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



On Krull dimension \dim M and \dim Supp(M)
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 $\dim({\rm Var}(0:\_{\mathbb{Z}}M))=1$.


