Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 2 (Commutative Algebra)

Put $K=\mathrm{Frac}(A)$ and let $B$ be the integral closure of $A$ in $L$. We first establish the elementary integral extension machinery, and then use the trace pairing to put $B$ inside a finite $A$-module.

An integral element $x$ satisfies a monic polynomial over the base ring. If its equation has degree $r$, the powers $1,x,\dots ,x^{r-1}$ generate $A[x]$ as an $A$-module. More generally, adjoining finitely many integral elements gives a finitely generated module: reduce the exponent of each generator using its monic equation. Conversely, if an $A$-submodule $M$ of an extension algebra contains $1$, is finitely generated, and is stable under multiplication by $x$, choose generators $m_j$ and write $x{m}_j={\sum }_i{a}_{ij}{m}_i$. Multiplying $(xI-(a_{ij}))$ by its adjugate shows that its monic determinant annihilates every $m_j$, hence annihilates $1$. This finite-module criterion for integrality is the determinant trick.

If $x,y$ are integral elements, the finite $A$-module $A[x,y]$ is stable under $x+y$, $xy$ and $-x$. The criterion proves that these are integral elements. Thus the integral closure really is a subring. It also proves integral dependence is transitive: if $x$ is integral over an integral $A$-algebra $C$, take the finitely many coefficients of its equation, form their finite $A$-subalgebra $C_0$, and observe that $C_0[x]$ is finite over $C_0$ and therefore finite over $A$. The criterion applies to $x$. In particular, an algebra generated by finitely many integral elements is a finite integral extension, even though an arbitrary integral extension need not be finite.

Choose a $K$-basis $v_1,\dots ,v_d$ of $L$. Each basis element is algebraic over $K$. Ifchoose a nonzero $a_i\in A$ clearing all coefficient denominators. Then $e_i=a_i{v}_i$ satisfies a monic equation with coefficients $a_i^j{c}_{ij}\in A$. Hence all $e_i$ belong to $B$ and still form a $K$-basis. This is integral field basis by denominator clearing.

We next prove the trace of an integral element over a normal domain property. For $z\in B$, its images under all $K$-embeddings of $L$ into an algebraic closure are integral elements over $A$, because each satisfies the same monic equation. Their sum is integral by the subring property just proved. Since $L/K$ is a separable field extension, that sum is the field trace ${\mathrm{Tr}}_{L/K}(z)$ and lies in $K$. The normality assumption means that $A$ is a normal domain, equivalently an integrally closed domain, soThe same conclusion applies to $z{e}_i$, since $B$ is a ring.

The trace pairing of a finite separable field extension is nondegenerate. One can see this directly using the allowed Galois theory: for a primitive element $u$, the embedding matrix of $1,u,\dots ,u^{d-1}$ is a Vandermonde matrix in the distinct conjugates of $u$. Its determinant is nonzero, and the trace Gram matrix is its transpose times itself. Nondegeneracy is unchanged by a change of basis.

Let $M={\sum }_{i}A{e}_i$. Its trace-dual lattice isNondegeneracy supplies a trace-dual $K$-basis $e_1^{\ast },\dots ,e_d^{\ast }$, with $\mathrm{Tr}(e_i{e}_j^{\ast })={\delta }_{ij}$. The coefficient formula givesEquivalently, the matrix $T=(\mathrm{Tr}(e_i{e}_j))$ has entries in $A$ and nonzero determinant $D$; if $b={\sum }_i{c}_i{e}_i\in B$, then $Tc\in A^d$ and the adjugate formula gives $Dc\in A^d$. Thus $B\subseteq D^{-1}M$ as well. Both descriptions exhibit a finite free ambient $A$-module.

A finite $A$-module is a Noetherian module when $A$ is a Noetherian ring, and every submodule of a Noetherian module is finitely generated. Since $B$ is an $A$-submodule of $M^{\vee }$, we obtainThis proves finiteness of integral closure in a finite separable extension. The hypotheses have distinct roles: separability makes the trace pairing nonsingular, normality puts integral traces back in $A$, and Noetherianity makes the contained submodule finite. Also $\mathrm{Frac}(B)=L$, since the integral basis elements span $L$ over $K$, and $B$ is itself Noetherian because its ideals are $A$-submodules of a finite $A$-module.

