Consider the associative algebra A with generators $T\_i$ and rule $T\_i\*T\_j=\Sigma\_kC^{ij}\_k\*T\_k$. Even if it makes no sense for a fusion ring (my momentary pet :-) to change basis it is still possible to linear transform the basis (via $T\_j=M\_{ij}E\_i$) into diagonal form: $E\_i\*E\_j=\delta\_{ij}\*E\_i$. (Or so I think.)  

1. Is commutativity (which I always assume tacitly) necessary? Is associativity necessary and/or sufficient? (For that A in fact can be "diagonalized" with some $M\_{ij}$)  

2. Has this M a name then? I compute-guessed a few cute relations of M and (assuming A is a fusion ring) the Verlinde matrix S of A (which then surely are known).



"Diagonalizing" an associative algebra
I misread your question. My comment alluded to the question whether you can have a basis such that $E\_i \* E\_j = $ *some* single $E\_k$, (as opposed to a linear combination of all of them) but what you ask is much stronger: does every commutative associative algebra have a basis of orthogonal idempotents. This implies commutativity: it is true if and only if your algebra is (as an algebra) isomorphic to a direct sum of copies of the ground field.


From here we find another necessary condition: no nilpotent elements. On the other hand we know from the Artin-Wedderburn structure theorem that if your algebra is commutative and semi-simple it must have this form. 


I'm not sure if commutative and without nilpotent elements are also sufficient conditions in the infinite dimensional case. In the finite dimensional case I believe they are: finite dimensional algebras are Artinian, which means that the Jacobson radical coincides with the nilradical, hence if the latter is zero, so is the former, which in turn implies semi-simplicity and we can apply Artin-Wedderburn. (I hope I got this right, otherwise please correct.)


