# Frobenius algebras from symmetric polynomials

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] ] 

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.