Telling group algebras apart It's a big, famous, hard problem in operator algebras to determine if the von Neumann algebras $L(F_2)$ and $L(F_3)$ are isomorphic, or not.  Here $F_n$ is the free group on n generators and $L(F_n)$ is the weak-operator-topology closure of the group algebra $\mathbb C[F_n]$ acting naturally on the Hilbert space $\ell^2(F_n)$.
I presume it must be known if the algebras $\mathbb C[F_2]$ and $\mathbb C[F_3]$ are isomorphic or not.  But from casually asking a few algebraists, I've never had any luck in finding this out (I admit to not working very hard on this!)  I'm guessing some (co)homology theories must help...?  What about for replacing $\mathbb C$ by a more general ring?
 A: One has $Hom({\mathbb C}[F_n],{\mathbb C}) = ({\mathbb C}^{\times})^n$ with the obvious topology. (Here, $Hom$ denotes the space of $\mathbb C$-linear homomorphisms.) This of course uses a little bit more than only the algebra structure, but every ${\mathbb C}$-linear isomorphism would preserve the topology on the space of $\mathbb C$-linear representations. Since the spaces $({\mathbb C}^{\times})^n$ are not homeomorphic for different $n$, the claim follows. The same applies to the maximal group $C^{\star}$-algebra of $F_n$. One has $Hom(C^{\star}(F_n),{\mathbb C}) = (S^1)^n$, where one considers only $\star$-homomorphisms.
A: Well, yes.  Imagine that you have an algebra $A$ over $\mathbb{C}$ and you want to find out whether it is $\mathbb{C}[F_2]$ or $\mathbb{C}[F_3]$.  Pick any one-dimensional $A$-module $M$ and compute $\operatorname{Ext}^1_A(M,M)$.  If $A=\mathbb{C}[F_2]$, you'll get a $2$-dimensional vector space over $\mathbb{C}$, while if $A=\mathbb{C}[F_3]$, you'll get a $3$-dimensional vector space.
