Koszul Complexes and Regular Sequences

Sort wedges using anticommutation and delete repetitions using $e_i\wedge e_i=0$, so the increasing wedges span.

The alternating determinant map to the free module on $p$-subsets kills the exterior relations and sends these wedges to distinct basis elements, proving independence.

Move $e_i$ across precisely the factors indexed below $i$; each adjacent swap contributes $-1$.

If $i$ already occurs, the wedge contains a square and is zero; this remains true in characteristic $2$.

Apply the graded Leibniz rule to the ordered wedge and use $d(e_{i_j})=x_{i_j}$.

Passing the differential through $j-1$ degree-one factors gives $(-1{)}^{j-1}$, yielding the formula.

Fix two deleted positions $a<b$. Deleting $a$ then $b$ has sign $(-1{)}^{a-1}(-1{)}^{b-2}$, while the reverse order has sign $(-1{)}^{b-1}(-1{)}^{a-1}$; these differ by $-1$ and have the same scalar $x_{i_a}{x}_{i_b}$.

Every summand of $d^2$ has a unique unordered pair of deleted indices, so the pairs cancel, including in characteristic $2$ where $-1=1$ and the two equal terms add to $0$.

The graded derivation sends $e_i\wedge e_i$ to $x_i{e}_i-e_i{x}_i=0$ and therefore respects the alternating quotient. Its coordinate action is the stated deletion formula.

The pairwise cancellation calculation proves $d^2=0$, so the graded modules and this differential satisfy the chain-complex axioms.

For the zero free module, only ${\bigwedge }^{0}0=R$ is nonzero.

Tensoring with $M$ gives $M$ in degree zero and zero elsewhere, with zero differential.

For one basis element, the only exterior powers are $Re$ and $R$.

The defining differential sends $e\otimes m$ to $xm$, producing the displayed two-term complex.

The sole differential is multiplication by $x$, whose cokernel is $M\mathrm{/}xM$.

Its kernel is $(0{:}_{M}x)$ and no other chain groups occur.

The image in degree zero is $(x_{1}M+\cdots +x_{n}M)$, so $H_0=M\mathrm{/}(\mathbf{x})M$.

There are no degrees above $n$, and the top differential kills $m$ exactly when every $x_i$ kills $m$.

The exterior algebra of the direct sum of the two based free modules is the graded tensor product of their exterior algebras.

Its total differential $d\otimes 1+(-1{)}^{\mathrm{deg}}1\otimes d$ agrees termwise with the Koszul deletion differential.

Split ${\bigwedge }^p(R^n\oplus Re)$ as ${\bigwedge }^p{R}^n\oplus e\wedge {\bigwedge }^{p-1}{R}^n$. Send the second summand to the shifted cone summand.

The coordinate formula gives $d(e\wedge z)=yz-e\wedge dz$, exactly the cone differential; hence the splitting is a chain isomorphism.

The cone identification gives the degreewise split short exact sequence $0\to K(\mathbf{x};M)\to K(\mathbf{x},y;M)\to K(\mathbf{x};M)[-1]\to 0$.

Its connecting map is multiplication by $y$ (check it on a cycle in the shifted summand), so the long exact homology sequence is the displayed one.

The direct-sum exterior identification supplies the signed concatenation chain isomorphism.

Splitting off the final basis vector gives its multiplication cone, whose long exact sequence gives the stated package.

Let $h_i(z)=e_i\wedge z$. The graded Leibniz rule gives $d{h}_i(z)+h_{i}d(z)=d(e_i)z=x_{i}z$.

Thus $h_i$ is a chain homotopy from multiplication by $x_i$ to zero, with no division or characteristic assumption.

For each generator, the contraction identity is $d{h}_i+h_{i}d=x_i\mathrm{id}$.

Thus each $x_i$ acts trivially on homology, hence so does every element of $(\mathbf{x})$.

Exterior multiplication by $e_i$ satisfies $d(e_i\wedge -)+(e_i\wedge -)d=x_i\mathrm{id}$.

This is a chain homotopy from multiplication by $x_i$ to zero, so its homology action vanishes.

The sequence ideal annihilates each homology module.

A prime in its support contains an annihilator of a nonzero element and therefore contains $(\mathbf{x})$.

Localization commutes with finite direct sums and with the finite free exterior bases, sending $e_I\otimes m\mathrm{/}s$ to $e_I\otimes (m\mathrm{/}s)$.

The coordinate formula is preserved term by term because $(x_{i}m)\mathrm{/}s=(x_i\mathrm{/}1)(m\mathrm{/}s)$, so this degreewise isomorphism is a chain isomorphism.

