TITLE: Algebra structure $Tor(A,A)$
QUESTION [7 upvotes]: 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?

REPLY [5 votes]: 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 $k$!)