I'm going to assume here that all rings have $1$.
Let $M$ be a left $R$-module. A subset $I$ of $M$ is linearly independent if $r_1m_1 + … + r_nm_n = 0$ implies $r_1 = … = r_n = 0$ for all $r_1,…,r_n \in R$ and $m_1,…,m_n \in I$. A basis of a left $R$-module $M$ is a linear independent subset which generates $M$.
However, I don't see this definition working well for modules (actually, for the module) over a zero ring. Indeed, it is known that $\varnothing$ generates a zero module and it is trivially linearly independent. But $\{0\}$ is also linearly independent if $R = \{0\}$ for $r_1m_1 + … + r_nm_n = 0$ implies trivially that $r_1 = … = r_n = 0$ for the only element of $R$ is $0$. Since $\{0\}$ is commutive, the IBN property of commutative rings breaks: free module over them no longer have a well-defined dimension: for an $\{0\}$-module $\{0\}$ has bases of cardinalities both $0$ (namely $\varnothing$) and $1$ (namely $\{0\}$).
So, what am I missing (if anything)?
Best Answer
You’re right that the zero ring does not have the IBN property. Note also that the usual proof of “the IBN property of commutative rings” relies on the existence of a maximal ideal, which fails for the zero ring.
(One takes a maximal ideal $\mathfrak{m}$ in $R$ and then considers for any free $R$-module $F$ the extension of scalars $(R/\mathfrak{m}) \otimes_R F$. Any $R$-basis of $F$ gives an $(R/\mathfrak{m})$-basis of $(R/\mathfrak{m}) \otimes_R F$ of the same cardinality. Now one uses that $R/\mathfrak{m}$ is a field to argue that the dimension of $(R/\mathfrak{m}) \otimes_R F$ is well-defined.)