Classification of commutative Frobenius algebras

Question

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.

Answer 1

Answering Question 2, I'll show that there are infinitely many $14$-dimensional commutative Frobenius algebras over an infinite field $k$ so that no two are isomorphic as algebras. Probably $14$ is not minimal.

It was shown by Suprunenko that there are infinitely many pairwise nonisomorphic $7$-dimensional commutative $k$-algebras, and in fact from the proof of this fact by Poonen in

Poonen, Bjorn, Isomorphism types of commutative algebras of finite rank over an algebraic closed field, Lauter, Kristin E. (ed.) et al., Computational arithmetic geometry. AMS special session, San Francisco, CA, USA, April 29–30, 2006. Providence, RI: American Mathematical Society (AMS) (ISBN 978-0-8218-4320-8/pbk). Contemporary Mathematics 463, 111-120 (2008). ZBL1155.13015. (Slightly updated version available online here.)

it follows that, among the following class of algebras parametrized by $\lambda\in k$, there are only finitely many in each isomorphism class. I'll assume that $\text{char}~k\neq2$, but an easy adjustment deals with the case $\text{char}~k=2$.

$B_\lambda$ will be the algebra with basis $\{1,x_1,x_2,x_3,x_4,s,t\}$, where $x_1^2=s$, $x_2^2=s+t$, $x_3^2=s+2t$ and $x_4^2=s+\lambda t$, and all other products of the last six basis elements are zero. The trivial extension algebras $TB_\lambda = B_\lambda\ltimes DB_\lambda$ are commutative Frobenius algebras, and I'll show, by a rather ad hoc argument, that $TB_\lambda\cong TB_\mu$ only if $B_\lambda\cong B_\mu$. I'm using $D$ here for the vector space dual.

Wakamatsu proved in

Wakamatsu, Takayoshi, Note on trivial extensions of Artin algebras, Commun. Algebra 12, 33-41 (1984). ZBL0537.16008.

that two finite dimensional algebras have isomorphic trivial extension algebras if and only if they are of the form $A\ltimes M$ and $A\ltimes DM$ for some algebra $A$ and $A$-bimodule $M$. Since our algebras are commutative, for us $M$ will just be an $A$-module (or a bimodule with the same action on both sides).

First, note that $\text{rad}^2B_\lambda=\text{soc}B_\lambda$ is two dimensional, spanned by $s$ and $t$.

If $TB_\lambda\cong TB_\mu$ but $B_\lambda\not\cong B_\mu$ then there must be a direct sum decomposition of vector spaces $B_\lambda=A\oplus M$ such that $A$ is a subalgebra, $M$ is a square zero ideal, and $B_\mu\cong A\ltimes DM$. Note that $M\not\cong DM$ or else $B_\lambda\cong B_\mu$. Also, $\text{soc}(A\ltimes DM)$ must be equal to $\text{rad}^2(A\ltimes DM)$ and be two-dimensional. I'll show that these things cannot happen.

Suppose they do.

If $M=S\oplus M'$ has a simple summand $S$, we can replace $A$ by $A\oplus S$ and $M$ by $M'$, so we can assume that $M$, which has radical length two, satisfies $\text{soc}M=\text{rad}M$. Since $\text{soc}(A\ltimes DM)=\text{rad}^2A\oplus\text{soc}DM$ is two dimensional, we must have $\dim\text{rad}M=\dim(M/\text{rad}M)$. It is easy to check that any two dimensional $B_\lambda$-module is self-dual, so $\dim\text{rad}M=\dim(M/\text{rad}M)=2$, and $A$ is three dimensional and $A\cap\text{soc}B_\lambda=0$.

So $\text{soc}A$ is (without loss of generality, as we can freely add multiples of $s$ and $t$ to a basis of $\text{soc}A$) a two dimensional subspace of the span of $\{x_1,x_2,x_3,x_4\}$ with square zero.

But there is no such subspace. If $x=ax_1+bx_2+cx_3+dx_4$ with $x^2=0$, then $a$ and $b$ are each determined up to two possibilities by $c$ and $d$. So in a two dimensional vector space of such elements, all pairs of values of $c$ and $d$ would have to occur, and it is easy to check that no choices of $a$ and $b$ would then give a subspace.

Answer 2

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.

Answer 3

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 $k$.