$\dim M_p+\dim R/p\leq\dim M$?

Question

Let $R$ be a commutative ring with unity. Then, $\dim R_p+\dim R/p\leq\dim R$ for $p\in\mathrm{Spec}(R)$.

Under the additional assumption that $R$ is Noetherian, can we replace $R$ by a nonzero finitely generated $R$-module $M$ with $p\supset annM$ where $annM$ is the annihilator of $M$?

$\dim M_p+\dim R/p\leq\dim M$

Answer 1

Let $M$ be a finitely generated $R$-module and $p \in {\rm Supp}\, M$. Set ${\dim}_{R_p} M_p=t$ and ${\dim} \, R/p=s$. Then there exists a strict chain of primes in ${\rm Supp }_{R_p} M_p$: $$p_0R_p \subset p_1R_p \subset \dots \subset p_tR_p=pR_p.$$ On the other hand we have a strict chain of primes in ${\rm Spec}\, R/p$: $$p/p=q_0/p \subset q_1/p \subset \dots \subset q_s/p. $$ Thus there exists a strict chain of prime of length $s+t$ in ${\rm Supp \, M}$: $$p_0 \subset p_1 \subset \dots \subset p_t \subset q_1 \subset \dots \subset q_s $$ which shows that $s+t \leq {\rm dim}\, M.$

Answer 2

Definition: Let $M$ be an $R$-module. Then $dim M = dim (R/ann(M) )$ if $M \ne 0$ and $dim M = -1$ if $M = 0$.

Set $R:=\mathbb{Z} $, $p=\{0\}$ and $M:=\mathbb{Z}\times\mathbb{Z}$. Then $ann(\mathbb{Z}\times\mathbb{Z})=0 $ and $(\mathbb{Z}\times\mathbb{Z})_p\cong \mathbb{Z}_p\times\mathbb{Z}_p=\mathbb{Q}\times\mathbb{Q}$. Hence, $ann((\mathbb{Z}\times\mathbb{Z})_p)=0$. And we have

$$dimM_p+dim(R/p)=dimR/annM_p+dimR/0=dim\mathbb{Z}+dim\mathbb{Z}=1+1=2\not\leq dimM=dimR=dim\mathbb{Z}=1.$$