(\mathcal{O}(-1))$ and $H^1(\mathcal{O}(-1))$ are zero by the Čech cohomology of twists on the projective line. The long exact sequence in sheaf cohomology therefore givesThe first is zero by the choice of $a$, so each $H^0(\mathcal{O}(b_j-a-1))$ vanishes. Thus $b_j\le a$ for every $j$. The extension class lies insince $a-b_j\ge 0$. The identification of this Ext functor with cohomology can be seen without additional duality: local splittings of a vector-bundle extension differ by a Čech cocycle in $\mathcal{H}om(Q,\mathcal{O}(a))$, and a coboundary changes them to compatible splittings. Vanishing therefore gives a global splitting. Consequentlycompleting the induction over any field $k$.

The degrees can be arranged in nonincreasing order and their multiset is unique. Indeed, for any such decomposition,so the dimensions of the twisted global sections recover the number of summands of every degree.

Since $M$ is a finitely generated module, choose a surjection from a finite free module, $F_0\twoheadrightarrow M$. A finite free module over a left Noetherian ring is a Noetherian module, so the kernel $K_1$ is finitely generated. Choose a finite free module $F_1$ surjecting onto $K_1$, then compose its map with $K_1\hookrightarrow F_0$. Its kernel is again finitely generated. Repeating constructs an exact sequencewith all $F_i$ finite free modules, hence finitely generated projective modules. Thus the requested projective resolution exists, without asserting that it has finite length.

Assume now that $M\ne 0$ and ${\mathrm{pd}}_{R}M=n<\infty$. We may arrange a length-$n$ projective resolution with all terms finite projective. Here is why the preceding construction gives one. Set $K_0=M$, with short exact sequences $0\to K_{j+1}\to F_j\to K_j\to 0$. Applying the Ext functor, whose positive degrees vanish on a projective first argument, gives by repeated dimension shiftingfor every left module $N$. The final equality follows from the existence of any length-$n$ projective resolution. Vanishing of these degree-one groups means that every short exact sequence ending in $K_n$ splits: its extension class is zero. In particular a free surjection onto $K_n$ splits, so $K_n$ is a projective module. It is finitely generated by the Noetherian kernel construction. Use $P_n=K_n$ and $P_j=F_j$ for $j<n$ to obtainFor a left finite projective module $P$, put $P^{\ast }={\mathrm{Hom}}_R(P,R)$, a right module under $(fr)(p)=f(p)r$. It is a right finite projective module, and its natural evaluation mapis an isomorphism. These claims hold for a finite free module by coordinates, and pass to direct summands, proving them for every finite projective module without any right-Noetherian hypothesis.

Suppose $n\ge 1$ and ${\mathrm{Ext}}_R^n(M,R)=0$. Computing the Ext functor from this resolution shows that its last dual mapis surjective. Since $P_n^{\ast }$ is projective as a right module, this map has a right-module section $h:P_n^{\ast }\to P_{n-1}^{\ast }$, with $d_n^{\ast }h=1$. Dualizing gives $h^{\ast }{d}_n^{\ast \ast }=1$. Under the natural double-dual identifications, $d_n^{\ast \ast }$ is $d_n$, so this gives a left-module retraction of $d_n$.

Therefore $P_{n-1}=d_n(P_n)\oplus Q$ for a finite projective module $Q$. The next differential has kernel $d_n(P_n)$, so restricting it to $Q$ gives an exact resolution with $Q$ replacing $P_{n-1}$ and the initial $P_n$ removed. This has length $n-1$, contradicting the definition of projective dimension. For $n=0$, the nonzero $M$ is a finite projective module and a summand of some $R^d$. A nonzero vector in that summand has a nonzero coordinate, so one coordinate projection restricts to a nonzero map $M\to R$. Thus the degree-zero Hom functor is nonzero as well. We have provedThis is top Ext detects finite projective dimension over a left Noetherian ring. The nonzero-module convention matters when $n=0$: if one assigns projective dimension zero to the zero module, then ${\mathrm{Hom}}_R(0,R)=0$ is an exception. With the convention ${\mathrm{pd}}_{R}0=-\infty$, it is outside the asserted finite nonnegative case.

