MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $S=k[x_1,...,x_n]$ be a polynomial ring over field $k$ with maximal ideal $m=(x_1,...,x_n)$. I wanna make a $3$-dimensional $S$-module $M$ such that $H^0_m(M)=H^1_m(M)=0$ and $H^2_m(M)\neq 0$ be finitely generated (or in general case: $H^i_m(M)$ be finite for all $i=0,1,2$ ). Is there a simple way to create similar examples(for any dimension)?


$H^i_m(M)$ means $i$'th local cohomology module of $M$.

share|cite|improve this question
Local cohomology vanishes for all $i < \text{depth}(M)$. Doesn't this give examples right away? – Thomas Kahle Aug 19 '13 at 6:27
I think the post asks $H^2_m (M)$ to be finitely generated. They are Artinian, but not Noetherian in general. Without this condition, taking a direct sum of Cohen-Macaulay modules of desired dimensions would give an answer to the post. – Youngsu Aug 19 '13 at 6:50
Ok, so for $n = 6$ (I think), you can do a cone over an elliptic curve cross $\mathbb{P^1}$. For $n \geq 6$, you can just add variables as appropriate. You may be able to project this guy down to handle perhaps $n = 5$. – Karl Schwede Aug 19 '13 at 14:13
up vote 9 down vote accepted

Take $M$ to be the second syzygy of $k$ over $S=k[x_1,x_2,x_3]$. Then a graded version of local duality tells us that $H^2_m(M)$ is dual to $Ext^1(M,R)= Ext^3(k,R)$, the last one is $k$ either by direct computation or duality again.

One can easily generalize this, the $j$ syzygy of $k$ in $n$ variables will have local cohomologies vanish up to degree $j-1$ and finitely generated up to $n-1$.

share|cite|improve this answer
Please explain more. – Angel Aug 21 '13 at 12:48
Dear Algel, what are the parts that are not clear to you? – Hailong Dao Aug 21 '13 at 14:44
How can generalize this example?(your claim in last paragraph)Thanks. – Angel Aug 21 '13 at 15:37
Let me elaborate Dao's answer. Let $E$ be a f.g. $S=k[x_1,...,x_n]$-module. Then $H^i_{\mathfrak m}(E)$ is f.g. for all $i<n$ if and only if the sheaf $\widetilde{E}$ on $\mathbb P^{n-1}$ is a vector bundle. Next, by Auslander-Buchsbaum, $H^i_{\mathfrak m}(E)=0$ for all $i<j$ if and only if the depth of $E$ is at least $j$. So you are looking for vector bundles $\widetilde{E}$ where depth(E)=j. The best way I know to do this is exactly what Dao proposes: any syzygy module of any finite length module is a vector bundle; and the j'th syzygy module has depth j. – Daniel Erman Aug 22 '13 at 14:20
My previous comment uses the fact that if $\widetilde{E}$ is a vector bundle then the intermediate local cohom modules have finite length. (The converse is true as well but not needed for your answer.) This fact follows by an elementary argument combining Serre Vanishing + Serre Duality + the relation b/w local cohom and sheaf cohom. – Daniel Erman Aug 28 '13 at 12:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.