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 $t \geq 2$.



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> Question 1: Is $A\_{n,t}$ for general $n,t$ a Frobenius algebra?
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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).



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> 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).
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> Question 3: Is there a generalisation of question 1 in case question 1 is true?
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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)*x*y], [(1)*x*z], [(1)*y^2], [(1)*y*z],
[(1)*x^3], [(1)*x^2*y], [(1)*x^2*z], [(1)*x*y^2], [(1)*x*y*z], [(1)*y^3], [(1)*y^2*z],
[(1)*x^3*z], [(1)*x^2*y^2], [(1)*x^2*y*z], [(1)*x^4+(2)*x^3*z+(-2)*x*y^3], [(1)*x*y^3],
[(1)*x^3*y+(1)*x*y^3], [(1)*x*y^2*z], [(1)*y^4], [(1)*x^4*z], [(1)*x^3*y^2], [(1)*x^3*y*z],
[(1)*x^5+(2)*x^4*z+(-4)*x^3*y^2+(-4)*x^3*y*z+(-2)*x^2*y^3],
[(1)*x^4*y+(2)*x^3*y^2+(2)*x^3*y*z+(1)*x^2*y^3], [(1)*x^2*y^2*z], [(1)*x^2*y^3+(1/2)*x*y^4],
[(1)*x*y^4], [(1)*x^4*y*z], [(1)*x^3*y^3], [(1)*x^5*y+(2)*x^4*y^2+(2)*x^4*y*z+(1)*x^3*y^3],
[(1)*x^6+(4)*x^5*y+(2)*x^5*z+(4)*x^4*y^2+(4)*x^4*y*z+(2)*x^3*y^3], [(1)*x^2*y^4],
[(1)*x^4*y^2+(2)*x^3*y^3+(1)*x^2*y^4], [(1)*x^5*z+(2)*x^2*y^4], [(1)*x^6*z+(-2)*x^5*y^2],
[(1)*x^5*y^2], [(1)*x^7+(3/2)*x^6*y+(3)*x^5*y^2], [(1)*x^4*y^3],
[(1)*x^6*y+(2)*x^5*y^2+(4)*x^4*y^3], [(1)*x^7*y], [(1)*x^6*y^2],
[(1)*x^8+(3/2)*x^7*y+(3)*x^6*y^2], [(1)*x^8*y] ]`



Frobenius algebras from symmetric polynomials
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 $\prod\_{1\leq
i<j\leq n}(x\_i-x\_j)^t$.