For an example of infinite projective dimension, take the dual numbers $R=k[\epsilon ]/({\epsilon }^2)$ over a field $k$, and $M=k=R/(\epsilon )$. The periodic resolution over dual numbers isIt is exact because both the kernel and image of multiplication by $\epsilon$ are the ideal $(\epsilon )$, and that is also the kernel of $R\to k$. Applying ${\mathrm{Hom}}_R(-,k)$ gives a copy of $k$ in every degree and zero differentials, since $\epsilon$ acts as zero on $k$. ConsequentlyA finite projective resolution would make these groups zero above its length, so

Choose a surjection $P_0=A^{r_0}\twoheadrightarrow M$. Since $A$ is a Noetherian ring, a finite free module is a Noetherian module, and its kernel $K_1$ is a finitely generated module. Choose a finite free surjection $P_1\twoheadrightarrow K_1$ and repeat. This constructs a resolution by finitely generated free moduleswhose terms are, in particular, finitely generated projective modules. It may be infinite; the request is finite generation of every term, not finite length.

Now suppose the nonzero finitely generated module $M$ has finite projective dimension $n$. For $n\ge 1$, its $n$th syzygy $K_n$ in this free resolution is projective. Indeed, dimension shifting along the successive kernel sequences givesfor every module $N$. The vanishing follows by computing the latter Ext functor from any length-$n$ projective resolution. Vanishing of ${\mathrm{Ext}}^1(K_n,-)$ makes every short exact sequence ending in $K_n$ split: apply the long exact sequence of Ext groups to a free surjection onto $K_n$ to lift its identity. Thus $K_n$ is a projective module. It is finitely generated by the preceding construction. Truncating giveswith every term finite projective.

Apply the Hom functor ${\mathrm{Hom}}_A(-,A)$ and write $P_i^{\ast }={\mathrm{Hom}}_A(P_i,A)$. At the top degree,If this group were zero, $d_n^{\ast }$ would be surjective. The dual module of a finite projective module is finite projective, so this surjection has a section $s:P_n^{\ast }\to P_{n-1}^{\ast }$. Finite projective modules identify with their double dual modules: the evaluation map is an isomorphism for a finite free module and passes to its direct summands. Dualizing $d_n^{\ast }s=\mathrm{id}$ therefore gives $s^{\ast }{d}_n={\mathrm{id}}_{P_n}$. Hence $d_n$ splits, and $P_{n-1}/d_n(P_n)$ is projective. Replacing the last two terms by this quotient shortens the projective resolution to length $n-1$, contradicting minimality. This proves top Ext detects finite projective dimension:For $n=0$, $M$ is a nonzero finite projective module and embeds as a summand of $A^r$. Some coordinate projection restricts nontrivially to $M$, proving ${\mathrm{Hom}}_A(M,A)\ne 0$ as required.

For infinite projective dimension, take the dual-number algebra $A=k[\epsilon ]/({\epsilon }^2)$ and $M=k=A/(\epsilon )$. Multiplication by $\epsilon$ has image and kernel both equal to $(\epsilon )$, giving the periodic resolution over dual numbersApplying ${\mathrm{Hom}}_A(-,k)$ makes every differential zero, so ${\mathrm{Ext}}_A^i(k,k)\cong k$ for every $i\ge 0$. A finite projective resolution would force all sufficiently high Ext functors to vanish. Thus ${\mathrm{pd}}_{k[\epsilon ]/({\epsilon }^2)}k=\infty$.

The finite-generation hypothesis is essential if the middle sentence's “a module” is interpreted as an unrestricted new module. A concrete vanishing top Ext without finite generation occurs for $A=k[[t]]$ and $K=A[1/t]=k((t))$. This Noetherian ring has the exact free presentationInjectivity of $d$ follows by examining the largest index of a finitely supported input, since $t$ is a non-zero-divisor. For exactness in the middle, use the relations to move a finite sum to its largest index $N$; its remaining coefficient is ${\sum }_{i=0}^N{a}_i{t}^{N-i}$, which is zero precisely when its image in $K$ is zero. The map onto $K$ is surjective because every Laurent series has a lower bound on its exponents.

