Let $K$ be a field of characteristic 0 (maybe it works for more general fields) and $K[x_1,...,x_n]$ the polynomial ring in $n$ variables. Let $e_1,e_2,...,e_n$ denote the elementary symmetric polynomials in $n$ variables (see for example https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial ). (One might also choose other symmetric polynomials such as https://en.wikipedia.org/wiki/Complete_homogeneous_symmetric_polynomial or https://en.wikipedia.org/wiki/Power_sum_symmetric_polynomial for this problem ).

For $n,t \geq 2$ let $A_{n,t}:=K[x_1,...,x_n]/(e_1^t,...,e_n^t)$. $A_{n,t}$ is by definition a Frobenius algebra if there is a unique polynomial up to scalars $p \in A_{n,t}$ with $p x_i=0$ for $i=1,...,n$. This is a purely combinatorical condition.

For example for $n=t=2$, we have $A_{2,2}$ has vector space basis $1,x_1,x_2,x_1^2,x_1 x_2,x_1^3,x_1^2 x_2,x_1^4$ and indeed the unique longest polynomial in $A_{2,2}$ is $p=x_1^4$. Note that we have in $A_{2,2}$ $x_1^4=-x_2^4$ and unique means here really in the algebra $A_{2,2}$ and not as a polynomial in the polynomial ring. I was able to prove by force that $A_{2,t}$ is a Frobenius algebra for $t \geq 2$.

Question 1: Is $A_{n,t}$ for general $n,t$ a Frobenius algebra?

In case this is true in general, there is probably a nice proof that avoid heavy computation (and might work for the other classes of symmetric polynomials such as the power sum symmetric polynomials).

Question 2: Did those algebras appear in the literature already? Do they have other nice properties such as being Hopf algebras in special cases (maybe allowing fields of certain characteristics).

Question 3: Is there a generalisation of question 1 in case question 1 is true?

For example small experiments suggest that even the algebras $K[x_1,...,x_n]/(e_1^{t_1},...,e_n^{t_n})$ are Frobenius as long as $t_1 \geq 2$ and $t_i \geq 1$. But maybe one can also do more, for example taking certain polynomials of the $e_i$.

The problem gets complicated quickly for larger $n$. Here is the case $n=3, t=2$ with the computer over the rationals($x_1=x,x_2=y,x_3=z$):

$A_{3,2}$ has vector space dimension 48 and longest polynomial for example $x^8y$. A vector space basis via the computer is given as follows: [ [(1)*v1], [(1)*x], [(1)y], [(1)z], [(1)x^2], [(1)xy], [(1)xz], [(1)y^2], [(1)yz], [(1)x^3], [(1)x^2y], [(1)x^2z], [(1)xy^2], [(1)xyz], [(1)y^3], [(1)y^2z], [(1)x^3z], [(1)x^2y^2], [(1)x^2yz], [(1)x^4+(2)x^3z+(-2)xy^3], [(1)xy^3], [(1)x^3y+(1)xy^3], [(1)xy^2z], [(1)y^4], [(1)x^4z], [(1)x^3y^2], [(1)x^3yz], [(1)x^5+(2)x^4z+(-4)x^3y^2+(-4)x^3yz+(-2)x^2y^3], [(1)x^4y+(2)x^3y^2+(2)x^3yz+(1)x^2y^3], [(1)x^2y^2z], [(1)x^2y^3+(1/2)xy^4], [(1)xy^4], [(1)x^4yz], [(1)x^3y^3], [(1)x^5y+(2)x^4y^2+(2)x^4yz+(1)x^3y^3], [(1)x^6+(4)x^5y+(2)x^5z+(4)x^4y^2+(4)x^4yz+(2)x^3y^3], [(1)x^2y^4], [(1)x^4y^2+(2)x^3y^3+(1)x^2y^4], [(1)x^5z+(2)x^2y^4], [(1)x^6z+(-2)x^5y^2], [(1)x^5y^2], [(1)*x^7+(3/2)x^6y+(3)x^5y^2], [(1)x^4y^3], [(1)x^6y+(2)x^5y^2+(4)x^4y^3], [(1)x^7y], [(1)x^6y^2], [(1)*x^8+(3/2)x^7y+(3)x^6y^2], [(1)x^8y] ]


1 Answer 1


In this commutative situation, a Frobenius algebra is the same as an artinian Gorenstein ring. In general, if $\theta_1,\dots,\theta_n$ are homogeneous elements of positive degree of $A=K[x_1,\dots,x_n]$ and if $K[x_1,\dots,x_n]/(\theta_1,\dots,\theta_n)$ is artinian (i.e., a finite-dimensional vector space in this situation), then $A/(\theta_1,\dots,\theta_n)$ is Gorenstein (in fact, a complete intersection, which is stronger). Since $A/(e_1,\dots,e_n)$ is artinian the same is true for $A/(e_1^t,\dots,e_1^t)$, so the Gorenstein property follows. In fact, the socle polynomial $p$ is given by $\prod_{1\leq i<j\leq n}(x_i-x_j)^t$.


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