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Postmodern Algebra

Sam Macdonald

Section 16.2 The Koszul Complex

Let \(C\) and \(D\) be chain complexes with differentials \(\partial^C\) and \(\partial^D\text{,}\) respectively. Recall the definition of the tensor product complex \(C\otimes_R D\) (see Grifo 915: Remark 6.14): For all \(n\text{,}\)
\begin{equation*} (C\otimes_R D)_n=\bigoplus_{p+q=n}C_p\otimes_R D_{q} \end{equation*}
and
\begin{equation*} \partial^{C\otimes_R D}_n(c\otimes d)=\partial^C_p(c)\otimes 1 + (-1)^p\otimes \partial_q^D(d) \end{equation*}
for \(c\in C_p\) and \(d\in D_q\) and \(p+q=n\text{.}\)
For ease of notation, we often supress the superscripts and subscripts on the differentials so long as they are clear from the context. For an element \(c\) of a complex \(C\text{,}\) we define the degree of \(c\text{,}\) denoted \(|c|\text{,}\) to be \(p\) if \(c\in C_p\text{.}\) Thus, the differential of \(C\otimes_R D\) can be expressed as
\begin{equation*} \partial(c\otimes d)=\partial(c)\otimes 1 + (-1)^{|c|} \otimes \partial(d). \end{equation*}

Definition 16.24.

Definition 70. Let \(R\) be a ring and \({\bf x}=x_1,\dots,x_n\) a sequence of elements in \(R\text{.}\) We define the Koszul complex \(K(\bf x)\) of \(\bf x\) (on \(R\)) inductively as follows: When \(n=1\text{,}\) we define \(K(x_1)\) to be the complex
\begin{equation*} 0\to R\xrightarrow{x} R\to 0, \end{equation*}
where the \(R\) on the right is in homological degree \(0\text{.}\) Suppose \(n>1\) and let \({\bf x'}=x_1,\dots,x_{n-1}\text{.}\) Assume that \(K({\bf x'})\) has been defined. Then we set \(K({\bf x}):=K({\bf x'})\otimes_R K(x_n)\text{.}\)

Example 16.25.

Example 71.
Let \(R\) be a ring and \(x,y\) elements of \(R\text{.}\) Let’s find \(K(x,y)=K(x)\otimes_R K(y)\text{.}\) So \(K(x,y)\) is the complex
\begin{equation*} (0\to R\xrightarrow{x} R\to 0) \otimes (0\to R\xrightarrow{y}R\to 0). \end{equation*}
To track degrees, let’s write \(K(x)\) as \(0\to K_1\xrightarrow{x} K_0\to 0\) and \(K(y)\) as \(0\to L_1\xrightarrow{y}L_0\to 0\text{,}\) where \(K_i=L_i=R\) for all \(i=0,1\text{.}\) Then \(K(x)\otimes K(y)\) has the form
\begin{equation*} 0\to K_1\otimes_R L_1 \xrightarrow{\partial_2} (K_0\otimes L_1)\oplus (K_1\otimes_R L_0)\xrightarrow{\partial_1} K_0\otimes_R L_0\to 0. \end{equation*}
Using the rule for the differential of a tensor product of complexes, we obtain
\begin{equation*} \begin{aligned} \partial_2(1\otimes 1)&=x\otimes 1 - 1\otimes y\\ \partial_1(1_{K_0}\otimes 1_{L_1})&= 0\otimes 1 + 1\otimes y=1\otimes y\\ \partial_1(1_{K_1}\otimes 1_{L_0})&= x\otimes 1 - 1\otimes 0= x\otimes 1\\ \partial_0(1\otimes 1)&= 0\otimes 1+1\otimes 0 = 0\otimes 0. \end{aligned} \end{equation*}
Now, each \(K_i\otimes L_j\cong R\) by virtue of the map sending \(1\otimes 1\) to \(1\text{.}\) Under this identification, the resulting complex is
\begin{equation*} 0\to R\xrightarrow{\begin{pmatrix} x\\ -y \end{pmatrix}} R^2\xrightarrow{\begin{pmatrix} y & x\end{pmatrix}}R\to 0, \end{equation*}
where again the rightmost \(R\) is in degree 0.

Definition 16.26.

Let \(\bf x\) be a (finite) sequence of elements of \(R\) and \(M\) an \(R\)-module. Then the Koszul complex of \(\bf x\) on \(M\) is defined to be \(K({\bf x};M):=K({\bf x})\otimes_R M\text{.}\)

Example 16.27.