Applying ${\mathrm{Hom}}_A(-,A)$ gives on products the map $(b_i)\mapsto (b_i-t{b}_{i+1})$. It is surjective: for any $(c_i)$, set $b_i={\sum }_{j\ge 0}{t}^j{c}_{i+j}$. These are well-defined formal power series, because each coefficient receives only finitely many contributions, and $b_i-t{b}_{i+1}=c_i$. Consequently ${\mathrm{Ext}}_A^1(K,A)=0$. But ${\mathrm{Hom}}_A(K,A)=0$: the image of every $t^{-i}$ must lie in ${\bigcap }_{r\ge 0}{t}^{r}A=0$, since it is divisible by every power of $t$. A nonzero projective module is a summand of a free module and therefore has a nonzero coordinate map to $A$. Thus $K$ is not projective, while its displayed resolution has length one. In particular,The nonvanishing conclusion proved above retains the finitely generated $M$ introduced at the start of the question.

Let $R$ be a commutative ring and $f=(f_1,\dots ,f_r)$. The Koszul complex packages these elements and their relations in a finite chain complex of free modules. With $E=R^r$ and basis $e_1,\dots ,e_r$, set $K_i(f;R)={\bigwedge }^{i}E$ for $0\le i\le r$, and defineDeleting two distinct basis elements in the two possible orders gives opposite signs and the same coefficient, so ${\partial }^2=0$. Equivalently $K(f;R)$ is the tensor product of chain complexes of $r$ two-term complexes $[R\stackrel{f_i}{\to }R]$ in degrees one and zero. The tensor differential is $\partial (u\otimes v)=\partial u\otimes v+(-1{)}^{\mathrm{deg}u}u\otimes \partial v$. This fixes the sign convention. On the exterior algebra, the differential is a graded derivation determined by $\partial e_i=f_i$.

For an $R$-module $M$, define the Koszul complex with module coefficients by $K(f;M)=K(f;R){\otimes }_{R}M$. Its degree-zero Koszul homology isIts higher Koszul homology measures the failure of these equations to form a regular sequence on a module. Each $f_j$ annihilates every homology group. Indeed the Koszul homotopy for multiplication by a generator is $h_j(u)=e_j\wedge u$, and the graded product rule givesFor coefficients in $M$ the same identity holds after tensoring. Thus multiplication by $f_j$ is zero on homology, and the homology groups are naturally modules over $R/(f)$. If $(f)=R$, choose ${\sum }_j{a}_j{f}_j=1$; then ${\sum }_j{a}_j{h}_j$ is a contracting homotopy, so the entire complex is contractible. These identities are also useful after localization of a ring: wherever one generator is a unit, the complex has zero homology.

The construction is functorial under a ring homomorphism, and base change gives $K(f;R){\otimes }_{R}S=K(\bar{f};S)$, since its terms are free with the displayed basis and differential. An invertible change of generators $f_i={\sum }_j{u}_{ij}{g}_j$ gives a chain isomorphism by sending $e_i$ to ${\sum }_j{u}_{ij}{e}_j$ and extending to exterior powers. In particular, permuting the generators changes only the exterior signs, not the isomorphism class of the complex.

The essential exactness theorem is that a regular sequence on a module gives zero positive Koszul homology. Recall that $f_i$ must act injectively on $M/(f_1,\dots ,f_{i-1})M$, with final quotient nonzero. Put $f^{\prime }=(f_1,\dots ,f_{r-1})$. Adding the final two-term complex identifies $K(f;M)$ with the mapping cone of multiplication by $f_r$ on $K(f^{\prime };M)$, with the chosen tensor signs. The long exact sequence in homology consequently has segmentsFor $r=1$, positive homology is exactly $\ker (f_1:M\to M)$. By induction the preceding homology vanishes above zero. The final injectivity assumption kills $H_1(f;M)$, while the exact sequence kills every $H_i$ for $i>1$. Thereforefor a regular sequence on a module. In particular $K(f;R)$ is a finite free resolution of $R/(f)$ when $f$ is a regular sequence, with ranks $\left(\genfrac{}{}{0ex}{}{r}{i}\right)$ and length $r$.

