# Classification of commutative Frobenius algebras

Are there attempts to classify commutative finite dimensional Frobenius algebras? They appear often in mathematics, such as in algebraic geometry and the famous category equivalence between commutative Frobenius algebras and 2-dimension topolocial quantum field theories. However, I have not yet seen attempts to classify this class of algebras (up to isomorphism of k-algebras). Recall that a commutative Frobenius algebra is a finite dimensional algebra $$A$$ with $$A \cong D(A)$$ or equivalently simple socle in case it is local.

Here are two questions related to such a classification (we can assume that commutative Frobenius algebra are connected):

Question 1: Is a commutative Frobenius algebra "field-independent"? This means that in its presentation $$KQ/I$$ by quiver and relations, there exists such $$I$$ which only contains the field element 1 or -1 so that a given commutative Frobenius algebra is defined over all fields.

In case question 1 has a positive answer, this would mean that a classification is independet of the field (maybe excluding characteristic 2).

Question 2: For a given integer $$d$$, are the only finitely many $$d$$-dimensional commutative Frobenius algebras of vector space dimension $$d$$?

(here we say that two algebras are isomorphic in case they are isomorphic as $$K$$-algebras)

A positive answer to question 2 would be surprising, but I think it should be true for $$d \leq 5$$ at least.

Question 2 has a negative answer, even in dimension $$d = 1$$. I'm not sure about question 1, but it's certainly not true that the classification is independent of the field.
For question 2: let $$k$$ be an infinite field and $$\lambda\in k^\times$$. Then the bilinear form $$\beta(a,b) := \lambda ab$$ is nondegenerate, defining a commutative Frobenius algebra structure on $$k$$. These Frobenius algebras are nonisomorphic for different $$\lambda$$, exhibiting infinitely many nonisomorphic one-dimensional commutative Frobenius algebras over $$k$$. These are the only Frobenius algebra structures on $$k$$: the space of bilinear forms $$k\times k\to k$$ is one-dimensional, and we've already used everything except the zero form, which doesn't define a Frobenius structure because it's not nondegenerate.
Question 1 is a little strange to me: the quiver-and-relations presentation I know of doesn't tell you a Frobenius form; it only defines an associative algebra. Because the Frobenius form is additional structure, a positive answer wouldn't imply that the classification of commutative Frobenius algebras is independent of $$k$$, as the possible Frobenius forms could depend on $$k$$. And indeed, that's what we saw above: fixing $$d = 1$$, the isomorphism classes of one-dimensional commutative Frobenius $$k$$-algebras are in bijection with $$k^\times$$, and this certainly depends on $$k$$.