Let $\overline{\mathbb{R}}_{\geq 0} = \mathbb{R}_{\geq 0} \cup \{\infty\}$.
Given a commutative ring with unity $R$, $R\operatorname{-Mod}$ denotes the category of $R$-modules.
Given $R$-modules $A$ and $B$, $A \cong B$ denotes isomorphism of $A$ and $B$ as $R$-modules.
Does there exist an example of the following?
- A commutative ring with unity $R$
- A mapping $\operatorname{d}: R\operatorname{-Mod} \to \overline{\mathbb{R}}_{\geq 0}$ with the following properties:
- If $A \cong B$, $\operatorname{d}(A) = \operatorname{d}(B)$
- $\operatorname{d}(\{0\}) = 0$
- $\operatorname{d}(R) = 1$
- $\operatorname{d}(A \oplus B) = \operatorname{d}(A) + \operatorname{d}(B)$
- $\operatorname{d}(A \otimes_R B) = \operatorname{d}(A) \cdot \operatorname{d}(B)$
- If $A \subseteq B$ is a submodule, then $\operatorname{d}(A) \leq \operatorname{d}(B)$
- If $A \subseteq B$ is a submodule, then $\operatorname{d}(B/A) \leq \operatorname{d}(B)$
- If $A \subseteq B$ is a submodule, then $\operatorname{d}(A) + \operatorname{d}(B/A) = \operatorname{d}(B)$
- $\operatorname{d}$ is surjective
The idea is that $\operatorname{d}$ would be analogous to the dimension $\operatorname{dim}$ in the case that $R$ is a field. However, fields do not satisfy the condition since then $\operatorname{dim}$ only takes values in $\overline{\mathbb{N}}_{\geq 0}$ and is therefore not surjective as a mapping to $\overline{\mathbb{R}}_{\geq 0}$.