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$I$ is an ideal of a local Noetherian ring $R$ and $i>0$ .

Clearly the height of primes in support of $H^i_I(R)$ is at least $i$

The question is if it contains a prime of height $i$, specially when $R$ is complete and unmixed?

PS What about the case $i$ equals the cohomological dimension of $I$, ie

$$i=\sup\lbrace j: H^j_I(R)\neq 0\rbrace $$

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If $R$ is Cohen-Macaulay and say $I$ is a maximal ideal, then most of these are zero except for $H^{\dim R}_I(R)$, which has support equal to $\text{Spec }R$. In particular, $H^{\dim R}_I(R)$ has primes of height zero, height one, etc. So I don't see why primes in the support of $H^i_I(R)$ have height at least $i$ either... Do you have a particular ring / ideal combination in mind? – Karl Schwede Jun 4 '13 at 19:19
@ Karl Schwede: If $p\in {\rm Supp}H^i_I(R)$ then $H^i_{IR_{P}}(R_p)\neq 0$ so by Vanishing theorem we have$i \leq \dim R_p={\rm ht}(p)$ – QED Jun 5 '13 at 9:06
@ Karl Schwede By the way, if $I$ is maximal then the support is contained in $V(I)$ hence equal to $\{I\}$. Please let me know if I'm wrong. – QED Jun 5 '13 at 9:13
I'm sorry, I was thinking the vanishing locus of the annihilator of the module, not the support. – Karl Schwede Jun 5 '13 at 15:12
up vote 1 down vote accepted

Let $(R, \mathfrak{m}) = k[[a, b, c, d]]$ where $k$ a field. Set $I_1 = (a, b),I_2 = (c, d)$ and $I = I_1 \cap I_2$. Consider $H^3_I(R)$ (use Mayer-Vietoris exact sequence, Hartshorne-Lichtenbaum vanising theorem) we can prove that $\text{supp}H^3_I(R) = \{\mathfrak{m}\}$. Notice that, by Lyubeznik (D-modules, F-modules) we have $H^3_I(R) \cong E(k)$ the injective hull of $k$.

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@Pham Hung Quy, Thanks a lot for good counter-example. I've edited the question. Please take a look because you may solve it again.:-) – QED Jun 8 '13 at 10:30
in my example $cd(I) = 3$. – Pham Hung Quy Jun 8 '13 at 11:08

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