The Noether normalization lemma is valid over every field $k$, including finite fields: a nonzero finitely generated algebra $C$ contains elements $y_1,\dots ,y_d$ with algebraic independence over $k$ such that $C$ is finite as a module over the polynomial ring $k[y_1,\dots ,y_d]$. We prove it without an infinite-field hypothesis.

Write $C=k[a_1,\dots ,a_n]$ and induct on $n$. If the generators have algebraic independence, they already give a polynomial ring and the assertion is immediate. Otherwise choose a nonzero relation $F(a_1,\dots ,a_n)=0$. Choose an integer $M$ larger than every exponent in the monomials of $F$, and setIn $F(Y_1+T^M,Y_2+T^{M^2},\dots ,Y_{n-1}+T^{M^{n-1}},T)$, the largest power of $T$ has a nonzero coefficient in $k$. Indeed, a monomial with exponents $(e_1,\dots ,e_n)$ contributes top weight $e_n+{\sum }_{i<n}{e}_i{M}^i$. These weights are distinct by uniqueness of base-$M$ expansion, so only one monomial contributes the highest power. After rescaling, the substituted relation is monic in $T$.

Consequently $a_n$ is an integral element over $C^{\prime }=k[b_1,\dots ,b_{n-1}]$, and $C=C^{\prime }[a_n]$ is finite over $C^{\prime }$. Apply the induction hypothesis to $C^{\prime }$ and compose the finite module extensions. This proves Noether normalization by weighted substitutions. The induction reaches $n=0$, where $C=k$. For an integral domain $C$, taking fraction fields makes the resulting extension finite algebraic, so $d$ is the transcendence degree of $\mathrm{Frac}(C)/k$. More generally $d=\dim C$: integral extensions preserve Krull dimension, and a polynomial algebra in $d$ variables has Krull dimension $d$.

A useful bridge to the Hilbert Nullstellensatz is the Zariski lemma. If a field $E$ is a finitely generated algebra over $k$, normalization makes it finite and integral over a polynomial ring $P=k[y_1,\dots ,y_d]$. A subring over which a field is integral is a field: for nonzero $a\in P$, an integral equation for $a^{-1}\in E$, multiplied by $a^{r-1}$, expresses $a^{-1}$ as an element of $P$. Therefore $P$ must be a field. A polynomial ring in a positive number of variables is not a field, since a variable has no polynomial inverse. Hence $d=0$ and $E/k$ is a finite field extension. This proves the Zariski lemma.

Now let $k$ be an algebraically closed field. The Weak Hilbert Nullstellensatz says that every maximal ideal of $k[X_1,\dots ,X_n]$ is uniquely of the formTo prove it, the residue field $k[X]/\mathfrak{m}$ is a field generated as a $k$-algebra by the images of the variables. The Zariski lemma makes it finite algebraic over $k$, and algebraic closedness makes it $k$. Thus each $X_i$ has an image $a_i\in k$. The evaluation map has kernel the displayed ideal: subtracting the constant value of a polynomial expresses its difference as a combination of $X_i-a_i$. That kernel is maximal and contained in $\mathfrak{m}$, so equality holds. Conversely every evaluation kernel is maximal because its quotient is $k$. Uniqueness follows from the variable images. Every proper ideal is contained in a maximal ideal, so it has a common zero; equivalently, an ideal with no common zero is the whole ring.

For an ideal $I\subseteq k[X_1,\dots ,X_n]$, let $V(I)$ be its common-zero set and let $I(V(I))$ be all polynomials vanishing on that set. The Strong Hilbert Nullstellensatz statesThe inclusion $\sqrt{I}\subseteq I(V(I))$ follows because a field has no nonzero nilpotent elements. For the other inclusion, take $f$ vanishing on $V(I)$, with $f\ne 0$, and form the Rabinowitsch trick idealIt has no common zero: at a zero of $I$ the second generator has value one. The Weak Hilbert Nullstellensatz in $n+1$ variables gives $J=(1)$. Hence a finite identity has the form $1={\sum }_j{h}_j(X,T)g_j(X)+h(X,T)(1-Tf)$, with $g_j\in I$. Substitute $T=f^{-1}$ in the localization of a ring $k[X{]}_f$. Clearing the finitely many powers of $f$ occurring in denominators gives $f^r\in I$ for some $r$, so $f\in \sqrt{I}$. The case $f=0$ is immediate. This completes all three proofs. The algebraically closed hypothesis belongs to the two forms of the Hilbert Nullstellensatz; it was not needed for the Noether normalization lemma or the Zariski lemma.

