Let $A$ be a finitely generated ${\mathbb Z}_{\geq 0}$-graded algebra over a field without zero divisors; assume that all graded components are finite-dimensional and that $Spec(A)$ is smooth outside of a subscheme of codimension 2. Are there conditions on the corresponding Hilbert series which imply that $Spec(A)$ is normal or that it has rational singularities? In particular, if two such algebras have the same Hilbert series and one of them is normal or has rational singularities, will the same be true for the other?

In fact, I would like to show that $A$ is Gorenstein; in this case the Hilbert series $H(t)$ satisfies $$ H(t)=(-1)^d t^q H(t^{-1}) $$ where $d$ is the dimension and $q$ is an integer. It is known that if $A$ is Cohen-Macaulay, then the above identity implies that $A$ is Gorenstein. I would like to know if the Cohen-Macaulayness assumptions can be dropped or weakened.


2 Answers 2


The general reason why Sam's examples work is that for a flat projective family $f:X\to B$ the Hilbert polynomial of the fibers is constant. So anything that can be read off from the Hilbert polynomial has to be invariant under flat deformation. In particular neither being normal or having rational singularities, or more generally any smoothable singularity have a chance for that.


1) Indeed the question was about $\mathrm{Spec}$ and not $\mathrm{Proj}$. Taking the cone over the corresponding projective variety gives you the $\mathrm{Spec}$ and the question becomes slightly different. But for instance a smooth projective family degenerating to a singular fiber gives you an example of having normal and non-normal cones of the fibers in the same family.

2) To respond to the additional question raised in the remarks regarding Gorenstein versus CM.

First of all, $X$ is Gorenstein = $X$ is CM + $\omega_X$ is a line bundle.

Let $f:X\to B$ be a flat family of CM schemes. Then $$\omega_{X_b}\simeq (\omega_{X/B})|_{X_b}$$ for every $b\in B$.

This implies that any small deformation of a Gorenstein fiber remains Gorenstein. On the other hand I am not entirely sure what exactly you mean by being able to tell if something is Gorenstein. I guess the question is which Hilbert series are you looking at?

Another possibility is that I completely misunderstood the question. :)

  • $\begingroup$ This is a good answer, but slight caveat: the question assumes that the algebra has no zero divisors, which is not preserved under deformation. $\endgroup$
    – Steven Sam
    Commented Apr 8, 2011 at 6:02
  • $\begingroup$ I understand that, but it wasn't clear to me how to find a graded deformation of a "bad" ring to a good "ring" (note that I am talking about Spec(A) not about Proj(A)). This is probably known to everybody, but I couldn't come up with an example. I know that if A is Cohen Macaulay, then the fact that A is Gorenstein can be read off from the Hilbert series -- is it completely clear a priori that Gorenstein (subject to Cohen-Macaulay) is stable under deformations? $\endgroup$ Commented Apr 8, 2011 at 6:08
  • $\begingroup$ It is not preserved necessarily, but there are flat deformations that preserves it. As far as a counter-example goes, that is enough. I did not say that all flat deformations provide counter-examples. By the same token you could say that flat deformations include smooth ones and those are both normal and have rational singularities. Cheers! $\endgroup$ Commented Apr 8, 2011 at 6:09
  • $\begingroup$ No, I am talking about graded rings and such a ring cannot be smooth unless it is a polynomial ring. Again, I am talking about singularities of A, not of $Proj(A)$. $\endgroup$ Commented Apr 8, 2011 at 6:14
  • $\begingroup$ And actually I should have assumed that $A$ is smooth outside codimension 2 -- otherwise it is obviously wrong (it is probably wrong anyway...) $\endgroup$ Commented Apr 8, 2011 at 6:18

Here's a counterexample: the Hilbert series of a hypersurface is determined by its degree, but hypersurfaces need not be normal. For example, take a hypersurface cut out by a hyperdeterminant of boundary format. For example the hyperdeterminant of a $3 \times 2 \times 2$ tensor (it is an irreducible polynomial, so it also satisfies the no nonzero divisor condition).

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    $\begingroup$ Even simpler, consider the cubic plane curves $y^2 z = x^3 - t z^3$. The Hilbert series are all the same, but of course at $t=0$ the curve is not normal. $\endgroup$ Commented Apr 8, 2011 at 6:12
  • $\begingroup$ You are right of course, I edited the question a little: is it obvious how to construct a counterexample if $A$ is smooth outside of codimension 2? $\endgroup$ Commented Apr 8, 2011 at 6:23
  • $\begingroup$ @Braverman: I don't think this is obvious at all... since now you are asking that $A$ is not Cohen-Macaulay. I know some families of non-normal varieties are that regular in codimension 1, but don't have a good way to get their equations / Hilbert series (precisely because they are not normal!). They have explicit descriptions, and if it would be of any help to have more such examples (to see if they do not apply to your situation or whatever), you can see my paper arxiv.org/abs/1101.4604 (Lemma 9.4; definition of varieties in Section 8). $\endgroup$
    – Steven Sam
    Commented Apr 8, 2011 at 13:36

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