# Degree bound for power of ideal

Let $I$ be an ideal in a commutative graded ring $R$, $M$ be a finitely generated graded $R$-module. Let $\varepsilon(M)$ be the smallest degree of a homogeneous element of $M$. An ideal $J$ is called an $M$-reduction of $I$ if $I^{n+1}M=JI^{n}M$ for some $n>0$. Define :

$$d(I):=max\lbrace \text{deg}f| f \text{ belongs to a minimal generating set of } I \rbrace$$ $$\rho_{M}(I)=min\lbrace d(J)|J\text{ is an M reduction of I}\rbrace$$

I faced the following problem : $d(I^{n}M)\ge \rho_{M}(I)+\varepsilon(M)$. This is a lemma in a paper, which can be found here :http://arxiv.org/abs/math/0212161v1

Here is the proof :

We may write $I = J + K$, where $J$ and $K$ denote the ideals generated by the homogeneous elements of$I$ of degree $< \rho_M(I)$ and $\geq \rho_M (I)$, respectively. Then $I^nM=JI^{n-1}M+K^nM$ Note that$K^n M$ is generated by homogeneous elements of degree $\geq \rho_M(I)n+ \varepsilon(M)$. If $d(I^{n}M)\ge \rho_{M}(I)+\varepsilon(M)$, then $I^{n}M=JI^{n-1}M$. Hence $J$ is an $M$-reduction of $I$. Since by definition of $J$, $d(J) <\rho_M (I)$, this gives a contradiction to the definition of $\rho_M (I)$

My question is why : $I^nM=JI^{n-1}M+K^nM$ ? How can we get it ?

I guess this follows from how to compute $I^n$. Since $I = J + K, \; I^n = J^n + J^{n-1}K + \cdots + K^n = J(J^{n-1} + \cdots + K^{n-1}) + K^n = JI^{n-1} + K^n$.
No problem! $\phantom{df}$ – Youngsu Dec 4 '12 at 8:48