Koszul complex

In mathematics, the Koszul complex was first introduced to define a cohomology theory for Lie algebras, by Jean-Louis Koszul (see Lie algebra cohomology). It turned out to be a useful general construction in homological algebra.

Definition

Let R be a commutative ring and E a free module of finite rank r over R. We write $\wedge ^{i}E$ for the i-th exterior power of E. Then, given an R-linear map $s:E\to R$ , the Koszul complex associated to s is the chain complex of R-modules:

K_{\bullet }(s):0\to \wedge ^{r}E{\overset {d_{r}}{\to }}\wedge ^{r-1}E\to \cdots \to \wedge ^{1}E{\overset {d_{1}}{\to }}R\to 0

where the differential d_k is given by: for any e_i in E,

d_{k}(e_{1}\wedge \dots \wedge e_{k})=\sum _{i=1}^{k}(-1)^{i+1}s(e_{i})e_{1}\wedge \cdots \wedge {\widehat {e_{i}}}\wedge \cdots \wedge e_{k}

The superscript ${\widehat {\cdot }}$ means the term is omitted. (Showing $d_{k}\circ d_{k+1}=0$ is straightforward; alternatively, this identity also follows using the #Self-duality of a Koszul complex.)

Note $\wedge ^{1}E=E$ and $d_{1}=s$ . Note also $\wedge ^{r}E\simeq R$ ; this isomorphism is not canonical (for example, a choice of a volume form in differential geometry is an example of such an isomorphism.)

If E = R^r (i.e., an ordered basis is chosen), then, giving an R-linear map s: R^r → R amounts to giving a finite sequence s₁, ..., s_r of elements in R (namely, a row vector) and then one sets $K_{\bullet }(s_{1},\dots ,s_{r})=K_{\bullet }(s).$

If M is a finitely generated R-module, then one sets:

K_{\bullet }(s,M)=K_{\bullet }(s)\otimes _{R}M

which is again a chain complex with the induced differential $(d\otimes 1_{M})(v\otimes m)=d(v)\otimes m$ .

The i-th homology of the Koszul complex

\operatorname {H} _{i}(K_{\bullet }(s,M))=\operatorname {ker} (d_{i}\otimes 1_{M})/\operatorname {im} (d_{i+1}\otimes 1_{M})

is called the i-th Koszul homology. For example, if E = R^r and $s=[s_{1}\cdots s_{r}]$ is a row vector with entries in R, then $d_{1}\otimes 1_{M}$ is

s:M^{r}\to M,\,(m_{1},\dots ,m_{r})\mapsto s_{1}m_{1}+\dots +s_{r}m_{r}

and so

\operatorname {H} _{0}(K_{\bullet }(s,M))=M/(s_{1},\dots ,s_{r})M=R/(s_{1},\dots ,s_{r})\otimes _{R}M.

Similarly,

\operatorname {H} _{r}(K_{\bullet }(s,M))=\{m\in M:s_{1}m=s_{2}m=\dots =s_{r}m=0\}=\operatorname {Hom} _{R}(R/(s_{1},\dots ,s_{r}),M).

Explanation

In commutative algebra, if x is an element of the ring R, multiplication by x is R-linear and so represents an R-module homomorphism x:R →R from R to itself. It is useful to throw in zeroes on each end and make this a (free) R-complex:

0\to R\xrightarrow{\ x\ }R\to0.

Call this chain complex K_•(x).

Counting the right-hand copy of R as the zeroth degree and the left-hand copy as the first degree, this chain complex neatly captures the most important facts about multiplication by x because its zeroth homology is exactly the homomorphic image of R modulo the multiples of x, H₀(K_•(x)) = R/xR, and its first homology is exactly the annihilator of x, H₁(K_•(x)) = Ann_R(x).

This chain complex K_•(x) is called the Koszul complex of R with respect to x, as in #Definition. For the case of two elements x and y, the Koszul complex can be written down quite succinctly as

0\to R{\xrightarrow {\ d_{2}\ }}R^{2}{\xrightarrow {\ d_{1}\ }}R\to 0,

with the matrices $d_{1}$ and $d_{2}$ given by

d_{1}={\begin{bmatrix}x&y\\\end{bmatrix}}

and

d_{2}={\begin{bmatrix}-y\\x\\\end{bmatrix}}.