There is a precise converse under local finiteness assumptions. Suppose $(R,\mathfrak{m})$ is a Noetherian local ring, $0\ne M$ is finitely generated, and all $f_i\in \mathfrak{m}$. If positive Koszul homology vanishes, the same exact sequence makes multiplication by $f_r$ surjective on every $H_i(f^{\prime };M)$ for $i>0$. These modules are finitely generated because the ring is Noetherian and the terms of the complex are finite modules. The Nakayama lemma gives $H_i(f^{\prime };M)=0$. Induction makes $f^{\prime }$ regular on $M$, and the degree-one part of the exact sequence makes $f_r$ injective on $H_0(f^{\prime };M)$. The final quotient is nonzero, again by the Nakayama lemma. This proves the Koszul acyclicity criterion in a Noetherian local ring:The hypotheses matter: a unit ideal makes the complex contractible but cannot be a proper regular sequence.

Examples make both sides visible. For $R=k[x,y]$ and $f=(x,y)$, the Koszul resolution isThe signs agree with $\partial (e_1\wedge e_2)=x{e}_2-y{e}_1$. For $R=k[x]$ and $f=(x,x)$, the degree-one cycles are $R(1,-1)$ and the boundaries are $xR(1,-1)$; hence $H_1\cong R/(x)$, while $H_2=0$. The repeated element has exposed a relation that the first element already kills.

The Koszul complex also computes derived functors. If $f$ is a regular sequence on $R$, its free resolution givesHere no regularity of $f$ on $M$ is assumed. For $S=k[x_1,\dots ,x_r]$, tensor the Koszul resolution on the variables with $k$. Every differential becomes zero, soThe top group is nonzero, showing that a free resolution of $k$ cannot be shorter than $r$.

Finally, wedging complementary exterior degrees gives a perfect pairing $K_i{\otimes }_R{K}_{r-i}\to {\bigwedge }^r{R}^r\cong R$. It identifies the dual cochain complex with the degree-reversed Koszul complex, after the appropriate signs. This self-duality of the Koszul complex shows, for a regular sequence, thatThus the same explicit complex simultaneously records quotient equations, regularity, relations, Tor functor computations and Ext functor computations. Its finiteness and exterior structure are what make it especially effective in commutative algebra.

Set $S=R\setminus P$, a multiplicative subset. The localization at a prime ideal is $R_P=S^{-1}R$, whose elements are fractions $r/s$. Equality of two fractions means that $u(r{s}^{\prime }-r^{\prime }s)=0$ for some $u\in S$. Likewise the localization of a module is $M_P=S^{-1}M$, withAddition uses a common denominator, and the module action isThe equivalence relations make these operations well-defined, so $M_P$ is an $R_P$-module.

If $M\ne 0$, choose $0\ne m\in M$. Its annihilator is proper, so it lies in a maximal ideal $P$. The element $m/1$ cannot vanish in $M_P$: vanishing would mean $sm=0$ for some $s\notin P$, contrary to ${\mathrm{Ann}}_R(m)\subseteq P$. The converse is immediate by localizing the zero module. ThusThis proof of localization detects zero elements actually applies to arbitrary modules, without finite-generation or Noetherian assumptions. We will use that extra generality for an Ext functor module below.

An injective module $E$ has the extension property: for each inclusion $A\hookrightarrow B$, every map $A\to E$ extends to a map $B\to E$. Equivalently, ${\mathrm{Hom}}_R(-,E)$ is exact. A projective module $Q$ has the lifting property: for each surjection $B\twoheadrightarrow C$, every map $Q\to C$ lifts to $Q\to B$. Equivalently, ${\mathrm{Hom}}_R(Q,-)$ is exact, or $Q$ is a direct summand of a free module.

