# Hilbert-Poincaré Series of Finite-Dimensional Graded Algebras

Suppose I have two finite-dimensional $\mathbb{Z}_{\geq 0}$-graded $\mathbb{C}$-algebras $A = \bigoplus_{k \geq 0} A_{k}$ and $B = \bigoplus_{k \geq 0} B_{k}$ with Hilbert-Poincaré series, $P_{A}(t) = \sum_{k \geq 0} \dim A_{k} \ t^{k}$ and $P_{B}(t) = \sum_{k \geq 0} \dim B_{k} \ t^{k}$, respectively.

When is it true that $A$ and $B$ are isomorphic as graded $\mathbb{C}$-algebras if $P_{A} = P_{B}$? Suppose that $P_{A} \neq P_{B}$ but $P_{A}(1) = P_{B}(1)$, what can be said about $A$ and $B$ in this case? Are they isomorphic as $\mathbb{C}$-algebras but not as graded $\mathbb{C}$-algebras?

The algebras that brought me to ask these questions are all of the form $\mathbb{C} \{ z_1, \dots, z_n \} / J$, where $J$ is a finitely generated ideal of partial derivatives of a complex analytic function $f$ with an isolated critical point at the origin.

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??? two algebras may have the same Hilbert series and not be isomorphic as algebras. Example: $A=k[x]/x^3$ with $x$ of degree one, $B=k[x,y]/(x^2,y^2,xy)$ with $x$ in degree one, $y$ in degree two. –  mt_ Aug 23 '11 at 20:47
Dear mt, I first thought you had made a mistake and that's why I decided to write an answer . But I hadn't seen that you slily took $y$ to have degree$2$ and so your remark is perfectly correct. Nice trick that ! –  Georges Elencwajg Aug 23 '11 at 21:47
This is the kind of questions that answers itself upon the consideration of a couple of examples :) –  Mariano Suárez-Alvarez Dec 28 '11 at 15:17
Of course, it is still very useful to know that two Poincare series are equal! Because if you have an actual (graded) map (of algebras if you want) $A \rightarrow B$ and you can show it's either injective or surjective, then it's an isomorphism by the equality of the Poincare polynomials. Very useful. –  Dylan Wilson Dec 28 '11 at 19:14

Your claim that equality of Hilbert-Poincaré series implies equality of algebras is false.

Take $A=\mathbb C [X,Y]/(X^2)$
and $B=\mathbb C [X,Y]/(X^2+Y^2)$

We have $P_{A}(t)=P_{B}(t) =1+2t+2t^2+2t^3+2t^4+\ldots=\frac{1+t}{1-t}$ .
But $A$ and $B$ are not isomorphic as rings (and thus certainly not as graded algebras), since $A$ has a non-zero nilpotent element (the class of $X$), whereas $B$ is reduced i.e. has zero as only nilpotent element.

Generalization If you consider a homogeneous polynomial of degree $d$ in $n$ variables $f(X_1,\ldots,X_n)\in \mathbb C [X_1,\ldots,X_n]$, then the algebra $A(f)=\mathbb C [X_1,\ldots,X_n]/\lt f(X_1,\ldots,X_n)\gt$ has a Hilbert-Poincaré series independent of $f$, namely (combinatorists: please combine!)

$$P(t)=\sum_{k\geq 0}\binom{k+n-1}{n-1}t^k-\sum_{k\geq d}\binom{k-d+n-1}{n-1}t^k$$

However the rings $A(f)$ strongly depend on $f$. This is evident geometrically because the hypersurfaces $Z(f)=\{ x \in \mathbb C^n|f(x)=0\}$ vary a lot with $f$: they may be irreducible or not, smooth or not, etc. And so their rings of functions $A(f)$ differ a lot too and are not in general isomorphic for different $f\;$'s.

Edit I had overlooked that the OP wanted finite-dimensional examples. It is easy to modify the above examples by killing off all monomials of sufficiently high degree. Geometrically, this means looking at the intersection or the hypersurface $Z(f)$ with a sufficiently large infinitesimal neighbourhood of the origin. Let me do this explicitly for the first example and kill all monomials of degree $\geq 3$:

Put $A'=\mathbb C [X,Y]/(X^2,XY^2,Y^3)=\mathbb C [x,y ]\;$ and $B'=\mathbb C [U,V]/(U^2+V^2,U^3,V^3)=\mathbb C [u,v]$ .
Then $P_{A'}(t)=P_{B'}(t) =1+2t+2t^2 \;$.
However the algebras $A'$ and $B'$are not isomorphic: for example, their spaces of solutions of the equation $\xi^2=0$ have dimensions $3$ and $2$ respectively over $\mathbb C$. Namely, these spaces are:
$Nil_2(A')=\mathbb C . x\oplus \mathbb C .xy \oplus \mathbb C .y^2$ and $Nil_2(B')=\mathbb C . u^2 \oplus \mathbb C .uv$

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Your example looks right, but the OP did ask for finite dimensional algebras. I left a finite dimensional counterexample in the comments. –  mt_ Aug 23 '11 at 21:56
Yes, I am aware of these examples. However, the original question is about finite dimensional graded algebras in which the Hilbert-Poincare series are actually polynomials. –  user02138 Aug 23 '11 at 23:10
Ah, yes, you are right @mt and user 02138. But the examples can easily be modified so as to make them finite-dimensional : just divide the algebras by the ideal generated by all monomials of sufficiently high degree . I've written an "Edit" detailing this procedure. –  Georges Elencwajg Aug 24 '11 at 0:05
I really like the geometric interpretation Georges -- I'd never seen that before. –  mt_ Aug 24 '11 at 13:21

Asking "does $P_A=P_B$ imply $A \cong B$?" is like asking "does $|G|=|H|$ imply $G \cong H$?": the answer is no, except in certain very special circumstances. If $P_A = 1$, or $1+t^{57}$, or $1+t^2+t^7$ etc. the answer is yes, but in general it is no: given a series $P_A$ simply construct an algebra with the same graded pieces as $A$ but with trivial multiplication. This will not normally be isomorphic to $A$.

If $P_A(1) = P_B(1)$ you have much less information. If they are both 2 you can say something, but I doubt you can get much more without restricting the class of algebras

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Thank you for the answer! –  user02138 Aug 24 '11 at 13:11

The simplest counterexample is, I think, this: $$A=A_0=\mathbb C\times\mathbb C,\quad B=B_0=\mathbb C[X]/(X^2).$$

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