Note that d_i is applied on the left. The cycles in degree 1 are then exactly the linear relations on the elements x and y, while the boundaries are the trivial relations. The first Koszul homology H₁(K_•(x, y)) therefore measures exactly the relations mod the trivial relations. With more elements the higher-dimensional Koszul homologies measure the higher-level versions of this.

In the case that the elements x₁, x₂, ..., x_n form a regular sequence, the higher homology modules of the Koszul complex are all zero.

Example

If k is a field and X₁, X₂, ..., X_d are indeterminates and R is the polynomial ring k[X₁, X₂, ..., X_d], the Koszul complex K_•(X_i) on the X_i's forms a concrete free R-resolution of k.

Properties of a Koszul homology

Let E be a finite-rank free module over R, s: E → R an R-linear map and t an element of R. Let $K(s,t)$ be the Koszul complex of $(s,t):E\oplus R\to R$ . Let M be a finitely generated module over R.

Using $\wedge ^{k}(E\oplus R)=\oplus _{i=0}^{k}\wedge ^{k-i}E\otimes \wedge ^{i}R=\wedge ^{k}E\oplus \wedge ^{k-1}E$ , there is the exact sequence of complexes:

0\to K(s;M)\to K(s,t;M)\to K(s;M)[-1]\to 0

where [-1] signifies the degree shift by -1 and $d_{K(s)[-1]}=-d_{K(s)}$ . One notes:^[1] for (x, y) in $\wedge ^{k}E\oplus \wedge ^{k-1}E$ ,

d_{K(s,t)}((x,y))=(d_{K(s)}x+ty,d_{K(s)[-1]}y).

(In the language of homological algebra, the above means that $K(s,t)$ is the mapping cone of $t:K(s)[-1]\to K(s)$ .)

Taking the long exact sequence of homologies, we get:

\cdots \to \operatorname {H} _{i}(K(s;M)){\overset {t}{\to }}\operatorname {H} _{i}(K(s;M))\to \operatorname {H} _{i}(K(s,t;M))\to \operatorname {H} _{i-1}(K(s;M)){\overset {t}{\to }}\cdots .

Here, the connecting homomorphism

\delta :\operatorname {H} _{i+1}(K(s;M)[-1])=\operatorname {H} _{i}(K(s;M))\to \operatorname {H} _{i}(K(s;M))

is computed as follows. By definition, $\delta ([x])=[d_{K(s,t)}(y)]$ where y is an element of $K(s,t)$ that maps to x. Since $K(s,t)$ is a direct sum, we can simply take y to be (0, x). Then the early formula for $d_{K(s,t)}$ gives $\delta ([x])=t[x]$ .

The above exact sequence can be used to prove the following.

Theorem — Let R be a ring, M a finitely generated module over R. If a sequence x₁, x₂, ..., x_r of elements of R is a regular sequence on M, then

\operatorname {H} _{i}(K(x_{1},\dots ,x_{r})\otimes M)=0

for all i ≥ 1. In particular, when M = R, this is to say

0\to \wedge ^{r}R^{r}{\overset {d_{r}}{\to }}\wedge ^{r-1}R^{r}\to \cdots \to \wedge ^{2}R^{r}{\overset {d_{2}}{\to }}R^{r}{\overset {[x_{1}\cdots x_{r}]}{\to }}R\to R/(x_{1},\cdots ,x_{r})\to 0

is exact; i.e., $K(x_{1},\dots ,x_{r})$ is an R-free resolution of $R/(x_{1},\dots ,x_{r})$ .

Proof by induction on r. If r = 1, then $\operatorname {H} _{1}(K(x_{1};M))=\operatorname {Ann} _{M}(x_{1})=0$ . Next, assume the assertion is true for r - 1. Then, using the above exact sequence, one sees $\operatorname {H} _{i}(K(x_{1},\dots ,x_{r};M))=0$ for any i ≥ 2. The vanishing is also valid for i = 1, since $x_r$ is a nonzerodivisor on $\operatorname {H} _{0}(K(x_{1},\dots ,x_{r-1};M))=M/(x_{1},\dots ,x_{r-1})M.$ $\square$

Corollary^[2] — Let R, M be as above and x₁, x₂, ..., x_n a sequence of elements of R. Suppose there are a ring S, an S-regular sequence y₁, y₂, ..., y_n in S and a ring homomorphism S → R that maps y_i to x_i. (For example, one can take S = Z[y₁, ..., y_n].) Then

\operatorname {H} _{i}(K(x_{1},\dots ,x_{n})\otimes _{R}M)=\operatorname {Tor} _{i}^{S}(S/(y_{1},\dots ,y_{n}),M).

where Tor denotes the Tor functor and M is an S-module through S → R.

