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Let $M$ be a module over Noetherian ring $R$ such that $\operatorname{H}^1_I(M)=0$ for every ideal $I$ of $R$. Show that $\operatorname{H}^1_J(\Gamma_I(M))=0$ for every ideal $J$.

I tried to prove it by Mayer-Vietoris sequence but I can't, unfortunately. Also applied the following sequences $$0‎\rightarrow \Gamma_I(M)‎\rightarrow M‎\rightarrow D_I(M)‎\rightarrow 0 $$ $$0‎\rightarrow \Gamma_I(M)‎\rightarrow M ‎\rightarrow M/\Gamma_I(M)‎\rightarrow0 .$$ Moreover, since $\operatorname{H}^1_I(M)=0$ for every ideal $I$ of $R$ then Proposition 4.1.3 from Brodmann-Sharp, Local Cohomology, suggests that $\operatorname{H}^i_I(M)=0$ for all $i\geq 0$.

Background: $D_I(M)$ means ideal transform with respect to $I$ and the first exact sequence obtains from Theorem 2.2.4(i)(c) in the same book.

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Thanks.Finally, I could prove it. –  Angel Jan 5 '13 at 8:58

1 Answer 1

up vote 1 down vote accepted

Sketch of proof:

Let $J=(x_1,\dots,x_n)$. If $J\subseteq I$ it's done. Let $J‎\nsubseteq‎ I$ therefore $\exists x\in J-I $. By 4.1.22 from the same book we have the following exact sequence $$0‎\rightarrow‎ \Gamma_{I+x}(M)‎\rightarrow‎\Gamma_I(M)‎\rightarrow‎ \Gamma_I(M_x)‎\rightarrow‎ 0=H^1_{I+x}(M).$$ Applying $\Gamma_J(-)$ we get $$ \cdots‎\rightarrow‎ \Gamma_J(\Gamma_I(M_x))‎\rightarrow‎ H^1_J(\Gamma_{I+x}(M))‎\rightarrow‎ H^1_J(\Gamma_I(M))‎\rightarrow‎ H^1_J(\Gamma_I(M_x))‎\rightarrow‎\cdots $$ Now use the independence Theorem $$\Gamma_J(\Gamma_I(M_x))\cong \Gamma_J(\Gamma_{I_x}(M_x))\cong \Gamma_J(\Gamma_I{(M )})_x\cong \Gamma_{J_x}(\Gamma_I(M))_x=0. ~ (Since~ x\in J)$$ $$H^1_J(\Gamma_I(M_x))\cong H^1_J(\Gamma_{I_x}(M_x))\cong H^1_J(\Gamma_I{(M )})_x\cong H^1_{J_x}(\Gamma_I(M))_x=0.~ (Since~ x\in J)$$ Therefore $H^1_J(\Gamma_{I+x}(M))‎\cong‎ H^1_J(\Gamma_I(M))$. Continuing as before we have $$H^1_J(\Gamma_I(M))\cong H^1_J(\Gamma_{I+J}(M))‎=0. ~(Since~ J\subseteq I+J)$$

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