Example 73. Let \(R=k[x,y]\) be a polynomial ring over a field \(k\) and let \(M=R/(y)\text{.}\) Then using Example Example 16.25 and that \(M\cong k[x]\text{,}\) we have \(K(x,y;M)=K(x,y)\otimes_R R/(y)\) is isomorphic to
\begin{equation*} 0\to k[x]\xrightarrow{\begin{pmatrix} x\\ 0 \end{pmatrix}} k[x]^2\xrightarrow{\begin{pmatrix} 0 & x\end{pmatrix}}k[x]\to 0. \end{equation*}

Definition 16.28.

Definition 74. Let \(R\) be a ring and \(C\) a complex of \(R\)-module. Then the shift or suspension  \(\mathsf{\Sigma}C\) of \(C\) is the complex defined by \((\mathsf{\Sigma}C)_i=C_{i-1}\) and \(\partial^{\mathsf{\Sigma}C}=-\partial^C\text{.}\) Note that \(\operatorname{H}_i(\mathsf{\Sigma}C)=\operatorname{H}_{i-1}(C)\text{.}\)
Construction 75. Let \(C\) be a complex of \(R\)-modules and \(x\in R\text{.}\) Let \(C(x)\) denote the complex \(C\otimes_R K(x)\text{.}\) Note that for all \(i\)
\begin{equation*} C(x)_i=(C_{i-1}\otimes_R R)\oplus (C_i\otimes_R R)\cong C_{i-1}\oplus C_{i}. \end{equation*}
Let \((u,v)\in C(x)_{i-1}\text{,}\) where \(u\in C_{i-1}\) and \(v\in C_i\text{.}\) Then
\begin{equation*} \begin{aligned} \partial^{C(x)}(u,v)&=\partial^{C(x)}(u\otimes 1 + v\otimes 1)\\ &=\partial(u)\otimes 1+ (-1)^{|u|}u\otimes x + \partial(v)\otimes 1+(-1)^{|v|}v \otimes 0\\ &=(\partial(u), \partial(v)+(-1)^{|u|}xu)\in C(x)_{i-1}. \end{aligned} \end{equation*}
There exists chain maps \(\alpha:C\to C(x)\) and \(\beta:C(x)\to \mathsf{\Sigma}C\) given by
\begin{equation*} \begin{aligned} \alpha(v)&:=(0,v)\in C(x)_i\\ \beta(u,v)&:=(-1)^{|u|}u\in (\mathsf{\Sigma}C)_i \end{aligned} \end{equation*}
for \(u\in C_{i-1}\) and \(v\in C_i\text{.}\) One can easily check that these maps commute with differentials, and so are indeed chain maps. (Keep in mind that the differential of \(\mathsf{\Sigma}C\) is \(-\partial^C\text{.}\)) Hence, we have a short exact sequence of complexes
\begin{equation*} 0\to C\xrightarrow{\alpha} C(x)\xrightarrow{\beta} \mathsf{\Sigma}C\to 0. \end{equation*}
This leads to the long exact sequence on homology
\begin{equation*} \cdots \to \operatorname{H}_i(C)\xrightarrow{\alpha_*}\operatorname{H}_i(C(x))\xrightarrow{\beta_*}\operatorname{H}_{i-1}(C) \xrightarrow{\delta_{i-1}} \operatorname{H}_{i-1}(C)\to \cdots, \end{equation*}
where here we have used the identification \(\operatorname{H}_i(\mathsf{\Sigma}C)\) with \(\operatorname{H}_{i-1}(C)\text{.}\)

Proof.

Proof. This is a classic diagram chase argument. Consider the commutative diagram
Let \(\overline{z}\in \operatorname{H}_i(\mathsf{\Sigma}C)=\operatorname{H}_{i-1}(C)\text{,}\) where \(z\in C_{i-1}\) is a cycle. Lift \(z\) via \(\beta\) to \(((-1)^{|z|}z,0)\in C_{i-1}\oplus C_{i}\text{.}\) Then \(\partial^{C(x)}((-1)^{|z|}z,0)=(0,xz)\) and \(\alpha^{-1}(0,xz)=xz\text{.}\) (Note: here we are using \(z\) is a cycle in \(C\text{.}\)) Hence, \(\delta_{i-1}(\overline{z})=\overline{xz}=x\overline{z}\text{,}\) where \(\overline{*}\) denotes image in homology. ◻

Proof.

