Let $R$ be a Noetherian commutative ring and $x$ and $y$ two elements in $R$. We 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 $CD:=C\otimes D$, $$CD_2=C_1\otimes D_1$$ $$CD_1=C_1\otimes D_0 \oplus C_0\otimes D_1 $$ $$CD_0=C_0\otimes D_0$$ and we get the chain complex $$CD_3=0 \to CD_2\xrightarrow{\ \partial_2\ } CD_1\xrightarrow{\ \partial_1\ } CD_0 \to CD_{-1} =0$$ where

$$\partial_2 (c_1\otimes d_1)=(-c_1\otimes (yd_1)\,,\,(xc_1)\otimes d_1)$$ and $$\partial_1 (c_1\otimes d_0\,,\,c_0\otimes d_1)=(xc_1)\otimes d_0+c_0\otimes (yd_1)$$ It is clear that $\partial_1\circ\partial_2=0$

Question: How do we define the multiplication $CD_i\times CD_j\to CD_{i+j}$ on $(CD,\partial)$ that will make it a graded differential algebra; for example if we take two elements in $CD_1=C_1\otimes D_0 \oplus C_0\otimes D_1$, $(c_1\otimes d_0,c_0\otimes d_1)$ and $(c'_1\otimes d'_0,c'_0\otimes d'_1)$ how do we multiply them to obtain an element in $CD_2=C_1\otimes D_1$ ?

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    $\begingroup$ This doesn't exactly answer your question, but it is maybe worth mentioning here that it is perhaps easier to understand the algebraic structure of the Koszul complex if one thinks of it as an exterior algebra equipped with a differential (as in section 1.6 of "Cohen-Macaulay Rings" by Bruns and Herzog). $\endgroup$
    – user55407
    Mar 2 '13 at 17:58
  • $\begingroup$ Well, given $A, B$ two dga over $R$, can you see that $A\otimes_R B$ is then a naturally a dga? In your case, $C_\bullet = R[x]/(x^2)$, $D_\bullet = R[y]/(y^2)$ are dga with differential $x, y$ respectively. $\endgroup$
    – Aaron
    Mar 4 '13 at 0:47
  • $\begingroup$ can you please be more specific and tell me how do we define the multiplication $CD_i\times CD_j\rightarrow CD_{i+j}$? Thanks $\endgroup$
    – palio
    Mar 4 '13 at 15:03
  • $\begingroup$ suppose $a_1 , a_2 \in C_\bullet; b_1,b_2\in D_\bullet$, then $(a_1\otimes b_1), (a_2\otimes b_2) \in C_\bullet\otimes D_\bullet = CD$, now multiplication is just $a_1a_2\otimes b_1b_2$. You can put all the degree back by assuming all the elements are homogeneous $\endgroup$
    – Aaron
    Mar 4 '13 at 23:06
  • $\begingroup$ If i understand, the multiplication of two elements of $CD_1=C_1\otimes D_0\oplus C_0\otimes D_1$ is $(c_1\otimes d_0\,,\,c_0\otimes d_1)*(c'_1\otimes d'_0\,,\,c'_0\otimes d'_1)=(c_1c'_1\otimes d_0d'_0\,,\, c_0c'_0\otimes d_1d'_1)$ but this is not in $CD_2=C_1\otimes D_1$ as is supposed to be a gradation on $CD$, am i missing something? $\endgroup$
    – palio
    Mar 5 '13 at 5:19

To my understanding you are trying to consider $2$ differential graded algebras (dgas) $(C,d_C)$ and $(D,d_D)$ and to find the associative product on the differential graded algebra $(C\otimes D, d_{C\otimes D})$. In this case, the formula written by Aaron in his comment is correct.

Your complexes $C$ and $D$ are no dgas, though for any associative ring $R$, though. The Leibniz rule fails. Check this example in $C$:

$d_C( t \star w)= d_C(t)\star w - t \star d_C(w)$, with $t,w\in C_1=R$ and associative product $\star: C \otimes C \rightarrow C$ on $C$ induced by the multiplication on $R$. Note that $t \star w=0=C_2$, though. The r.h.s. of the above relation is $(xt)w-t(xw) \in C_1$, instead.

Consider that the Koszul complex is in general a free resolution of the given ring $R$ as left $R$-module; what is important in the applications is the module structure. You can even consider the above Koszul complex as an $R$-$R$-bimodule.


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