Let $S=R[x,y,u,v]$ be polynomial ring over noetherian ring $R$.Set $M=S/(xu+yv)$. I guess $H^2_{(x,y)}(M)=0$ . for example $H^2_{(x,y)}(\frac{\Bbb Z[x,y]}{(5x+4y)})=0$. We know that $Supp H^i_I(M)\subseteq V(I)\cap Supp(M)$ then $Supp H^2_{(x,y)}(\frac{\Bbb Z[x,y]}{(5x+4y)})\subseteq V((x,y))\cap V((5x+4y))=V((x,y))=\lbrace (x,y) \rbrace\cup \lbrace(x,y,p); \rbrace$ where $p$ is prime number. We must show that for every $P\in V((x,y)), \big(H^2_{(x,y)}(\frac{\Bbb Z[x,y]}{(5x+4y)})\big)_P=0?$. **background:** $H^i_I(M)$ means $i$-th local cohomology module of $M$ with respect to ideal $I$. $V(I)=\lbrace P\in Spec(R); I\subseteq P\rbrace $ and $Supp(M)=\lbrace P\in Spec(R) ; M_p\neq 0\rbrace$and $M_P$ means $M$ localized at prime ideal $P$.Furthermore $Supp(R/I)=V(I)$