Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
    
Duplicate of mathoverflow.net/questions/115483/vanishing-of-local-cohomology –  user26857 Dec 6 '12 at 15:55
add comment

1 Answer 1

up vote 2 down vote accepted

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.