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This is a question i asked on math.stackexchange but i didn't get any answer.

Let $A$ be algebra over commutative ring $k$ and $P_{\bullet}=(P_i,d_i)\rightarrow A$, $k$ projective resolution. Then we have obvious lift of multiplication $f:A\otimes A\rightarrow A$ to $F:P\otimes P \rightarrow P$. Of course there is no reason for $F$ to be associative so we can't claim that $P_{\bullet}$ has algebra structure. Consider the map

$Tor_*(A,A)\otimes Tor_*(A,A)=H_*(P_{\bullet}\otimes P_{\bullet})\otimes H_*(P_{\bullet}\otimes P_{\bullet})\rightarrow H_*(P_{\bullet}\otimes P_{\bullet}\otimes P_{\bullet}\otimes P_{\bullet})\rightarrow H_*(P_{\bullet}\otimes P_{\bullet})=Tor_*(A,A)$

I was wondering if this map gives graded algebra structure on $Tor_*(A,A)$. Do i need any assumptions?

Edit: Can you give me references to any papers about those algebras?

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    $\begingroup$ $F$ is a lift of an associative map so is associative up to chain homotopy. $\endgroup$ – Qiaochu Yuan Nov 15 '13 at 18:29
  • $\begingroup$ I believe the Tor algebra has been studied as an $A_\infty$-algebra, can't find the paper right this moment. $\endgroup$ – John Salvatierrez Nov 15 '13 at 18:34
  • $\begingroup$ This is essentially done in Cartan-Eilenberg (!) $\endgroup$ – Mariano Suárez-Álvarez Nov 15 '13 at 19:22
  • $\begingroup$ @MarianoSuárez-Alvarez - Sorry I'm not really familiar with this book and can't find it. Could you tell me where exactly should i look. $\endgroup$ – user35861 Nov 15 '13 at 19:43
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    $\begingroup$ The book has a chapter entitled «Products.» You'll find there indications about the proof of associativity. $\endgroup$ – Mariano Suárez-Álvarez Nov 16 '13 at 0:53
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More generally, if $A,B,C,D $ are $k$-algebras, there's a multiplication $$Tor_m^k(A,B)\otimes_k Tor_n^k(C,D)\rightarrow Tor^k_{m+n}(A\otimes_k C,B\otimes_k D)\qquad(1)$$ If you take $C=A$ and $B=D$, this becomes $$Tor_m^k(A,B)\otimes_k Tor_n^k(A,B)\rightarrow Tor^k_{m+n}(A\otimes_A A,B\otimes_k B)$$ and you can compose with the maps induced by multiplication on $A$ and $B$ to get $$Tor_m^k(A,B)\otimes_k Tor_n^k(A,B)\rightarrow Tor_{m+n}^k(A,B)$$

The resulting map is commutative up to a sign (namely $(-1)^{m+n}$).

To get (1), you can represent elements of $Tor_m^k(A,B)$ and $Tor_n^k(A,B)$ as cycles in he appropriate complexes, then tensor these cycles together to get a cycle in the total tensor product complex.

You can find the details, among other places, in Eisenbud's book on commutative algebra, where he mentions that a lot of work has been done on the structure of this algebra in the case where $k$ is a local ring and $A=B$ is the residue field. (Here's where I really wish you'd called your ring $R$ instead of $k$!)

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  • $\begingroup$ Thank you for the answer. For me R is not commutative, k commutative and A is an algebra. $\endgroup$ – user35861 Nov 15 '13 at 20:15

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