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.



finiteness of local cohomology
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.