All rings below are nonzero and commutative with identity; the zero ring is trivially Artinian. A primary ideal is proper. An ideal $Q$ is primary precisely when every zero divisor of $R/Q$ is a nilpotent element. Its radical of an ideal $\mathfrak{p}=\sqrt{Q}$ is prime: if $ab\in \sqrt{Q}$ and $a\notin \sqrt{Q}$, apply the primary property to $(ab{)}^r\in Q$ to obtain a power of $b$ in $Q$. In a Noetherian ring, finite generation of $\mathfrak{p}$ gives ${\mathfrak{p}}^s\subseteq Q$ for some $s$. For generators $u_i$ with $u_i^{s_i}\in Q$, one can take $s=1+{\sum }_i(s_i-1)$.

We first prove existence of primary decomposition. The ascending chain condition implies that every proper ideal is a finite intersection of irreducible ideals. Otherwise choose a maximal counterexample $I$ under inclusion. It is not irreducible, so $I=J\cap K$ with both $J,K$ strictly larger; their finite irreducible decompositions give one for $I$, a contradiction.

An irreducible ideal in a Noetherian ring is primary. Pass to the quotient and suppose zero is irreducible. If $xy=0$ with $x\ne 0$, the ascending chain of annihilators of $y^j$ stabilizes, say at $j=r$. Any element of $(x)\cap (y^r)$ can be written $z=ax=b{y}^r$. Then $yz=0$, so $b\in \mathrm{Ann}(y^{r+1})=\mathrm{Ann}(y^r)$ and $z=0$. Irreducibility of zero implies $(y^r)=0$, since $(x)\ne 0$. Thus $y$ is nilpotent. This proves irreducible ideals are primary in Noetherian rings, and hence the Lasker–Noether theorem.

An intersection of finitely many primary ideals with the same radical $\mathfrak{p}$ is again primary: if $ab$ belongs to all of them and $b\notin \mathfrak{p}$, the primary property forces $a$ into every component. Combining such components and removing redundant ones yields a minimal primary decompositionThe radicals are uniquely determined, but the components need not all be unique. Here is a proof identifying the invariant radicals as the associated primes of a module $R/I$.

Every nonzero module over a Noetherian ring has a nonzero element with a prime annihilator: maximize the annihilator of a nonzero element using the ascending chain condition. If $abm=0$ and $bm\ne 0$, maximality gives $\mathrm{Ann}(bm)=\mathrm{Ann}(m)$ and hence $am=0$. This is the maximal annihilator of a module element is prime argument. For a $\mathfrak{p}$-primary quotient $R/Q$, the radical of the annihilator of every nonzero element is $\mathfrak{p}$: it is contained in $\mathfrak{p}$ by the primary property and contains a power of $\mathfrak{p}$ because ${\mathfrak{p}}^s\subseteq Q$. Thus the only possible associated prime is $\mathfrak{p}$, and it does occur.

The diagonal injection $R/I\hookrightarrow {⨁}_{i}R/Q_i$ shows that every associated prime of $R/I$ is one of the ${\mathfrak{p}}_i$. Indeed, for an element whose annihilator $P$ is prime, that annihilator is the intersection of the finitely many component annihilators, all containing $P$. Their product is contained in $P$, so primality forces one of them to equal $P$; its component element is nonzero and has associated prime ${\mathfrak{p}}_i$. Conversely, irredundancy gives $b\in {\bigcap }_{j\ne i}{Q}_j\setminus Q_i$. The nonzero cyclic submodule generated by $b+I$ embeds in $R/Q_i$. It has an associated prime, necessarily ${\mathfrak{p}}_i$, which is then an associated prime of $R/I$. We have proved the first uniqueness theorem for primary decomposition:Also the zero divisors on $R/I$ are exactly ${\bigcup }_i{\mathfrak{p}}_i$. For a scalar killing a nonzero element, extend its element annihilator to a maximal element annihilator containing it; the preceding argument gives an associated prime containing that scalar. The converse is immediate from the definition of an associated prime.

