On comparing dimension of $(0:_M I)$ and $M/IM$

Question

Let $(R, \mathfrak m)$ be a commutative Noetherian local ring of dimension $d$. For a finitely generated $R$-module $M$ we define $\dim M:=\dim R/\mathrm{ann}_R(M)$ and for an ideal $I\subseteq \mathfrak m$ of $R$, let $(0:_M I):=\{m\in M : Im=0\}$. Let $s=\dim M >0$.

How to show that

(1) $\dim\, (0:_M I)\le \dim M/IM$ ?

(2) If $x_1,...,x_s\in \mathfrak m$ is such that $l_R(M/(x_1,...,x_s)M)<\infty$, then how to show that $\dim\, (0:_M (x_1,...,x_t))<\dim M$, for every $1\le t\le s$ ?

Here $l_R(-)$ denotes length as $R$-module.

Here we note that $l_R(M/(x_1,...,x_s)M)<\infty$ is same as requiring $\mathrm{Supp}(M/(x_1,...,x_s)M)=\{\mathfrak m\}$, i.e. $\mathrm{Supp}(M ) \cap V(x_1,...,x_s)=\{\mathfrak m\}$ i.e. $\sqrt {\mathrm{ann}_R(M)+(x_1,...,x_s)}=\mathfrak m$ .

Answer 1

(1) Recall that $\operatorname{Supp}(M/IM)=\operatorname{Supp}(M)\cap V(I)$. If $P\notin\operatorname{Supp}(M/IM)$, then $P\notin\operatorname{Supp}(M)$ or $P\notin V(I)$. If $P\notin\operatorname{Supp}(M)$, then $M_P=0$ and $(0:_MI)_P=(0:_{M_P}I_P)=0$. If $P\notin V(I)$, then $I_P=R_P$ and $(0:_{M_P}R_P)=0$. We proved that $\operatorname{Supp}(0:_MI)\subseteq\operatorname{Supp}(M/IM)$, and therefore $\dim(0:_MI)\leq\dim(M/IM)$.

(2) Use (1) and $\dim M/(x_1,\dots,x_t)M=\dim M-t$.