Exact localization commutes with the kernel/image quotient defining homology.

The termwise localization chain isomorphism identifies the result with the claimed localized Koszul homology.

In degree $p$, base change sends $(e_I\otimes m)\otimes a$ to $e_I\otimes (m\otimes a)$; the exterior basis makes this an isomorphism.

The two differentials agree on every basis wedge by $x_i(m)\otimes a=(x_{i}a)(m\otimes 1)$, which proves the chain claim.

Flat tensoring is exact and therefore commutes with homology quotients.

The base-changed complex is the Koszul complex on $\mathbf{x}A$ with coefficient $M{\otimes }_{R}A$.

Let $E_{\mathbf{y}}\to E_{\mathbf{x}}$ send the $i$th basis vector to ${\sum }_j{a}_{ij}{e}_j$. The assumed identity $y_i={\sum }_j{a}_{ij}{x}_j$ makes the degree-one squares commute.

Extending by exterior powers respects wedge products, and the derivation rule then makes the resulting map commute with the differentials in every degree.

The matrix chain map is defined by exterior powers of the generator map.

The inverse matrix induces its inverse in every exterior degree, so the complexes are isomorphic.

Termwise localization and flat-base-change maps are chain isomorphisms and exactness supplies their homology conclusions.

An invertible generator matrix has an inverse exterior chain map; all maps commute with coefficient-module maps by construction.

At the first stage the definition requires injectivity of $x_1$ on $M$.

It also requires all stage quotients, especially $M$ and the terminal quotient, to be nonzero; units and $M=0$ therefore fail.

Separate the first condition: $x_1$ must be injective on $M$.

Every remaining condition is precisely the regularity condition for the tail on $M\mathrm{/}{x}_{1}M$, including the nonzero terminal quotient.

Each injectivity requirement for an initial segment occurs among those of the full sequence.

Its terminal quotient is an already-required nonzero intermediate quotient.

Localization preserves injectivity, so each regularity condition survives; the stated nonzero localized terminal quotient prevents the convention from failing at the end.

Faithfully flat tensoring preserves and reflects injectivity and nonzero modules. Apply this successively to the quotient stages to obtain the base-change assertion.

The one-element computation identifies positive homology with the kernel of multiplication by $x$.

Injectivity makes that kernel zero and degree zero is the stated quotient.

Put $C=K(\mathbf{x};M)$. By hypothesis $H_i(C)=0$ for $i>0$, while $H_0(C)=M\mathrm{/}(\mathbf{x})M$ by the basic Koszul-homology calculation.

The mapping-cone exact sequence for $K(\mathbf{x},y;M)$ identifies its $H_1$ with the kernel of multiplication by $y$ on $H_0(C)$, because $H_1(C)=0$; this kernel is zero by hypothesis. For $i>1$, the adjacent groups $H_i(C)$ and $H_{i-1}(C)$ both vanish, so exactness gives $H_i(K(\mathbf{x},y;M))=0$.

For the empty sequence the Koszul complex is $M$ in degree zero, so the conclusion is immediate. For length one it is the one-element calculation.

Suppose the result holds for an initial segment $\mathbf{x}$. Regularity says that the next element $y$ acts injectively on $M\mathrm{/}(\mathbf{x})M$. The induction lemma applied to the already acyclic $K(\mathbf{x};M)$ therefore makes $K(\mathbf{x},y;M)$ acyclic in positive degrees. Induction on the length proves the claim.

Regularity gives zero positive homology and the degree-zero calculation gives $M\mathrm{/}(\mathbf{x})M$.

Each term ${\bigwedge }^p{R}^n{\otimes }_{R}M$ is finite free because both factors are finite free, so this is a finite free resolution.

Let ${\mathbf{x}}^{\mathrm{\prime }}=(x_1,\dots ,x_{j-1})$ and $N=M\mathrm{/}({\mathbf{x}}^{\mathrm{\prime }})M$. The preceding prefix is regular, so $K({\mathbf{x}}^{\mathrm{\prime }};M)$ has zero positive homology and $H_0=N$. The failure at $x_j$ gives $0\mathrm{\ne }{\mathrm{Ann}}_N(x_j)$. The cone exact sequence therefore identifies $H_1(K({\mathbf{x}}^{\mathrm{\prime }},x_j;M))$ with this nonzero annihilator.