Proof. We proceed by induction on \(n\text{.}\) In the case \(n=1\text{,}\) \(K^M\) is the complex
\begin{equation*} 0\to M\xrightarrow{x_1}M\to 0. \end{equation*}
It is clear part (a) holds and that \(H_0(K^M)\cong M/x_1M\text{,}\) \(H_1(K^M)=(0:_Mx_1)\) and \(\partial(K^M_1)=x_1M\text{.}\) If \(x_1\) is regular on \(M\) then \(\operatorname{H}_1(K^M)=0\text{.}\)
Suppose \(n>1\) and all parts hold for smaller values. Let \(\bold x' =x_1,\dots,x_{n-1}\) and \(C=K(\bold x';M)\text{.}\) Note that \(K^M=C(x_n)\text{.}\) From Construction [cross-reference to target(s) "const" missing or not unique] we have that \(K^M_i=C(x_n)_i\cong C_{i-1}\oplus C_i\text{.}\) By the \(n-1\) case, we have \(C_{i-1}\cong M^{\binom{n}{i-1}}\) and \(C_i\cong M^{\binom{n}{i}}\) for all \(i\text{.}\) Part (a) now follows. Furthermore, \(\partial^C(C_i)\subseteq (\bold x')C_i\) for all \(i\text{.}\) From Construction [cross-reference to target(s) "const" missing or not unique], we have
\begin{equation*} \partial(K^M_i)=\partial(C(x)_i)\subseteq ((\bold x')C_{i-2}, (\bold x')C_{i-1} + x_nC_{i-1})\subseteq (\bold x)C(x)_i. \end{equation*}
For the remaining parts, we have by Construction [cross-reference to target(s) "const" missing or not unique] and Lemma [cross-reference to target(s) "connect" missing or not unique] the long exact sequence
\begin{equation*} \cdots \to \operatorname{H}_i(C)\to \operatorname{H}_i(K^M)\to \operatorname{H}_{i-1}(C) \xrightarrow{x_n} \operatorname{H}_{i-1}(C)\to \cdots \end{equation*}
By the induction hypothesis, \(\operatorname{H}_{n-1}(C)=(0:_M (\bold x'))\) and \(\operatorname{H}_0(C)\cong M/(\bold x')M\text{.}\) From the l.e.s. above, we have the exact sequences
\begin{equation*} 0\to H_n(K^M)\to (0:_M (\bold x'))\xrightarrow{x_n} (0:_M (\bold x')) \end{equation*}
and
\begin{equation*} M/(\bold x')M \xrightarrow{x_n} M/(\bold x')M\to \operatorname{H}_0(K^M)\to 0. \end{equation*}
This proves (b) and (c).
Suppose \(\bold x\) is an \(M\)-sequence. By induction, we have that \(\operatorname{H}_i(C)=0\) for all \(i\geqslant 1\text{.}\) From the long exact sequence above we obtain \(\operatorname{H}_i(K^M)=0\) for \(i\geqslant 2\text{.}\) It also yields the exact sequence
\begin{equation*} 0\to \operatorname{H}_1(K^M)\to M/(\bold x')M\xrightarrow{x_n} M/(\bold x')M. \end{equation*}
Since \(x_n\) is a regular element on \(M/(\bold x')M\) we conclude \(\operatorname{H}_1(K^M)=0\text{.}\) ◻
The converse of part (d) of Proposition [cross-reference to target(s) "kos-basic" missing or not unique] holds in strengthened form under certain conditions:

Proof.

