# Do commutative rings without unity have the IBN property?

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

Does $$R$$ still have the Invariant Basis Number (IBN) property?

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.

• Simply adjoint a unit to $R$ to obtain $R_1$ (e.g., $R_1 = R[x]/(r x - r : r \in R)$). Then an isomorphism $R^m \cong R^n$ automatically extends to an isomorphism $R_1^m \cong R_1^n$, and you can use the IBN property for $R_1$. – LSpice Oct 17 '18 at 18:54
• The answer is "not necessarily". For a counterexample, let $R = \oplus^{\omega} \mathbb Z$ with zero multiplication. – Keith Kearnes Oct 17 '18 at 19:08
• It should be pointed out that $R^n$ doesn't have a basis either when $R$ is non-unital. – Benjamin Steinberg Oct 17 '18 at 19:50
• If $R$ is a rng, then $R^n$ doesn't have a basis! (The usual basis would be the columns of an identity matrix, but we cannot make sense of the entry $1$!) What you've written still makes mathematical sense, of course, but it's still a little funny to call it the "invariant basis number" in this context. – John Wiltshire-Gordon Oct 18 '18 at 0:24
• @LSpice Your notation is ambiguous. An isomorphism $R^m\simeq R^n$ lifts to an isomorphism $(R^m)_1\simeq(R^n)_1$, but not necessarily to an isomorphism $(R_1)^m\simeq(R_1)^n$. – Emil Jeřábek Oct 18 '18 at 9:01

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 $$R^n$$.
• Apologies for the delay! Thank you for editting and clearing up what goes on with both definitions and for including the functorial point of view! I have to admit that after getting used to objects having a $1$ for long time, suddenly removing the unit can be a bit confusing and difficult to track down all the properties one is accustomed to that fail in the non-unital setting. – M.G. Oct 19 '18 at 23:02