For the local criterion for injectivity over a Noetherian ring, recall the Baer criterion: $E$ is injective precisely when every map from an ideal $J\subseteq R$ into $E$ extends to $R$. Through the short exact sequence $0\to J\to R\to R/J\to 0$, this is equivalent toWe also need localization of Ext over a Noetherian ring. Because $R$ is Noetherian, the module $R/J$ has a free resolution with every term finitely generated: all successive kernels are finitely generated, so this can be built recursively. For a finite free term $F$, the natural mapis an isomorphism, as is clear from a finite basis. Exactness of localization lets us pass to cohomology of the Hom complex. ThereforeThis explains the finiteness hypothesis needed for the localization argument, rather than assuming that localization preserves injectivity automatically.

If $M$ is injective, the left-hand side vanishes for every $J$ and $P$. Every ideal $L$ of $R_P$ is $J{R}_P$, where $J$ is its contraction to $R$: if $r/s\in L$, then $r/1=s(r/s)\in L$, and conversely localization of a member of the contraction stays in $L$. Hence all ideal tests for $M_P$ vanish, and the Baer criterion over $R_P$ makes $M_P$ injective.

Conversely, suppose every $M_P$ is injective. For each ideal $J$, the displayed Ext functor localization is zero at every prime. The zero-detection argument above gives ${\mathrm{Ext}}_R^1(R/J,M)=0$, without needing this Ext module to be finitely generated. Applying the Baer criterion over $R$ provesIn fact, under the Noetherian ring hypothesis this equivalence holds for arbitrary $M$; the printed finite-generation assumption is more than is needed.

The global dimension iswhere projective dimension is the smallest length of a projective resolution, or infinity if there is no finite one. If $\mathrm{gldim}R=0$, every module is projective. Given an inclusion $A\hookrightarrow B$, the quotient $B/A$ is then projective, so the exact sequence $0\to A\to B\to B/A\to 0$ splits. A retraction $\rho :B\to A$ exists. For any module $E$ and any map $f:A\to E$, the composition $f\rho :B\to E$ extends $f$. Thus every module is also injective. Global dimension zero makes all modules both projective and injective, as recorded by global dimension zero and split exact sequences.

Choose a projective presentation $0\to K\stackrel{j}{\to }P\stackrel{\pi }{\to }C\to 0$, with $P$ a projective module. Continuing it to a projective resolution showsIndeed a degree-one cocycle in the Hom functor applied to the projective resolution descends to $K$, while the degree-one coboundaries are exactly restrictions of maps from $P$. This establishes the description of the Ext functor without first assuming the extension classification.

For a module extension with inclusion $i:A\to B$ and quotient map $r:B\to C$, projectivity lifts $\pi$ to $s:P\to B$. Since $rsj=0$, there is a unique map $h:K\to A$ with $ih=sj$. A different lift changes $h$ by $\ell j$ for some $\ell :P\to A$. An equivalence of module extensions, which is the identity on both end modules, also preserves this class. We have therefore defined a map from extension classes to ${\mathrm{Ext}}_R^1(C,A)$.

Conversely, for $h:K\to A$, form the pushout of a module extensionThe map $A\to B_h$ sends $a$ to $[a,0]$, and $B_h\to C$ sends $[a,p]$ to $\pi (p)$. The first is injective because $j$ is injective. If $\pi (p)=0$, write $p=j(k)$; then $[a,p]=[a+h(k),0]$, proving exactness in the middle. The last map is surjective. Thus $B_h$ is a module extension. If $h^{\prime }=h+\ell j$, the map $[a,p]\mapsto [a-\ell (p),p]$ is an equivalence of module extensions from $B_h$ to $B_{h^{\prime }}$. Finally $[a,p]\mapsto i(a)+s(p)$ identifies the pushout of a module extension built from an existing extension with its middle module. These two constructions are inverse, proving the classification:The zero class corresponds to a split short exact sequence. Fixed end modules matter: equivalence does not permit arbitrary automorphisms of $A$ or $C$.