The minimal members of the set $\{{\mathfrak{p}}_i\}$ are the isolated primes of a primary decomposition. They are exactly the primes minimal over $I$: if a prime contains ${\bigcap }_i{\mathfrak{p}}_i$, it contains one ${\mathfrak{p}}_i$ by the product argument. A component belonging to an isolated prime is unique. Localize at ${\mathfrak{p}}_i$. Every other component becomes the whole ring, since its radical contains an element outside ${\mathfrak{p}}_i$ whose power lies in that component. A ${\mathfrak{p}}_i$-primary ideal contracts unchanged from this localization, because $sa\in Q_i$ with $s\notin {\mathfrak{p}}_i$ implies $a\in Q_i$. Hence the second uniqueness theorem for primary decomposition givesAn embedded primary component can vary. For example, in $k[x,y]$,Both second components are $(x,y)$-primary, since their quotients are dual-number algebras. For the second equality, reducing modulo $(x^2,y+x)$ makes $y=-x$; an element $xg$ vanishes exactly when $g$ has zero constant term, giving $(x^2,xy)$. The isolated component is $(x)$ and the embedded associated prime is $(x,y)$.

The relevance to Artinian rings is particularly sharp in Krull dimension zero. Suppose $R$ is Noetherian and all its primes are maximal. Apply a minimal primary decomposition to zero. Its radicals are distinct maximal ideals, so the components are pairwise comaximal ideals: each contains a power of its radical, and expanding $1=(a+b{)}^N$ for $a+b=1$ shows that powers of comaximal ideals remain comaximal. The Chinese remainder theorem gives the Artinian decomposition into local factorsEach factor has one maximal ideal, and that ideal is nilpotent. Its finite radical filtration has successive layers finitely generated over its residue field, hence of finite vector-space dimension. It follows that the factor, and therefore $R$, has a finite composition series of a module, so $R$ is Artinian. This proves the Noetherian dimension-zero criterion for an Artinian ring in this direction. Applied to $R/I$, it says that the quotient is Artinian exactly when all primes over $I$ are maximal. In that case no primary component is embedded, so all components are unique.

For completeness, the converse does not require initially assuming Noetherianity. In an Artinian ring, an Artinian domain is a field: stabilization of $(a^j)$ and cancellation gives an inverse to every nonzero $a$. Thus every prime is maximal. There are finitely many maximal ideals, since infinitely many distinct ones would give strictly descending finite intersections; comaximality guarantees strictness by the Chinese remainder theorem. Its nilradical $N$ is nilpotent. Indeed, its powers stabilize at $J=N^r$, with $J^2=J$. If $J\ne 0$, choose an ideal $C$ minimal among those satisfying $JC\ne 0$. Then $JC=C$, since $J(JC)=JC\ne 0$. Choose $x\in C$ with $Jx\ne 0$; minimality also gives $Rx=C$. Therefore $x=ax$ for some $a\in J\subseteq N$. But $1-a$ is a unit, contradicting $x\ne 0$. Hence $J=0$.

Now $R/N$ is a finite product of residue fields. Each $N^j/N^{j+1}$ is an Artinian module over that product, and therefore a finite direct sum of finite-dimensional vector spaces: an infinite-dimensional vector space would have a strictly descending chain of subspaces. The finite filtration by powers of $N$ makes $R$ a module of finite composition length, in particular Noetherian. Combining both directions yieldsThus primary decomposition separates the local pieces of a zero-dimensional ring, while their nilpotent maximal ideals record the multiplicities that the reduced set of primes alone does not detect.

Let $k\to A$ be a map of commutative rings. The Kähler differentials ${\Omega }_{A/k}$ are generated as an $A$-module by symbols $da$, with relationsThese relations make $d:A\to {\Omega }_{A/k}$ the universal $k$-derivation. Every $k$-derivation $D:A\to M$ into an $A$-module factors uniquely through the map $da\mapsto D(a)$. Thus the universal property of Kähler differentials isThis constructs the module and proves its universal property, rather than merely listing a formal derivative rule.

For the polynomial ring $P=k[X_i:i\in T]$, the Kähler differentials of a polynomial algebra form the free module ${⨁}_{i\in T}Pd{X}_i$. The usual formal partial derivatives prove that assigning arbitrary images to $d{X}_i$ defines a derivation, and every polynomial involves only finitely many variables. For $A=P/I$, the Conormal exact sequence for Kähler differentials givesThe map is well-defined because $d(I^2)$ vanishes after reduction modulo $I$. Its cokernel has exactly the universal property of derivations on $P$ that kill $I$, so is ${\Omega }_{A/k}$. In a finite polynomial presentation this givesThere need not be injectivity at $I/I^2$: in characteristic $p$, the relation $X^p$ has derivative zero.