Proof. From Proposition [cross-reference to target(s) "kos-basic" missing or not unique], we have (a) implies (b), and the implication (b) to (c) is obvious. We prove (c) implies (a) by induction on \(n\text{.}\) Suppose \(n=1\text{.}\) Since \((0:_M x_1)=\operatorname{H}_1 K(x_1;M)=0\) we have that \(x_1\) is a non-zero-divisor on \(M\text{.}\) And as \(x_1M\neq M\) by Nakayama’s lemma, we see that \(x_1\) is \(M\)-regular.
Suppose that \(n>1\text{.}\) As in the proof of Proposition [cross-reference to target(s) "kos-basic" missing or not unique], we have the long exact sequence
\begin{equation*} \cdots \to \operatorname{H}_i(C)\to \operatorname{H}_i(K^M)\to \operatorname{H}_{i-1}(C) \xrightarrow{x_n} \operatorname{H}_{i-1}(C)\to \cdots, \end{equation*}
where \(C=K(\bold x';M)\) and \(\bold x'=x_1,\dots,x_{n-1}\text{.}\) From the assumption \(\operatorname{H}_1(K^M)=0\) we obtain the exact sequence
\begin{equation*} \operatorname{H}_1(C)\xrightarrow{x_n} \operatorname{H}_1(C)\to 0. \end{equation*}
Thus, \(x\operatorname{H}_1(C)=\operatorname{H}_1(C)\text{.}\) Since \(x_n\in J(R)\) and \(\operatorname{H}_1(C)\) is finitely generated (since \(C_1\cong M^{n-1}\) is Noetherian), we obtain that \(\operatorname{H}_1(C)=0\text{.}\) By the induction hypothesis, \(\bold x'\) is an \(M\)-sequence. From the exact sequence
\begin{equation*} 0\to \operatorname{H}_0(C)\xrightarrow{x_n} \operatorname{H}_0(C) \end{equation*}
we conclude that \(x_n\) is a non-zero-divisor on \(M/(\bold x')M\text{.}\) Since \((\bold x)M\neq M\) by Nakayama, we conclude that \(\bold x\) is an \(M\)-sequence. ◻

Proof.

Proof. This is clear from parts (a) and(e) of Proposition [cross-reference to target(s) "kos-basic" missing or not unique].

Proof.

Proof. Note that \(K(\bold x)\) is a free resolution of \(R/(\bold x)\) be the previous resolution. From part (b) of Proposition [cross-reference to target(s) "kos-basic" missing or not unique], \(\partial(K(\bold x))\subseteq (\bold x)K(\bold x)\subseteq mK(\bold x)\text{.}\) Hence, \(K(\bold x)\) is minimal. ◻

Proof.

Proof. Let \(F_{\bullet}\) be a minimal free resolution of \(M\text{.}\) Let \(F_{\bullet}\) be
\begin{equation*} 0\to F_t\xrightarrow{\phi_t} F_{t-1}\to \cdots \to F_0\to 0, \end{equation*}
where \(\phi_t(F_t)\subseteq mF_{t-1}\text{.}\) Applying \(\operatorname{Hom}_R(-,N)\text{,}\) we have the complex
\begin{equation*} \operatorname{Hom}_R(F_{t-1},M)\xrightarrow{\phi_t^*} \operatorname{Hom}_R(F_{t},M)\to 0. \end{equation*}
It is easy to see that \(\phi_t^*(\operatorname{Hom}_R(F_{t-1},N))\subseteq m\operatorname{Hom}_R(F_t,N)\text{.}\) Since \(\operatorname{Hom}_R(F_t,N)\neq 0\text{,}\) we see that \(\operatorname{Ext}^t_R(M,N)=\operatorname{coker}\phi_t^*\neq 0\) by Nakayama. ◻

Proof.

Proof. Let \(p=\operatorname{depth}R\) and \(r=\operatorname{id}_R M\text{.}\) Let \(\bold x=x_1,\dots,x_p\in m\) be a maximal \(R\)-sequence. Then \(\operatorname{pd}_R R/(\bold x)=p\) by Corollary [cross-reference to target(s) "minl" missing or not unique]. Hence, by Lemma [cross-reference to target(s) "pd" missing or not unique], we obtain that \(\operatorname{Ext}^p_R(R/(\bold x),M)\neq 0\text{.}\) Thus, \(r=\operatorname{id}_R M\geqslant p\text{.}\) Also note \(\operatorname{depth}R/(\bold x)=0\) since \(\bold x\) is a maximal \(R\)-sequence. Thus, \(m\in \operatorname{Ass}_R R/(\bold x)\text{,}\) so there exists an injection \(R/m\hookrightarrow R/(\bold x)\text{.}\) Applying \(\operatorname{Hom}_R(-,M)\) to the resulting short exact sequence,
\begin{equation*} \cdots\to \operatorname{Ext}^r_R(R/(\bold x), M)\to \operatorname{Ext}^r_R(R/m,M)\to 0, \end{equation*}
where we have used that \(\operatorname{Ext}^i(A,M)=0\) for all \(i>r=\operatorname{id}_RM\) for all \(R\)-modules \(A\text{.}\) Since \(\operatorname{id}_R M=r\text{,}\) we have \(\operatorname{Ext}^r_R(R/m,M)\neq 0\) by Proposition [cross-reference to target(s) "bass-id" missing or not unique]. Thus, \(\operatorname{Ext}^r_R(R/(\bold x),M)\neq 0\text{,}\) which implies \(r\leqslant p\text{.}\) ◻
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