On Krull dimension \dim M and \dim Supp(M)

Question

Let $R$ be a commutative Noetherian ring and $M$ an $R$-module (not finitely generated). Are $\dim M$ and $\dim Supp(M)$ same?

Answer 1

Let $R$ be a ring and let $M$ be an $R$-module. We consider the topological space ${\rm Spec}(R)$. For a subset $Z\subseteq{\rm Spec}(R)$ we denote by $\dim(Z)$ its Krull dimension. In particular, we may consider the Krull dimensions of ${\rm Var}(0:_RM)$ and of ${\rm Supp}_R(M)$.

Fact 1: ${\rm Supp}_R(M)\subseteq{\rm Var}(0:_RM)$, and hence $\dim({\rm Supp}_R(M))\leq\dim({\rm Var}(0:_RM))$.

Fact 2: If $M$ is of finite type, then $\dim({\rm Supp}_R(M))=\dim({\rm Var}(0:_RM))$.

The question probably intended by the OP is:

Question: Does the conclusion of Fact 2 hold if $M$ is not of finite type?

Answer: No, not even if $R$ is noetherian. For a counterexample we consider $R=\mathbb{Z}$, a prime number $p$, and the $\mathbb{Z}$-module $M=\bigoplus_{k\in\mathbb{N}^*}\mathbb{Z}/p^k\mathbb{Z}$.

For $k\in\mathbb{N}^*$ we have ${\rm Supp}_{\mathbb{Z}}(\mathbb{Z}/p^k\mathbb{Z})={\rm Var}(p^k\mathbb{Z})=\{p\mathbb{Z}\}$. It follows ${\rm Supp}_{\mathbb{Z}}(M)=\bigcup_{k\in\mathbb{N}^*}{\rm Supp}_{\mathbb{Z}}(\mathbb{Z}/p^k\mathbb{Z})=\{p\mathbb{Z}\}$. In particular we see that ${\rm Supp}_{\mathbb{Z}}(M)$ is closed and that $\dim({\rm Supp}_{\mathbb{Z}}(M))=0$.

Furthermore, it is readily checked that $(0:_{\mathbb{Z}}M)=0$, hence ${\rm Var}(0:_{\mathbb{Z}}M)={\rm Spec}(\mathbb{Z})$, and therefore $\dim({\rm Var}(0:_{\mathbb{Z}}M))=1$.