When is an algebra defined by generators and relations finite-dimensional and satisfies Poincaré duality?


**EDIT:** My question is not very concrete. Rather I am wondering if there is anything known in the following direction.


Assume we are given a commutative algebra $A$ (say over complex numbers) which is graded and is given by its homogeneous generators and homogeneous relations:
$$A=\mathbb{C}[x\_1,\dots,x\_k]/\langle f\_1,\dots,f\_l\rangle,$$
where $f\_i$ are homogenous polynomials in $x\_1,\dots,x\_k$.



> 
> 1) Are there sufficient conditions on $f\_i$'s to guarantee $A$ to be finite-dimensional?
> 
> 
> 2) Are there sufficient conditions on $f\_i$'s to guarantee $A$ satisfies Poincaré duality? (Poincaré duality means that $A\_j=0$ for $j>N$, $A\_0=\mathbb{C}, A\_N\simeq \mathbb{C}$, and the product $A\_j\times A\_{N-j}\to A\_N\simeq \mathbb{C}$ is a perfect pairing for any $j$.)
> 
> 
> 



When is an algebra defined by generators and relations finite-dimensional and satisfies Poincaré duality?
Say $I = (f\_1,\dotsc,f\_l)$ is the ideal generated by the $f\_i$. The $f\_i$ are homogeneous; let’s add an assumption that none of the $f\_i$ are constant (degree zero). The following conditions are equivalent:


1. $A$ is finite-dimensional (as a vector space over $\mathbb{C}$).
2. The [radical](https://en.wikipedia.org/wiki/Radical_of_an_ideal) of $I$ is the maximal graded ideal $(x\_1,\dotsc,x\_k)$.
3. For each $i$ there’s an $n\_i$ so that $x\_i^{n\_i} \in I$, or there’s an $n$ so that for each $i$, $x\_i^n \in I$.


This list can be extended. Perhaps some of the experts can suggest additional, simpler, conditions that are sufficient but not necessary.


Now suppose that $A$ is finite-dimensional. The following conditions are equivalent:


1. $A$ has Poincaré duality as you described it ($A\_N \cong \mathbb{C}$ and for every $n$, the multiplication map $A\_n \times A\_{N-n} \to A\_N \cong \mathbb{C}$ is a perfect pairing).
2. The [socle](https://en.wikipedia.org/wiki/Socle_(mathematics)#Socle_of_a_module) of $A$ is one-dimensional (as a $\mathbb{C}$ vector space).
3. $A$ is Gorenstein; $I$ is a Gorenstein ideal.
4. There is a homogeneous polynomial $F$ so that $I$ is the ideal of polynomials $g$ such that $g(\partial/\partial x\_1,\dotsc,\partial/\partial x\_k)(F) = 0$ (the ideal of annihilators of $F$).


If $I$ is a complete intersection ($l=k$ and $A$ is finite-dimensional) then $A$ is Gorenstein, has Poincaré duality, etc. This is an example of a sufficient, but not necessary, condition.


There are various sources for this, including:


* Meyer and Smith, [*Poincaré duality algebras, Macaulay's dual systems, and Steenrod operations*](https://mathscinet.ams.org/mathscinet-getitem?mr=2177162)
* Iarrobino and Kanev, [*Power sums, Gorenstein algebras, and determinantal loci*](https://mathscinet.ams.org/mathscinet-getitem?mr=1735271)
* Chapter 21 of Eisenbud's *Commutative Algebra*


It's not sufficient to only have a symmetric Hilbert function ($\dim A\_n = \dim A\_{N-n}$). For example $A = \mathbb{C}[x,y]/(x^3,xy,y^4)$ has Hilbert function $1,2,2,1$. But the multiplication map $A\_1 \times A\_2 \to A\_3$ is not a perfect pairing because multiplication by $x \in A\_1$ is zero on $A\_2$.


