Krull dimension of a module and isomorphisms

Question

I've been working with the Krull dimension of an $R$-module $M$ defined as the deviation of the lattice of submodules of $M$, i.e. $\operatorname{Kdim}(M)=\operatorname{dev}(\delta(M))$. I have been unable to prove or present a counterexample for the following statement: "The Krull dimension of a module is invariant under isomorphism". I would appreciate any help or ideas.

Answer 1

Since the definition only depends on the lattice of submodules and submodules and their structure are preserved under isomorphisms, you get that the krull dimension is invariant. Otherwise it would have been a bad notion either way.

The krull dimension can also be defined as $\text{dim}(M) = \text{dim}(R/\text{Ann}_R(M))$, where $\text{Ann}_R(M) = \lbrace a \in R \mid am = 0 \text{ for all } m \in M \rbrace$ is the annihilator of $M$. In that way you can also see the invariance by the same argument.