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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.

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    $\begingroup$ 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 $R_1$. $\endgroup$
    – LSpice
    Commented Oct 17, 2018 at 18:54
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    $\begingroup$ The answer is "not necessarily". For a counterexample, let $R = \oplus^{\omega} \mathbb Z$ with zero multiplication. $\endgroup$
    – Keith Kearnes
    Commented Oct 17, 2018 at 19:08
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    $\begingroup$ It should be pointed out that $R^n$ doesn't have a basis either when $R$ is non-unital. $\endgroup$
    – Benjamin Steinberg
    Commented Oct 17, 2018 at 19:50
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    $\begingroup$ 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. $\endgroup$
    – John Wiltshire-Gordon
    Commented Oct 18, 2018 at 0:24
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    $\begingroup$ @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 $(R_1)^m\simeq(R_1)^n$. $\endgroup$
    – Emil Jeřábek
    Commented Oct 18, 2018 at 9:01

1 Answer 1

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(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 $R^n$.

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  • $\begingroup$ 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. $\endgroup$
    – M.G.
    Commented Oct 19, 2018 at 23:02

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