Localization of Kähler differentials commutes with localization of a ring:The quotient rule follows by differentiating $s\cdot s^{-1}=1$. It extends every derivation uniquely and proves the isomorphism by the universal property. Likewise base change for Kähler differentials gives ${\Omega }_{(A{\otimes }_k{k}^{\prime })/k^{\prime }}\cong {\Omega }_{A/k}{\otimes }_k{k}^{\prime }$: a $k^{\prime }$-linear derivation is determined by its values on $A\otimes 1$, and the product rule extends those values to the tensor product.

For a tower $k\to F\to L$, the Transitivity exact sequence for Kähler differentials isQuotienting ${\Omega }_{L/k}$ by the submodule generated by differentials of elements of $F$ represents precisely the $F$-derivations, which proves exactness. The first map need not be injective in general. To relate this to a transcendence basis, we need the stronger property supplied by a separable field extension.

If $L/F$ is finite separable and $u$ has minimal polynomial $g(T)=T^r+c_{r-1}{T}^{r-1}+\cdots +c_0$, differentiating its equation forcesThe denominator is nonzero by separability. Conversely this formula extends an arbitrary $k$-derivation $F\to M$, for an $L$-module $M$, to $F[u]$; it kills the relation $g$ and hence descends to $L$. A tower of simple separable extensions proves unique extension for all finite separable $L/F$. Therefore Kähler differentials under a separable field extension satisfyThis proves injectivity in this case, which would not follow from right exactness alone.

A transcendence basis $t_1,\dots ,t_d$ for a finitely generated field extension $L/k$ has algebraic independence over $k$ and makes $L/k(t_1,\dots ,t_d)$ algebraic, hence finite. A separating transcendence basis additionally makes that finite extension separable. Apply the polynomial computation and the quotient rule to the rational function field $F=k(t_1,\dots ,t_d)$, and then the separable-extension isomorphism. We obtain the central link:when the basis is separating. The dual statement says that any prescribed values of the $D(t_i)$ extend uniquely to a $k$-derivation of $L$ with values in $L$. In characteristic zero every transcendence basis of a finitely generated field extension is separating, so Kähler differentials measure transcendence degree exactly.

There is also a characteristic-zero test for algebraic independence. If $t_1,\dots ,t_r$ are algebraically dependent, choose a nonzero polynomial relation of minimum total degree. Some formal partial derivative is nonzero in characteristic zero, and it cannot also vanish on the tuple, since it has smaller degree. Differentiation therefore gives a nontrivial linear relation among the $d{t}_i$. Conversely, an algebraically independent tuple extends to a transcendence basis, whose differentials form a basis as just proved. Thus differentials detect algebraic independence in characteristic zero: a finite tuple is algebraically independent exactly when its differentials are linearly independent. Its differentials form a basis of ${\Omega }_{L/k}$ exactly when the tuple is a transcendence basis.

The separability qualification is essential in positive characteristic. For $k={\mathbb{F}}_p(s)$ and $L=k(u)$ with $u^p=s$, the extension is purely inseparable of degree $p$, with transcendence degree zero. The polynomial presentation yields ${\Omega }_{L/k}=Ldu\ne 0$, because the defining relation has derivative zero. Also, in ${\mathbb{F}}_p(t)/{\mathbb{F}}_p$, the tuple $\{t^p\}$ is a transcendence basis but $d(t^p)=0$, while $dt$ is a basis of the one-dimensional differential module. A chosen arbitrary transcendence basis therefore need not give differential coordinates in characteristic $p$.

Finally, the presentation makes the relation useful geometrically. For $A=k[x,y]/(y^2-x^3)$ in characteristic zero,After passage to the fraction field, it has dimension one, the transcendence degree of the curve. At the origin, tensoring with its residue field leaves both $dx,dy$ independent, so the differential fibre has dimension two. For a $k$-rational point with maximal ideal $\mathfrak{m}$, that fibre is the cotangent space of a local ring $\mathfrak{m}/{\mathfrak{m}}^2$: write elements as their constant value plus an element of $\mathfrak{m}$, and note that derivations into $k$ kill ${\mathfrak{m}}^2$. This explains how Kähler differentials record both generic transcendental parameters and the extra tangent direction at a singular point.

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.