Proof: By the theorem applied to S and S as an S-module, we see K(y₁, ..., y_n) is an S-free resolution of S/(y₁, ..., y_n). So, by definition, the i-th homology of $K(y_{1},\dots ,y_{n})\otimes _{S}M$ is the right-hand side of the above. On the other hand, $K(y_{1},\dots ,y_{n})\otimes _{S}M=K(x_{1},\dots ,x_{n})\otimes _{R}M$ by the definition of the S-module structure on M. $\square$

Corollary^[3] — Let R, M be as above and x₁, x₂, ..., x_n a sequence of elements of R. Then both the ideal I = (x₁, ..., x_n) and the annihilator of M annihilate

\operatorname {H} _{i}(K(x_{1},\dots ,x_{n})\otimes M)

for all i.

Proof: Let S = R[y₁, ..., y_n]. Turn M into an S-module through the ring homomorphism S → R, y_i → x_i and R an S-module through y_i → 0. By the preceding corollary, $\operatorname {H} _{i}(K(x_{1},\dots ,x_{n})\otimes M)=\operatorname {Tor} _{i}^{S}(R,M)$ and then

\operatorname {Ann} _{S}(\operatorname {Tor} _{i}^{S}(R,M))\supset \operatorname {Ann} _{S}(R)+\operatorname {Ann} _{S}(M)\supset (y_{1},\dots ,y_{n})+\operatorname {Ann} _{R}(M)+(y_{1}-x_{1},...,y_{n}-x_{n}).

\square

For a Noetherian local ring, the converse of the theorem holds. More generally,

Theorem — Let R be a Noetherian ring and M a nonzero finitely generated module over R . If x₁, x₂, ..., x_r are elements of the Jacobson radical of R, then the following are equivalent:

The sequence $x_{1},\dots ,x_{r}$ is a regular sequence on M,
$\operatorname {H} _{1}(K(x_{1},\dots ,x_{r})\otimes M)=0$ ,
$\operatorname {H} _{i}(K(x_{1},\dots ,x_{r})\otimes M)=0$ for all i ≥ 1.

Proof: We only need to show 2. implies 1., the rest being clear. We argue by induction on r. The case r = 1 is already known. Let x' denote x₁, ..., x_r-1. Consider

\cdots \to \operatorname {H} _{1}(K(x';M)){\overset {x_{r}}{\to }}\operatorname {H} _{1}(K(x';M))\to \operatorname {H} _{1}(K(x_{1},\dots ,x_{r};M))=0\to M/x'M{\overset {x_{r}}{\to }}\cdots .

Since the first $x_r$ is surjective, $N=x_{r}N$ with $N=\operatorname {H} _{1}(K(x';M))$ . By Nakayama's lemma, $N=0$ and so x' is a regular sequence by the inductive hypothesis. Since the second $x_r$ is injective (i.e., is a nonzerodivisor), $x_{1},\dots ,x_{r}$ is a regular sequence. (Note: by Nakayama's lemma, the requirement $M/(x_{1},\dots ,x_{r})M\neq 0$ is automatic.) $\square$

Tensor products of Koszul complexes

In general, if C, D are chain complexes, then their tensor product C ⊗ D is the chain complex given by

(C\otimes D)_{n}=\sum _{i+j=n}C_{i}\otimes D_{j}

with the differential: for any homogeneous elements x, y,

d_{C\otimes D}(x\otimes y)=d_{C}(x)\otimes y+(-1)^{|x|}x\otimes d_{D}(y)

where |x| is the degree of x.

This construction applies in particular to Koszul complexes. Let E, F be finite-rank free modules, s: E → R, t: F → R R-linear maps. Let $K(s,t)$ be the Koszul complex of the linear map $(s,t):E\otimes F\to R$ . Then, as complexes,

K(s,t)\simeq K(s)\otimes K(t).

