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. As a tool, its homology can be used to tell when a set of elements of a (local) ring is an M-regular sequence, and hence it can be used to prove basic facts about the depth of a module or ideal which is an algebraic notion of dimension that is related to but different from the geometric notion of Krull dimension. Moreover, in certain circumstances, the complex is the complex of syzygies, that is, it tells you the relations between generators of a module, the relations between these relations, and so forth.

Definition

Let ''R'' be a commutative ring and ''E'' a free module of finite rank ''r'' over ''R''. We write $\bigwedge^i E$ for the ''i''-th exterior power of ''E''. Then, given an ''R''-linear map $s\colon E \to R$,
the Koszul complex associated to ''s'' is the chain complex of ''R''-modules:
:$K_(s)\colon 0 \to \bigwedge^r E \overset \to \bigwedge^ E \to \cdots \to \bigwedge^1 E \overset\to R \to 0$,
where the differential $d_k$ is given by: for any $e_i$ in ''E'',
:$d_k (e_1 \wedge \dots \wedge e_k) = \sum_^k (-1)^ s(e_i) e_1 \wedge \cdots \wedge \widehat \wedge \cdots \wedge e_k$.
The superscript $\widehat$ means the term is omitted. To show that $d_k \circ d_ = 0$, use the self-duality of a Koszul complex.

Note that $\bigwedge^1 E = E$ and $d_1 = s$. Note also that $\bigwedge^r E \simeq R$; this isomorphism is not canonical (for example, a choice of a volume form in differential geometry provides an example of such an isomorphism.)

If $E = R^r$ (i.e., an ordered basis is chosen), then, giving an ''R''-linear map $s\colon R^r\to R$ amounts to giving a finite sequence $s_1, \dots, s_r$ of elements in ''R'' (namely, a row vector) and then one sets $K_(s_1, \dots, s_r) = K_(s).$

If ''M'' is a finitely generated ''R''-module, then one sets:
:$K_(s, M) = K_(s) \otimes_R M$,
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
:$\operatorname_i(K_(s, M)) = \operatorname(d_i \otimes 1_M)/\operatorname(d_ \otimes 1_M)$
is called the ''i''-th Koszul homology. For example, if $E = R^r$ and s = _1 \cdots s_r/math> is a row vector with entries in ''R'', then $d_1 \otimes 1_M$ is
:$s \colon M^r \to M, \, (m_1, \dots, m_r) \mapsto s_1 m_1 + \dots + s_r m_r$
and so
:$\operatorname_0(K_(s, M)) = M/(s_1, \dots, s_r)M = R/(s_1, \dots, s_r) \otimes_R M.$
Similarly,
:$\operatorname_r(K_(s, M)) = \ = \operatorname_R(R/(s_1, \dots, s_r), M).$

Koszul complexes in low dimensions

Given a commutative ring ''R'', an element ''x'' in ''R'', and an ''R''-module ''M'', the multiplication by ''x'' yields a homomorphism of ''R''-modules,
:$M \to M.$
Considering this as a chain complex (by putting them in degree 1 and 0, and adding zeros elsewhere), it is denoted by $K(x, M)$. By construction, the homologies are
:$H_0(K(x, M)) = M/xM, H_1(K(x,M)) = \operatorname_M(x) = \,$
the annihilator of ''x'' in ''M''.
Thus, the Koszul complex and its homology encode fundamental properties of the multiplication by ''x''. This chain complex $K_(x)$ is called the Koszul complex of ''R'' with respect to ''x'', as in #Definition.

The Koszul complex for a pair $(x, y) \in R^2$ is
:$0 \to R \xrightarrow R^2 \xrightarrow R\to 0,$
with the matrices $d_1$ and $d_2$ given by

:$d_1 = \begin x & y\\ \end$ and
:$d_2 = \begin -y\\ x\\ \end.$
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 $x_1, x_2, \dots, x_n$ form a regular sequence, the higher homology modules of the Koszul complex are all zero.

