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?



Algebra structure $Tor(A,A)$
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 $k$!) 