To see this, it is more convenient to work with an exterior algebra (as opposed to exterior powers). Define the graded derivation of degree -1

d_{s}:\wedge E\to \wedge E

by requiring: for any homogeneous elements x, y in ΛE,

$d_{s}(x)=s(x)$ when $|x|=1$
$d_{s}(x\wedge y)=d_{s}(x)\wedge y+(-1)^{|x|}x\wedge d_{s}(y)$

One easily sees that $d_{s}\circ d_{s}=0$ (induction on degree) and that the action of d_s on homogeneous elements agrees with the differentials in #Definition.

Now, we have $\wedge (E\oplus F)=\wedge E\otimes \wedge F$ as graded R-modules. Also, by the definition of a tensor product mentioned in the beginning,

d_{K(s)\otimes K(t)}(e\otimes 1+1\otimes f)=d_{K(s)}(e)\otimes 1+1\otimes d_{K(t)}(f)=s(e)+t(f)=d_{K(s,t)}(e+f).

Since $d_{K(s)\otimes K(t)}$ and $d_{K(s,t)}$ are derivations of the same type, this implies $d_{K(s)\otimes K(t)}=d_{K(s,t)}.$

Note, in particular,

K(x_{1},x_{2},\dots ,x_{r})\simeq K(x_{1})\otimes K(x_{2})\otimes \cdots \otimes K(x_{r})

The next proposition shows how the Koszul complex of elements encodes some information about sequences in the ideal generated by them.

Proposition — Let R be a ring and I = (x₁, ..., x_n) an ideal generated by some n-elements. Then, for any R-module M and any elements y₁, ..., y_r in I,

\operatorname {H} _{i}(K(x_{1},\dots ,x_{n},y_{1},\dots ,y_{r};M))\simeq \bigoplus _{i=j+k}\operatorname {H} _{j}(K(x_{1},\dots ,x_{n};M))\otimes \wedge ^{k}R^{r}.

where $\wedge ^{k}R^{r}$ is viewed as a complex with zero differential. (In fact, the decomposition holds on the chain-level).

Proof: (Easy but omitted for now)

As an application, we can show the depth-sensitivity of a Koszul homology. Given a finitely generated module M over a ring R, by (one) definition, the depth of M with respect to an ideal I is the supremum of the lengths of all regular sequences of elements of I on M. It is denoted by $\operatorname {depth} (I,M)$ . Recall that an M-regular sequence x₁, ..., x_n in an ideal I is maximal if I contains no nonzerodivisor on $M/(x_{1},\dots ,x_{n})M$ .

The Koszul homology gives a very useful characterization of a depth.

Theorem (depth-sensitivity) — Let R be a Noetherian ring, x₁, ..., x_n elements of R and I = (x₁, ..., x_n) the ideal generated by them. For a finitely generated module M over R, if, for some integer m,

\operatorname {H} _{i}(K(x_{1},\dots ,x_{n})\otimes M)=0

for all i > m,

while

\operatorname {H} _{m}(K(x_{1},\dots ,x_{n})\otimes M)\neq 0,

then every maximal M-regular sequence in I has length n - m (in particular, they all have the same length). As a consequence,

\operatorname {depth} (I,M)=n-m

Proof: To lighten the notations, we write H(-) for H(K(-)). Let y₁, ..., y_s be a maximal M-regular sequence in the ideal I; we denote this sequence by $\underline{y}$ . First we show, by induction on $l$ , the claim that $\operatorname {H} _{i}({\underline {y}},x_{1},\dots ,x_{l};M)$ is $\operatorname {Ann} _{M/{\underline {y}}M}(x_{1},\dots ,x_{l})$ if $i=l$ and is zero if $i>l$ . The basic case $l = 0$ is clear from #Properties of a Koszul homology. From the long exact sequence of Koszul homologies and the inductive hypothesis,

\operatorname {H} _{l}\left({\underline {y}},x_{1},\dots ,x_{l};M\right)=\operatorname {ker} \left(x_{l}:\operatorname {Ann} _{M/{\underline {y}}M}(x_{1},\dots ,x_{l-1})\to \operatorname {Ann} _{M/{\underline {y}}M}(x_{1},\dots ,x_{l-1})\right)

which is $\operatorname {Ann} _{M/{\underline {y}}M}(x_{1},\dots ,x_{l}).$ Also, by the same argument, the vanishing holds for $i>l$ . This completes the proof of the claim.