In the second calculation the acting group is the infinite cyclic group. Its modules are modules over the group ring $S=\mathbb{Z}[z,z^{-1}]$, not merely over the underlying ring $\mathbb{Z}$. The trivial representation has $z$ acting as the identity. Its projective resolution isMultiplication by $z-1$ is injective, and the augmentation quotient is $\mathbb{Z}$. Applying the Hom functor into the trivial module makes the differential zero, so ${\mathrm{Ext}}_S^1(\mathbb{Z},\mathbb{Z})\cong \mathbb{Z}$. Taking two copies gives $\boxed{\mathcal{E}({\mathbb{Z}}^2,\mathbb{Z})\cong {\mathbb{Z}}^2}$.

An explicit representative of this extension of trivial modules for an infinite cyclic group is the abelian group $B_{m,n}=\mathbb{Z}u\oplus \mathbb{Z}v\oplus \mathbb{Z}w$, with injection $1\mapsto u$, quotient $v\mapsto (1,0)$, $w\mapsto (0,1)$, and actionThis is an invertible action: the inverse subtracts the same multiples of $u$. Every underlying abelian-group extension splits because ${\mathbb{Z}}^2$ is free, so any group-module extension has this form after choosing lifts of the quotient basis. Replacing those lifts by multiples of $u$ does not alter $m,n$. An equivalence fixing the ends has exactly such changes of lifts, so two representatives are equivalent precisely when their ordered pairs agree. Only $(0,0)$ splits as a group-module extension; overlooking the group action would incorrectly give a single class.

For central $t_1,\dots ,t_n$, define the Koszul complex on central ring elements using formal exterior symbols:with zero terms outside $0\le k\le n$ and differentialThe terms are free bimodules with the formal symbols commuting with coefficients. Centrality makes $d$ a bimodule map, and the terms in $d^2$ cancel in pairs because the $t_i$ commute. This defines the Koszul complex over a possibly noncommutative ring without using an undefined exterior algebra of arbitrary one-sided modules. In particular $K(\mathbf{t}){\otimes }_{R}M$ is a well-defined chain complex of left modules.

A regular sequence on a module means that multiplication by $t_i$ is injective on $M/(t_1,\dots ,t_{i-1})M$ for each $i$, usually with the additional convention that the final quotient is nonzero. The homology vanishing below uses only the injectivity conditions. For one element, the chain complex is $M\stackrel{t_1}{\to }M$, with zero first homology and zeroth homology $M/t_{1}M$.

For the induction let $C=K(t_1,\dots ,t_{n-1}){\otimes }_{R}M$. Adjoining the last generator identifies the new Koszul complex with the mapping cone of multiplication by $t_n$ on $C$. In the convention $\mathrm{Cone}(t_n{)}_k=C_k\oplus C_{k-1}$ its differential is $(a,b)\mapsto (da+t_{n}b,-db)$; the identification is $a+e_n\wedge b$. The long exact sequence in homology of this mapping cone givesThe induction hypothesis is $H_k(C)=0$ for $k>0$ and $H_0(C)=N=M/(t_1,\dots ,t_{n-1})M$. Regularity makes $t_n:N\to N$ injective. Hence the new positive homology is zero and its zeroth homology is $N/t_{n}N$. The augmentation to this quotient induces these homology isomorphisms, giving the quasi-isomorphismwhere the right side is placed in degree zero. The results used are the long exact sequence in homology of a degreewise short exact sequence of complexes and its mapping cone form, together with the stated induction; no unmentioned acyclicity criterion is needed.

