Let $R$ be a commutative rng, i.e. a commutative ring without an identity element.



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> Does $R$ still have the Invariant Basis Number (IBN) property?
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Recall that a ring is said to have the IBN property if $R^m \cong R^n \Rightarrow m=n$.


All commutative rings have the IBN property, but the standard proof I know makes an essential use of the existence of a maximal ideal by Zorn's Lemma and passing to the residue field, which depends on the presence of a unit.



Do commutative rings without unity have the IBN property?
(I will write my comment as an answer.)


The answer is "not necessarily" for the way IBN is defined in the problem. For a counterexample, let $R=\oplus^{\omega} \mathbb Z$
with zero multiplication. 


But the definition of IBN in the problem is not the right one. It IS true that when $R$ is a commutative nonunital ring, the f.g. free $R$-modules have uniquely determined rank. The reason is that, when $R\_1$ is the ring obtained from $R$ be formally adding a unit element, then the forgetful functor from the category of $R\_1$-modules to the category of $R$-modules is an equivalence. The free $n$-generated $R$-module is $R\_1^n$, not $R^n$.