Example

If ''k'' is a field and $X_1, X_2,\dots, X_d$ are indeterminates and ''R'' is the polynomial ring k _1, X_2,\dots, X_d/math>, 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'', let $s\colon E\to R$ be an ''R''-linear map, and let ''t'' be an element of ''R''. Let $K(s, t)$ be the Koszul complex of $(s, t)\colon E \oplus R \to R$.

Using $\wedge^k(E \oplus R) = \oplus_^k \bigwedge^ E \otimes \bigwedge^i R = \bigwedge^k E \oplus \bigwedge^ E$,
there is the exact sequence of complexes:
:0 \to K(s) \to K(s, t) \to K(s) 1\to 0
where 1signifies the degree shift by -1 and $d_ = -d_$. One 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
:$d_((x, y)) = (s(e_1) + t) e_2 \wedge \cdots \wedge e_ + \sum_^k (-1)^ s(e_i) (e_1 + \epsilon) \wedge e_2 \wedge \cdots \widehat \cdots \wedge e_k$,
which is $(d_x + ty, -d_ y)$. for $(x, y)$ in $\wedge^k E \oplus \bigwedge^ E$,
:$d_((x, y)) = (d_ x + ty, d_ y).$
In the language of homological algebra, the above means that $K(s, t)$ is the mapping cone of $t\colon K(s) \to K(s)$.

Taking the long exact sequence of homologies, we obtain:
:$\cdots \to \operatorname_i(K(s)) \overset \to \operatorname_i(K(s)) \to \operatorname_i(K(s, t)) \to \operatorname_(K(s)) \overset\to \cdots.$
Here, the connecting homomorphism
:\delta: \operatorname_(K(s) 1 = \operatorname_(K(s)) \to \operatorname_(K(s))
is computed as follows. By definition, \delta( = _(y)/math> 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 $d_$ gives \delta( = t /math>.

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

Proof by induction on ''r''. If ''$r=1$'', then $\operatorname_1(K(x_1;M)) = \operatorname_M(x_1) = 0$. Next, assume the assertion is true for ''r'' - 1. Then, using the above exact sequence, one sees $\operatorname_i(K(x_1, \dots, x_r; M)) = 0$ for any $i \geq 2$. The vanishing is also valid for $i=1$, since $x_r$is a nonzerodivisor on $\operatorname_0(K(x_1, \dots, x_; M)) = M/(x_1, \dots, x_)M.$ $\square$

Proof: By the theorem applied to ''S'' and ''S'' as an ''S''-module, we see ''K''(''y''₁, ..., ''y''_''n'') is an ''S''-free resolution of ''S''/(''y''₁, ..., ''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''. $\square$

Proof: Let ''S'' = ''R'' 'y''₁, ..., ''y''_''n'' Turn ''M'' into an ''S''-module through the ring homomorphism ''S'' → ''R'', ''y''_''i'' → ''x''_''i'' and ''R'' an ''S''-module through . By the preceding corollary, $\operatorname_i(K(x_1, \dots, x_n) \otimes M) = \operatorname_i^S(R, M)$ and then
:$\operatorname_S\left(\operatorname_i^S(R, M)\right) \supset \operatorname_S(R) + \operatorname_S(M) \supset (y_1, \dots, y_n) + \operatorname_R(M) + (y_1 - x_1, ..., y_n - x_n).$ $\square$

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

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 ''x''₁, ..., ''x''_''r''-1. Consider
:$\cdots \to \operatorname_1(K(x'; M)) \overset \to \operatorname_1(K(x'; M)) \to \operatorname_1(K(x_1, \dots, x_r; M)) = 0 \to M/x'M \overset\to \cdots.$
Since the first $x_r$ is surjective, $N = x_r N$ with $N = \operatorname_1(K(x'; M))$. 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 $M/(x_1, \dots, x_r)M \ne 0$ is automatic.) $\square$

Tensor products of Koszul complexes

In general, if ''C'', ''D'' are chain complexes, then their tensor product $C \otimes D$ is the chain complex given by

:$(C \otimes D)_n = \sum_ C_i \otimes D_j$
with the differential: for any homogeneous elements ''x'', ''y'',
:$d_ (x \otimes y) = d_C(x) \otimes y + (-1)^ x \otimes d_D(y)$
where , ''x'', is the degree of ''x''.