For $R=\mathbb{Z}[X]$, $(p,X)$ is a regular sequence: $p$ is a non-zero-divisor in $R$, and multiplication by $X$ is injective in $(\mathbb{Z}/p\mathbb{Z})[X]$, even when $p$ is composite. The Koszul resolution of $R/(p,X)$ isWith $N=\mathbb{Z}/q\mathbb{Z}$ and $X$ acting as zero, applying the Hom functor gives the cochain complexThus, more generally, the Ext groups between polynomial-ring residue modules areand vanish in all degrees above two. For the equal-modulus case,These are $R$-modules through $X=0$ and reduction modulo $p$. If the multiplicative self-Ext functor is desired, the Koszul self-Ext algebra is the exterior algebra on two degree-one generators over $\mathbb{Z}/p\mathbb{Z}$. The two contractions on the Koszul resolution lift these classes, square to zero and anticommute, so this description also holds for composite $p$ and characteristic two. For coprime $p,q$, multiplication by $p$ is invertible on $N$, so both its kernel and cokernel vanish:

Choose the rational point $x=[1:0:\cdots :0]$ and the coherent sheaf $\mathcal{F}=i_{\ast }k$ supported there. By skyscraper sheaf cohomology, $H^0(X,\mathcal{F})=k$.

Its dual of a sheaf ${\mathcal{F}}^{\vee }=\mathcal{H}om(\mathcal{F},{\mathcal{O}}_X)$ is zero. Away from $x$ its source is zero. At $x$, a homomorphism $k\to {\mathcal{O}}_{X,x}$ must send one to an element annihilated by the maximal ideal. Since $n>0$, that ideal contains a nonzero coordinate parameter; the local ring is an integral domain, so the image must vanish. This proves that every local homomorphism vanishes. ConsequentlyThere is no contradiction with Serre duality: the ordinary sheaf dual suffices for locally free sheaves, but general coherent sheaves require an Ext functor. This example illustrates failure of ordinary sheaf-dual Serre duality for a skyscraper sheaf.

Put $A^e=A{\otimes }_k{A}^{\mathrm{op}}$, so an $A$-bimodule is a left $A^e$-module through $(a\otimes b^{\mathrm{op}})m=amb$. The Hochschild cohomology isEquivalently, the Hochschild cochain complex has $C^n(A,M)={\mathrm{Hom}}_k(A^{\otimes n},M)$ and coboundary mapIts cohomology agrees with the displayed Ext functor because the bar resolution of an associative algebra is free over $A^e$ when $k$ is a field. The Hochschild cohomological dimension isThe supremum uses all bimodules, not merely the finitely generated one supplied in the question; an unbounded projective dimension is infinity.

An extension in this classification is a square-zero extension of an algebra: a short exact sequence $0\to M\to E\to A\to 0$, where $E$ and $A$ are unital algebras, $E\to A$ is unital, and $M$ is its two-sided ideal with $M^2=0$ and induced $A$-bimodule structure equal to the prescribed one. An equivalence of extensions is an algebra isomorphism of the middle terms commuting with the maps and inducing the identity on both $A$ and $M$. Arbitrary isomorphisms of middle algebras, or extensions without the square-zero ideal requirement, are not classified by this cohomology group.

Choose a $k$-linear map section $s:A\to E$ with $s(1)=1$. Its multiplication defectis a normalized Hochschild cocycle: $\mu (1,a)=\mu (a,1)=0$, and associativity in $E$ gives $\delta \mu =0$. Changing $s$ to $s+g$, where $g(1)=0$, changes the defect to $\mu +\delta g$, since $M^2=0$.

Conversely, for a normalized Hochschild cocycle $\mu$, put $E_{\mu }=A\oplus M$ as a vector space and defineThe Hochschild cocycle equation is exactly associativity, and $(1,0)$ is the identity. If ${\mu }^{\prime }=\mu +\delta g$, the map $E_{{\mu }^{\prime }}\to E_{\mu }$, $(a,m)\mapsto (a,m+g(a))$, is an equivalence. Conversely, every equivalence has this form after choosing sections. The normalized Hochschild cochain complex computes the same cohomology as the full complex: in the bar resolution of an associative algebra, the degenerate terms containing an inserted identity form a contractible subcomplex. Passing to the normalized bar resolution of an associative algebra, then applying the Hom functor, gives the same cohomology. Thus every class has a normalized representative. We obtain a bijection between $H{H}^2(A,M)$ and equivalence classes of square-zero extensions.