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Edited:

I guess

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

We know that if $\operatorname{Supp} H^i_I(M)‎\subseteq V(I)\cap \operatorname{Supp}(M)$, then $$\operatorname{Supp} H^2_{(x,y)}\frac{\Bbb Z[x,y]}{(5x+4y)})\subseteq V((x,y))\cap V((5x+4y))=V((x,y))=‎\lbrace(x,y) \rbrace\cup ‎\lbrace‎‎(x,y,p) \rbrace‎‎$$ 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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Do you mean $M=S/(xu+yv)$ ? Also I don't understand, why your example fits into this pattern. –  Ralph Dec 5 '12 at 10:35
    
Yes.$M=S/(xu+yv)$. –  Angel Dec 5 '12 at 10:39
    
In your example, the vanishing takes place because of a general vanishing result of Grothendieck. See 6.1.2 in "Local cohomology..." by Brodmann and Sharp (Cambdridge Univ. Press). Like Ralph, I don't understand how your example is a special case. –  Damian Rössler Dec 5 '12 at 12:39
    
How my example be solved by 6.1.2? –  Angel Dec 5 '12 at 13:47
    
I am sorry, I miscalculated: 6.1.2 implies only that $H^2_{(x,y)}({\bf Q}[x,y]/(5x+4y))$. –  Damian Rössler Dec 5 '12 at 15:44

2 Answers 2

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 $$H^2_{(x,y)}\left(\frac{\mathbb{Z}[x,y]}{(5x+4y)}\right)_P\simeq H^2_{(x,y)}\left(\frac{\mathbb{Z}[x,y]_P}{(5x+4y)}\right).$$ 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).

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Dear YACP: As you can see on the upper margin of this page: "put backticks around any math that contains underscores or asterisks." –  Mahdi Majidi-Zolbanin Dec 8 '12 at 18:30

I wasn't able to use the method you suggest. Here is a different proof. Let $M$ be any ${\bf Z}[x,y]$-module. We have (see Brodmann and Sharp, Th. 1.3.8), $$ H^2_{(x,y)}(M)=\varinjlim_n \operatorname{Ext}^2({\bf Z}[x,y]/(x,y)^n,M) $$ where the limit is an inductive limit. Now we have $$ \varinjlim_n \operatorname{Ext}^2({\bf Z}[x,y]/(x,y)^n,M)=\varinjlim_n \operatorname{Ext}^2({\bf Z}[x,y]/(x^n,y^n),M) $$ because for any $n$, we have $(x^n,y^n)\subseteq (x,y)^n$ and $(x,y)^{2n}\subseteq (x^n,y^n)$. Now the ideal $(x^n,y^n)$ defines a regular closed immersion into $\operatorname{Spec} {\bf Z}[x,y]$, so that we have the "fundamental local isomorphism" (see Hartshorne, Residues and duality, Prop. 7.2): ... EDIT: the end of the original argument was wrong. I don't know how to make it work.

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We should use property of $(5x+4y)$ under localization at $P$. Lichtenbaum,-Hartshorne vanishing Theorem may be useful. –  Angel Dec 7 '12 at 14:20
    
I don't know what are you talking about, but what I said referred to you comment, that is, $H_{(x,y)}^2(\mathbb Z[x,y])=0$ implies, by localizing to $Z−\{0\}$, that the similar local cohomology over $\mathbb Q$ is $0$ and this is false! –  user23950 Dec 7 '12 at 19:02
    
@YACP: I understood Angel's comment as meaning exactly what you are saying; ie Angel explained with his terse vanishing statement that my answer was flawed (which is true), because my answer implied that (incorrect) vanishing. –  Damian Rössler Dec 8 '12 at 10:22

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