TITLE: Classification of commutative Frobenius algebras
QUESTION [8 upvotes]: 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.

REPLY [3 votes]: I believe the answer to question 1 is no. But the question is slightly ambiguous. I can give a quiver presentation with only $\pm 1$ appearing as coefficients in the relations and where being Frobenius depends on the characteristic of the field.  Maybe you are asking if you can always find some quiver presentation of a commutative Frobenius algebra with $\pm 1$ coefficients that remains Frobenius over all fields. I suspect that the examples I give would fail this somewhat different question but I will stick with my interpretation.
I'll give a construction via a quiver presentation but the quiver just has loops at one vertex so this is basically a monoid presentation.  Let $A$ be an $n\times n$ symmetric matrix with 0/1 entries with $n\geq 2$.  We have a quiver $Q$ with $n$-loops at a single vertex labeled $x_1,\ldots,x_n$. We take as relations that all paths of length $3$ are zero, $x_ix_j =0$ if $A_{ij}=0$ and make all $x_ix_j$ with $A_{ij}=1$ equal to each other.  This gives an admissible ideal $I$ and all the generators of $I$ are either monomial or a difference of monomials.     The algebra is commutative since $A$ is symmetric. It’s essentially an algebera of a finite $3$-nilpotent commutative semigroup with adjoined identity. Notice all paths of length two which are not zero in $KQ/I$ are equal.  Also $KQ/I$ is local since there is one vertex in the quiver.
I claim that if $K$ is a field, $KQ/I$ is Frobenius iff $\det A\neq 0$ in the field $K$, i.e., the characteristic of $K$ does not divide $\det A$.
The radical of $KQ/I$ is spanned by $x_1,\ldots, x_n$ (or rather their cosets but I ignore the distinction).
If $A=0$, then the socle is spanned by $x_1,\ldots, x_n$ and hence is not simple.
So assume $A\neq 0$. So there is at least one nonzero product of two loops. All these nonzero products are the same element $z$ of $KQ/I$.  Clearly $z$ is in the socle.  So we need that nothing else is in the socle.  First note that the $1,x_1,\ldots, x_n$ and the element $z$ form a basis for $KQ/I$ and I shall identify the former elements with their coset.  To be in the socle is to be annihilated by $x_1,\ldots, x_n$.  Such an element must not be a unit, so it belongs to the radical.  Therefore, it must be of the form $a=c_1x_1+\cdots+c_nx_n+cz$.  But then it is straightforward from the defining relations to see $x_ia=\sum_{j=1}^nA_{ij}c_jz$ and so $x_ia=0$ if and only if row $i$ of $A$ dot product with $(c_1,\ldots, c_n)=0$.  Thus $a$ is in the socle iff $(c_1,\ldots, c_n)$ is in the nullspace of $A$ and so the socle is just $Kz$ iff $\det A\neq 0$.
For a concrete example let $J_n$ be the $n\times n$ all ones matrix and $B_n=J_n-I_n$.  So all diagonal entries are zero and all other entries are $1$.  So our defining relations are $x_i^2=0$ for all $i$ and $x_ix_j=x_kx_\ell$ whenever $i\neq j$ and $k\neq \ell$.  It is well known that $\det B_n =\pm (n-1)$.  This is because the characteristic polynomial of $J_n$ is $p(\lambda)=(\lambda -n)\lambda^{n-1}$ and so $\det B_n = (-1)^np(1)=\pm (n-1)$. If $p$ is a prime, taking $n=p+1$ you get an algebra which is Frobenius in all characteristics except $p$. More generally, for any finite set of primes, by taking $n$ one larger than the product of those primes you get an algebra which is Frobenius in all characteristics except those in your finite set.