# Explicit example of Koszul complex

Let $R$ be a commutative ring and $x$ and $y$ two elements in $R$. I want to construct the Koszul complex on $x$ and $y$. We start by the following two chain complexes

$$C_2=0\to C_1=R\xrightarrow{\ x\ } C_0= R\to C_{-1}=0$$ $$D_2=0\to D_1=R\xrightarrow{\ y\ } D_0= R\to D_{-1}=0$$ Now we construct the tensor product chain complex which we denote $CD:=C\otimes D$: $$CD_2=C_1\otimes D_1=R\otimes R$$ $$CD_1=C_1\otimes D_0 \oplus C_0\otimes D_1 =R\otimes R \oplus R\otimes R$$ $$CD_0=C_0\otimes D_0=R\otimes R$$ and we get the chain complex $$CD_3=0 \to CD_2=R \otimes R\xrightarrow{\ \partial_2\ } CD_1= R \otimes R \oplus R \otimes R \xrightarrow{\ \partial_1\ } CD_0= R \otimes R \to CD_{-1} =0$$ We now compute $\partial_1$ and $\partial_2$:

$$\partial_2 (c_1\otimes d_1)=(xc_1)\otimes d_1-c_1\otimes (yd_1)$$ and $$\partial_1 (c_1\otimes d_0+c_0\otimes d_1)=(xc_1)\otimes d_0+c_0\otimes (yd_1).$$

Now I want to move from here to express $\partial_1$ and $\partial_2$ in the way expressed in the wikipedia page (section Introduction). I don't understand the notation $R^2$ and the matrix expression of the differentials and where did the tensor product disappear from the final result.

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$R^2 = R \oplus R$, which explains the matrices. The tensor product disappeared because $R\otimes_R R\cong R$. – m_t_ Dec 14 '12 at 9:57
I still don't see how to get matrix expression for $\partial_1$ and $\partial_2$, can we say for example that $$\partial_2(c_1\otimes d_1)=\mathbf (x, -y) \begin{bmatrix} c_1\otimes d_1 & 0\\0 & c_1\otimes d_1 \end{bmatrix}$$ – palio Dec 14 '12 at 10:19
and $$\partial_1 (c_1\otimes d_0+c_0\otimes d_1)=\mathbf (x, y) \begin{bmatrix} c_1\otimes d_0\\ c_0\otimes d_1 \end{bmatrix}$$ – palio Dec 14 '12 at 10:23
Also why is that $R\otimes_R R \cong R$? – palio Dec 14 '12 at 10:25
@palio: as for your last question: convince yourself that the map $R \rightarrow R \otimes_R R$, $r \mapsto 1 \otimes r$ is an isomorphism. – Nils Matthes Dec 14 '12 at 10:58

As preliminary computation, I suggest you to consider the classical example of Koszul complex $(K,d):=(S(X^{*})\otimes \wedge (X^{*}), d)$, where $X$ is a finite dim. vector space over $\mathbb R$ or $\mathbb C$ and $S(X^{*})$, resp. $\wedge (X^{*})$ denote the symmetric resp. exterior algebra over $X$.

It is helpful because you can consider the symmetric and the exterior algebras as suitable algebra of polynomials and the construction is really explicit.

Let us specify that the Koszul complex is negatively graded, i.e.

$K^{-p}:= S(X^{*})\otimes \wedge^{p} (X^{*}) \rightarrow S(X^{*})\otimes \wedge^{p-1} (X^{*}) \rightarrow \dots \rightarrow S(X^{*})\otimes \wedge^{0} (X^{*}) \rightarrow \mathbb K$

in order to have a degree $+1$ differential $d$, which is given (on monomials in $\wedge^{p} (X^{*})$) by

$d(q \otimes (x_1 \wedge\dots\wedge x_p)):=\sum_{i=1}^{p}(-1)^{i-1} q\cdot x_i \otimes (x_i\wedge\dots \wedge\hat{x}_i\wedge\dots\wedge x_p))$, where $\hat{\cdot}$ denotes omission and $q\cdot x_i$ is the product in the symmetric algebra of the polynomial $q$ and the monomial $x_i$. The sign $(-1)^{i-1}$ comes from the fact that $x_i$ "surpasses" $x_1\wedge\dots \wedge x_{i-1}$ to reach the polynomial $q$ on the left.

This formula for the differential is exactly the one which appears in the Wiki-article on the Koszul complex you mention in your question. The reduction to your case is now straightforward.

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Your formulae seem correct to me; one short comment. The unital ring $R$ is an $R$-$R$-bimodule, where the bimodule structure is induced by the associative product in $R$ (check it). Then the tensor product $R\otimes_R R$ is well defined (because the left copy of $R$ in the product is seen as an $R$-right module, while the copy on the right as an $R$-left module) and it is an $R$-$R$-bimodule as well. If we focus only on the left $R$-module strucure on both $R$ and $R\otimes_R R$, then the map in Nils Matthes' comment above is an isomorphism of left $R$-modules (with $\otimes_R$!) – Avitus Jun 4 '13 at 9:09
Do you have a reference for the content of your post? Thanks. – lentic catachresis Mar 31 '15 at 17:01
Some references: Weibel's book on Homological algebra, ,this paper archive.org/details/arxiv-1002.2561 and Loday & Vallette's book on Operads – Avitus Mar 31 '15 at 19:04

As you were told in the comments $\partial_2:R=R\otimes R\to (R\otimes R)\oplus (R\otimes R)=R^2$, and $$\partial_2(1)=\partial_2(1\otimes 1)=(-1\otimes y,x\otimes 1)=(-y,x).$$

Similarly, $\partial_1:R^2=(R\otimes R)\oplus (R\otimes R)\to R\otimes R=R$, and $$\partial_1(1,0)=\partial_1(1\otimes 1,0\otimes 0)=x\otimes 1=x$$ $$\partial_1(0,1)=\partial_1(0\otimes 0,1\otimes 1)=1\otimes y=y.$$

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