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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.

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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 $k$.

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    $\begingroup$ Thanks for your answer but by isomorphic I mean isomorphic as k-algebras. So any 1-dimensional k-algebras are isomorphic. I think you mean isomorphic in a more strict sense (as algebra and coalgebra?). $\endgroup$ – Mare Apr 26 at 8:28
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    $\begingroup$ The Frobenius form can be read of from the socle of the algebra which is determined by the quiver and relations (the socle is the "largest non-zero" path). $\endgroup$ – Mare Apr 26 at 8:31
  • $\begingroup$ @Mare indeed, I meant isomorphic as Frobenius algebras; sorry for the confusion. Since a Frobenius algebra is additional structure, rather than a condition, that seemed more natural to me, but what you're thinking about is still an interesting question. $\endgroup$ – Arun Debray Apr 26 at 14:38
  • $\begingroup$ I didn't realize you could get the Frobenius form from the quiver and relations! That's neat; I'll have to look into it. $\endgroup$ – Arun Debray Apr 26 at 14:39

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