Now, it follows from the claim and the early proposition that $\operatorname {H} _{i}(x_{1},\dots ,x_{n};M)=0$ for all i > n - s. To conclude n - s = m, it remains to show that it is nonzero if i = n - s. Since $\underline{y}$ is a maximal M-regular sequence in I, the ideal I is contained in the set of all zerodivisors on $M/{\underline {y}}M$ , the finite union of the associated primes of the module. Thus, by prime avoidance, there is some nonzero v in $M/{\underline {y}}M$ such that $I\subset {\mathfrak {p}}=\operatorname {Ann} _{R}(v)$ , which is to say,

0\neq v\in \operatorname {Ann} _{M/{\underline {y}}M}(I)\simeq \operatorname {H} _{n}\left(x_{1},\dots ,x_{n},{\underline {y}};M\right)=\operatorname {H} _{n-s}(x_{1},\dots ,x_{n};M)\otimes \wedge ^{s}R^{s}.

\square

Self-duality

There is an approach to a Koszul complex that uses a cochain complex instead of a chain complex. As it turns out, this results essentially in the same complex (the fact known as the self-duality of a Koszul complex).

Let E be a free module of finite rank r over a ring R. Then each element e of E gives rise to the exterior left-multiplication by e:

l_{e}:\wedge ^{k}E\to \wedge ^{k+1}E,\,x\mapsto e\wedge x.

Since $e\wedge e=0$ , we have: $l_{e}\circ l_{e}=0$ ; that is,

0\to R{\overset {1\mapsto e}{\to }}\wedge ^{1}E{\overset {l_{e}}{\to }}\wedge ^{2}E\to \cdots \to \wedge ^{r}E\to 0

is a cochain complex of free modules. This complex, also called a Koszul complex, is a complex used in (Eisenbud 1995). Taking the dual, there is the complex:

0\to (\wedge ^{r}E)^{*}\to (\wedge ^{r-1}E)^{*}\to \cdots \to (\wedge ^{2}E)^{*}\to (\wedge ^{1}E)^{*}\to R\to 0

Using an isomorphism $\wedge ^{k}E\simeq (\wedge ^{r-k}E)^{*}\simeq \wedge ^{r-k}(E^{*})$ , the complex $(\wedge E,l_{e})$ coincides with the Koszul complex in #Definition.

Use

The Koszul complex is essential in defining the joint spectrum of a tuple of commuting bounded linear operators in a Banach space.

Notes

↑ Indeed, by linearity, we can assume $(x,y)=(e_{1}+\epsilon )\wedge e_{2}\wedge \cdots \wedge e_{k}\in \wedge ^{k}(E\oplus R)$ where $R\simeq R\epsilon \subset E\oplus R$ . Then
$d_{K(s,t)}((x,y))=(s(e_{1})+t)e_{2}\wedge \cdots \wedge e_{k}+\sum _{i=2}^{k}(-1)^{i+1}s(e_{i})(e_{1}+\epsilon )\wedge e_{2}\wedge \cdots {\widehat {e_{i}}}\cdots \wedge e_{k}$ ,
which is $(d_{K(s)}x+ty,-d_{K(s)}y)$ .
↑ Eisenbud, Exercise 17.10.
↑ Serre, Ch IV, A § 2, Proposition 4.

References

David Eisenbud, Commutative Algebra. With a view toward algebraic geometry, Graduate Texts in Mathematics, vol 150, Springer-Verlag, New York, 1995. ISBN 0-387-94268-8
William Fulton (1998), Intersection theory, Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge., 2 (2nd ed.), Berlin, New York: Springer-Verlag, ISBN 978-3-540-62046-4, MR 1644323
Serre, Jean-Pierre (1975), Algèbre locale, Multiplicités, Cours au Collège de France, 1957–1958, rédigé par Pierre Gabriel. Troisième édition, 1975. Lecture Notes in Mathematics (in French), 11, Berlin, New York: Springer-Verlag

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