# Vanishing of a local cohomology module

I guess

$$H^2_{(x,y)}\left(\frac{\Bbb Z[x,y]}{(5x+4y)}\right)=0$$

It is well known $\operatorname{Supp} H^i_I(M)‎\subseteq V(I)\cap \operatorname{Supp}(M)$, therefore $$\operatorname{Supp} H^2_{(x,y)}\frac{\Bbb Z[x,y]}{(5x+4y)})\subseteq V((x,y))\cap V((5x+4y))=V((x,y))=‎\lbrace(x,y) \rbrace\cup ‎\lbrace‎‎(x,y,p) \rbrace‎‎$$ where $p$ is prime number.

If we could show that for every $P\in V((x,y))$, $$\left(H^2_{(x,y)}\left(\frac{\Bbb Z[x,y]}{(5x+4y)}\right)\right)_P=0$$

then it is done.

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 $\operatorname{Supp}(M)=\lbrace P\in \operatorname{Spec}(R) ; M_p\neq 0\rbrace$and $M_P$ means $M$ localized at prime ideal $P$. Furthermore $\operatorname{Supp}(R/I)=V(I)$.

• You can calculate local cohomology over a noetherian ring by tensoring your module with the Cech complex (also known as the infinite Koszul complex). Have you tried computing this? – the L Dec 5 '12 at 22:49
• Can you computing this by Cech complex? – Angel Dec 5 '12 at 23:26
• Yes, just tensor your module with the complex $0\to R \to R_x\oplus R_y \to R_{xy} \to 0$ and take cohomology. – the L Dec 5 '12 at 23:32
• The method of calculation is very complicated. – Angel Dec 5 '12 at 23:50
• – user26857 Dec 6 '12 at 15:55

$P=(x,y,p)$ implies $P\cap\mathbb Z=p\mathbb Z$ (here $p$ is a prime or $0$). As localization commutes with local cohomology $$H^2_{(x,y)}\left(\frac{\Bbb Z[x,y]}{(5x+4y)}\right)_P\simeq H^2_{(x,y)}\left(\frac{\Bbb Z[x,y]_P}{(5x+4y)}\right).$$ But $\mathbb Z[x,y]_P\simeq\mathbb Z_{(p)}[x,y]_{\overline{P}}$, where $\overline{P}$ is the extension of $P$ to $\mathbb Z_{(p)}[x,y]$. Now let's see what happens with the involved ideals via this isomorphism: $(5x+4y)$ goes to $(5x+4y)\mathbb Z_{(p)}[x,y]_{\overline{P}}$ and note that $5$ or $4$ (or both) are invertible in $\mathbb Z_{(p)}$, so the factor ring $\mathbb Z_{(p)}[x,y]_{\overline{P}}/(5x+4y)\mathbb Z_{(p)}[x,y]_{\overline{P}}$ is isomorphic to $\mathbb Z_{(p)}[t]_Q$, where $Q\cap \mathbb Z_{(p)}=p\mathbb Z_{(p)}$. Local cohomology is independent of base ring, so finally we arrive to $H^2_{(t)}(\mathbb Z_{(p)}[t]_Q)=0$ (since the local cohomology in a principal ideal is zero from $2$ onwards).