What is the relation between $H^i_I()$ and $H^i_J()$ (cohomological functors) when $I\subset J$ are ideals of a (local) noetherian ring?

$\begingroup$ What kinds of relations are you looking for, maps between them, containment or etc? $\endgroup$ – Youngsu May 20 '13 at 22:02

$\begingroup$ Any nontrivial relation! $\endgroup$ – QED May 21 '13 at 13:29

2$\begingroup$ Well, I don't think you can get a good relation unless you specify something specific. For instance, let $(R,m)$ be a local ring and $I \subseteq m$ an ideal. Then $H^0_m(R) \subseteq H^0_I (R)$. However if you let $I = aR$ and $\dim R = 2$. Then $H^2_m(R) \neq 0$ whereas $H^2_I(R) = 0$. $\endgroup$ – Youngsu May 21 '13 at 16:37

$\begingroup$ An obvious special case is when $I$ and $J$ have the same radical. In that case the local cohomology groups are isomorphic. $\endgroup$ – Mahdi MajidiZolbanin May 22 '13 at 0:48
There's a map between them, and these do fit into a long exact sequence together. This is explained in the book on local cohomology by Hartshorne, see Lemma 1.8: You can even download this book if your institution has access...
I will sketch it briefly. Let $V(I) = Y \supseteq Z = V(J)$ and set $W = V(I) \setminus V(J) = Y \setminus Z$, this $W$ is only locally closed in $\text{Spec} R$.
We define $H^i_W(\bullet)$ to be the right derived functors of $\Gamma_W(\bullet)$. Here $\Gamma_W(M)$ for any $R$module $M$ is defined to be the set of elements of $M$ which are zero in $M_{p}$ for all $p \in (\text{Spec }R) \setminus W$.
With this notation, for any $R$module $M$ we have a long exact sequence $$ ... \to H^i_Z(M) \to H^i_Y(M) \to H^i_W(M) \to H^{i+1}_Z(M) \to ... $$ Note $H^i_Z(M) = H^i_J(M)$ and $H^i_Y(M) = H^i_I(M)$. I doubt this gives you much information unless $Y$ and $Z$ have some special relationship.
There are related things you can do too. If for example $Y = Z \cup X$ where $X$ is some other closed subset of $\text{Spec }R$, there's another long exact sequence but this is a special case. Can you tell me what $Y$ and $Z$ are in your case?

$\begingroup$ Thanks for typing such a detailed answer. I was just curious if (and how) the inclusion of zeroth cohomologies affects the upper ones. $\endgroup$ – QED May 22 '13 at 13:53