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Well-known Theorem:

Let $a$ be an ideal of the noetherian ring $R$ and let $M$ be a finitely generated $R$-module. Let $i \in \Bbb N_0$ be such that $H^j_a (M)$ is finitely generated for all $j < i$. Then the set $Ass_R((H^i_a (M))$ is finite.

Now my questions:

Is there an example that $H^0_a(M), H^1_a(M), H^2_a(M)$ be non-zero finitely generated modules. Is there an example that $H^0_a(M), H^1_a(M),..., H^j_a(M)$ be non-zero finitely generated modules for $j>2$.

Example for $\bf{H^0_a(M), H^1_a(M)}$:

Let $S=k[x_1,x_2,x_3,x_4]$ be a polynomial ring over field $k$ and $I=P_1\cap P_2$ where $P_1=(x_1,x_2), P_2=(x_3,x_4)$ and $m=(x_1,x_2,x_3,x_4)$. Set $R=S/I$. From the exact sequence $$0\rightarrow R\rightarrow S/P_1\oplus S/P_2\rightarrow S/m\rightarrow 0$$ we have the exact sequence $$0\rightarrow H^0_m(R)\rightarrow H^0_m(S/P_1)\oplus H^0_m(S/P_2)\rightarrow H^0_m(S/m)\rightarrow H^1_m(R)\rightarrow H^1_m(S/P_1)\oplus H^1_m(S/P_2)\rightarrow H^1_m(S/m)\rightarrow H^2_m(R)\rightarrow H^2_m(S/P_1)\oplus H^2_m(S/P_2)\rightarrow H^2_m(S/m)$$ Since $S/m$ is $m$-torsion then $H^i_m(S/m)=0$ for all $i\neq 0$. on the other hand $H^i_m(P_j)=0$ for $i=0,1 ,j=1,2$. Hence $H^0_m(R)=0,~ H^1_m(R)\cong H^0_m(S/m)=S/m$ are finitely generated.

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1 Answer

Ok, let $X$ by a smooth Abelian surface (for example, the segre product of elliptic curves) and let $R_0$ be section ring of some projectively normal embedding.

For example, to form this ring $R_0$, take the Segre product of the two standard graded rings $$ k[x,y,z]/\langle x^3 + y^3 + z^3 \rangle \text{ with } k[a,b,c]/\langle a^3 + b^3 + c^3 \rangle. $$

We know $R_0$ is a 3-dimensional ring with $H^2_m(R)$ a $1$-dimensional $k$-vector space (but $H^1_m$ and $H^0_m$ are zero). This follows from basic facts about Abelian varieties. You can also make $R_0$ the segre product of $k[x,y,z]/\langle x^3 + y^3 + z^3 \rangle$ with $k[s,t]$.

Let $R_1$ denote the gluing of $R_0$ to itself at the origin. In other words $$ R_1 = \mathrm{Ker}\big( R_0 \oplus R_0 \xrightarrow{-} R_0/m \big) $$ where the map is the is just the subtraction map. This is a ring, it is the set of all pairs $\{ (f, g) \in R \times R | f({\bf 0}) = g({\bf 0}) \}$.

From the fact that we have a short exact sequence $0 \to R_1 \to R_0 \oplus R_0 \to R_0/m \to 0$ we see that $R_1$ will have nonzero $H^1_m$ and $H^2_m$. But $H^0_m = 0$ since it is still reduced. Ok, now we just need to make it have non-zero $H^0_m$. Here's one way.

Add some non-reducedness (embedded prime) ONLY at the origin of $R_1$ to get $R_2$. For example, let $R_2$ denote the kernel of $$ R_1 \oplus k[\varepsilon]/\varepsilon^2 \xrightarrow{-} k $$ where the map on $R_1$ is evaluation at the origin.

Then the short exact sequence $$ 0 \to K \to R_2 \to R_1 \to 0$$ should guarantee the non-vanishing you want (since $K$ will have vanishing higher $H^i_m$). The same technique should let you construct the nonvanishing $H^j_m$ you want by using cones over higher dimensional Abelian or irregular varieties.

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I think we can give a simple example of a polynomial ring –  Angel Jul 10 '13 at 12:11
I'm not sure i see what you mean. –  Karl Schwede Jul 10 '13 at 16:13
In particular, a polynomial ring is always reduced. If you mean you want generators and relations, you can certainly use what I did above to construct the generators and relations (if I had to do it, I'd use Macaulay2 or something to help). –  Karl Schwede Jul 10 '13 at 21:13
Thanks. Please explain more? I need an example over polynomial ring $k[x_1,...x_n]$. –  Angel Jul 11 '13 at 11:00
I've explained more. You'll have to work out the explicit generators and relations yourself, or use a computer to do it for you though. –  Karl Schwede Jul 11 '13 at 14:27
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