This construction applies in particular to Koszul complexes. Let ''E'', ''F'' be finite-rank free modules, and let $s\colon E\to R$ and $t\colon F\to R$ be two ''R''-linear maps. Let $K(s, t)$ be the Koszul complex of the linear map $(s, t)\colon E \oplus F \to R$. 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 Λ''E'',
*$d_s(x) = s(x)$ when $, x, = 1$
*$d_s(x \wedge y) = d_s(x) \wedge y + (-1)^x \wedge d_s(y)$
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,
:$d_(e \otimes 1 + 1 \otimes f) = d_(e) \otimes 1 + 1 \otimes d_(f) = s(e) + t(f) = d_(e + f).$
Since $d_$ and $d_$ are derivations of the same type, this implies $d_ = d_.$

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.

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 $\operatorname(I, M)$. Recall that an ''M''-regular sequence ''x''₁, ..., ''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.

Proof: To lighten the notations, we write H(-) for H(''K''(-)). Let ''y''₁, ..., ''y''_''s'' be a maximal ''M''-regular sequence in the ideal ''I''; we denote this sequence by $\underline$. First we show, by induction on $l$, the claim that $\operatorname_i(\underline, x_1, \dots, x_l; M)$ is $\operatorname_(x_1, \dots, x_l)$ 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,
:$\operatorname_l\left(\underline, x_1, \dots, x_l; M \right) = \operatorname\left(x_l: \operatorname_(x_1, \dots, x_) \to \operatorname_(x_1, \dots, x_) \right)$,
which is
$\operatorname_(x_1, \dots, x_l).$ 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 $\operatorname_i(x_1, \dots, x_n; M) = 0$ for all ''i'' > ''n'' - ''s''. To conclude ''n'' - ''s'' = ''m'', it remains to show that it is nonzero if ''i'' = ''n'' - ''s''. Since $\underline$ is a maximal ''M''-regular sequence in ''I'', the ideal ''I'' is contained in the set of all zerodivisors on $M/\underlineM$, the finite union of the associated primes of the module. Thus, by prime avoidance, there is some nonzero ''v'' in $M/\underlineM$ such that $I \subset \mathfrak = \operatorname_R(v)$, which is to say,
:$0 \ne v \in \operatorname_(I) \simeq \operatorname_n\left(x_1, \dots, x_n, \underline; M\right) = \operatorname_(x_1, \dots, x_n; M) \otimes \wedge^s R^s.$ $\square$

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'':
:$l_e: \wedge^k E \to \wedge^ E, \, x \mapsto e \wedge x.$
Since $e \wedge e = 0$, we have: $l_e \circ l_e = 0$; that is,
:$0 \to R \overset\to \wedge^1 E \overset\to \wedge^2 E \to \cdots \to \wedge^r E \to 0$
is a cochain complex of free modules. This complex, also called a Koszul complex, is a complex used in . Taking the dual, there is the complex:
:$0 \to (\wedge^r E)^* \to (\wedge^ E)^* \to \cdots \to (\wedge^2 E)^* \to (\wedge^1E)^* \to R \to 0$.
Using an isomorphism $\wedge^k E \simeq (\wedge^ E)^* \simeq \wedge^ (E^*)$, the complex $(\wedge E, l_e)$ coincides with the Koszul complex in the definition.

Use

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

See also

* Koszul–Tate complex
*Syzygy (mathematics)

Notes

References

* David Eisenbud, ''Commutative Algebra. With a view toward algebraic geometry'', Graduate Texts in Mathematics, vol 150, Springer-Verlag, New York, 1995.
*
*
* {{Citation , last1=Serre , first1=Jean-Pierre , author1-link = Jean-Pierre Serre , title=Algèbre locale, Multiplicités , publisher=Springer-Verlag , location=Berlin, New York , series=Cours au Collège de France, 1957–1958, rédigé par Pierre Gabriel. Troisième édition, 1975. Lecture Notes in Mathematics , year=1975 , volume=11 , language=French

External links

* Melvin Hochster
Math 711: Lecture of October 3, 2007
(especially the very last part).

Homological algebra