Append the remaining entries one at a time. If $L$ is the current nonzero first homology and the next entry is $y\in \mathfrak{m}$, the cone exact sequence injects $L\mathrm{/}yL$ into the new first homology. The module $L$ is finite because it is homology of a bounded complex of finite modules over a Noetherian ring, and Nakayama gives $L\mathrm{/}yL\mathrm{\ne }0$. Thus first homology remains nonzero through every later entry, proving $H_1(K(\mathbf{x};M))\mathrm{\ne }0$.

Put $C=K(x_1,\dots ,x_{n-1};M)$ and $y=x_n$. For every $j>0$, the cone exact sequence and $H_{j+1}(K(\mathbf{x};M))=0$ show that multiplication by $y$ on $H_j(C)$ is surjective. Each $H_j(C)$ is finite because $C$ is a bounded complex of finite modules over the Noetherian ring $R$.

Since $y\in \mathfrak{m}$, Nakayama applied to the surjections in step 1.1 gives $H_j(C)=0$ for every $j>0$. The segment of the same exact sequence ending in $H_1(K(\mathbf{x};M))=0$ then shows that multiplication by $y$ on $H_0(C)=M\mathrm{/}(x_1,\dots ,x_{n-1})M$ is injective. These are the two asserted conclusions.

If $\mathbf{x}$ is empty, its Koszul complex is concentrated in degree zero and the assumed nonzero terminal quotient is exactly the nonzero-module condition in the definition of an empty regular sequence. Thus the equivalence holds in this case. For nonempty $\mathbf{x}$, regularity implies acyclicity by the forward theorem.

Conversely, suppose $\mathbf{x}$ is nonempty and the Koszul complex is acyclic in positive degrees. The local converse lemma makes the last element injective on the quotient by the preceding entries and makes the shorter Koszul complex acyclic. Its terminal quotient is nonzero because it surjects onto $M\mathrm{/}(\mathbf{x})M$. Iterating proves all ordered injectivity conditions, while the final nonzero quotient is assumed; hence $\mathbf{x}$ is regular.

The forward direction is regular-sequence Koszul acyclicity.

The reverse direction is the local converse; together with the terminal quotient hypothesis it gives regularity.

Koszul regularity says every positive homology group vanishes.

Taking degree one is exactly $H_1$-regularity.

If $M=0$, every term of $K(\mathbf{x};M)$ is zero and the conclusion is immediate. Suppose $M\mathrm{\ne }0$. Because $(\mathbf{x})\subseteq \mathfrak{m}$, Nakayama shows that $M\mathrm{/}(\mathbf{x})M\mathrm{\ne }0$; the same argument applies to every prefix quotient.

If $\mathbf{x}$ were not regular, it would therefore have a first injectivity failure. The detection lemma would give $H_1(K(\mathbf{x};M))\mathrm{\ne }0$, contradicting $H_1$-regularity. Thus $\mathbf{x}$ is regular, and the local regularity criterion gives vanishing of every positive Koszul homology group.

The local criterion makes the original regular sequence Koszul-acyclic. Interchanging two adjacent generators is multiplication by an invertible permutation matrix, so invariance under invertible generator change gives an isomorphic Koszul complex for the swapped sequence.

The swapped sequence still lies in $\mathfrak{m}$ and generates the same ideal, so its terminal quotient is the same nonzero module. Applying the reverse direction of the local criterion to its acyclic Koszul complex proves that it is regular.

Every permutation is a product of adjacent transpositions.

Apply the adjacent-swap result successively under the unchanged local finite hypotheses.

For one non-zero-divisor $x$, if $x^{a}m=0$, repeated injectivity of $x$ gives $m=0$; its quotient remains nonzero by the regularity convention.

Use adjacent swaps to place one entry at a time, replace it by its positive power, and induct through the quotient stages.

Regularity implies Koszul regularity, which implies $H_1$-regularity.

The local $H_1$ implication and local acyclicity converse close the cycle.

The preceding corollary identifies the three notions in the stated local setting.

The permutation corollary applies to ordinary regularity and equivalence transfers the conclusion.

In the exterior bases, every matrix coefficient of $d$ is $0$ or $\pm x_i$, hence belongs to $\mathfrak{m}$.

Since $M$ is finite free, every term $M{\otimes }_R{\bigwedge }^p{R}^n$ is finite free. The assumed acyclicity and degree-zero quotient therefore make the Koszul complex a finite free resolution, and the containment from step 1.1 is precisely the local minimality criterion.

The minimality lemma identifies the Koszul ranks with the Koszul Betti numbers. In degree $i$ the free module has basis indexed by $i$-subsets of an $n$-set.

There are $\left(\genfrac{}{}{0px}{}{n}{i}\right)$ such subsets and none in degrees outside $0,\dots ,n$, giving the stated table.