2

Let $S=k[x_1,\dots,x_m]$ be the polynomial ring over a field $k$ and let $I$ be a homogeneous saturated ideal of $S$. Let $h$ be a linear form of $S$ such that it is $S/I$-regular. Consider the exact sequence \begin{align} 0\rightarrow I(-1) \stackrel{h}{\rightarrow} I \rightarrow J \rightarrow 0, \, \, \, (1) \end{align} where $J=I/hI$. Suppose that $J^{\text{sat}}$ is $r$-regular; hence $H_{\mathfrak m}^i(J^{\text{sat}})_{j-i}=0, \, \forall i, \, \forall j>r$, where $H_{\mathfrak m}^i(\cdot)$ denotes the $i^{th}$ local cohomology functor. Note also that $H_{\mathfrak m}^i(J) = H_{\mathfrak m}^i(J^{\text{sat}}), \, \forall i>1$; this follows from the fact that $J^{\text{sat}}/J$ has finite length and that $H_{\mathfrak m}^i(J^{\text{sat}}/J)=0, \, \forall i>0$. Then our exact sequence yields \begin{align} H_{\mathfrak m}^{i-1}(J) \rightarrow H_{\mathfrak m}^i(I(-1)) \stackrel{h}{\rightarrow} H_{\mathfrak m}^i(I) \rightarrow H_{\mathfrak m}^i(J), \, \, \, (2) \end{align} from which we get \begin{align} H_{\mathfrak m}^{i-1}(J)_{j-i} \rightarrow H_{\mathfrak m}^i(I)_{j-i-1} \stackrel{h}{\rightarrow} H_{\mathfrak m}^i(I)_{j-i} \rightarrow H_{\mathfrak m}^i(J)_{j-i}. \, \, \, (3)\end{align} Hence for $i>2$ and $j>r+1$ the leftmost and rightmost terms become zero, yielding \begin{align} H_{\mathfrak m}^i(I)_{j-i} = h H_{\mathfrak m}^i(I)_{j-i-1}, \, \forall i>2, \forall j>r+1. \, \, \, (4) \end{align}

Question: Why does this last relation imply that $H_{\mathfrak m}^i(I)_{j-i}=0, \, \forall i>2, \, \forall j>r$?

Remark: This shows up in the proof of Theorem 4.3.2 in Bruns & Herzog (CM Rings).

PS: I will be happy to provide further explanations upon request.

user26857
  • 52,094
Manos
  • 25,833
  • Isn't $H_m^i(J^{sat}){j-i}=0, , \forall i, , \forall j>r$ equivalent to $H_m^i(J^{sat}){j-1-i}=0, , \forall i, , \forall j>r+1$? Then you have zero on the RHS of your equation which is right above the question. I guess the rest follows by induction. – Youngsu Nov 17 '15 at 03:47
  • @Youngsu: I don't see your point. I agree with what you wrote, but the equation above the question (equation (4)) involves only $I$, not $J^{sat}$. – Manos Nov 17 '15 at 04:34
  • You are right. I misread I and J. – Youngsu Nov 17 '15 at 06:37
  • Part I is here. – user26857 Nov 18 '15 at 21:47

1 Answers1

1

As you noticed $H_{\mathfrak m}^i(I)_{j-i-1} \simeq H_{\mathfrak m}^i(I)_{j-i}$ for $i>2$ and $j>r+1$. But for all large degrees these local cohomology modules are zero, so they are zero for $i>2$ and $j>r$.

user26857
  • 52,094