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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 $\operatorname{Supp}(R/I)=V(I)$.

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    Duplicate of http://mathoverflow.net/questions/115483/vanishing-of-local-cohomology2012-12-06

1 Answers 1

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$P=(x,y,p)$ 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).