# Error in a "proof" that all rings have IBN (invariant basis number)

It is known that there exist rings $R$ for which $R^{\oplus n}\cong R^{\oplus m}$ (as $R$-modules) even if $n\neq m$. For instance, take $R = \text{End}_F(V)$ where $F$ is a field and $V$ is an infinite-dimensional vector space over $R$ (for details, see here). Consider therefore, the following argument that all rings have IBN (invariant basis number).

Let $R$ be any ring, and suppose $R^{\oplus n}\cong R^{\oplus m}$ as left $R$-modules. Let $V$ be any simple left $R$-module (for instance, take $V = R/I$ where $I$ is any maximal left ideal of $R$). Then we have $$V^{\oplus n}\cong R^{\oplus n}\otimes_RV\cong R^{\oplus m}\otimes_RV \cong V^{\oplus m}.$$ We may write down composition series for $V^{\oplus n}$ and $V^{\oplus m}$: $$V^{\oplus n}\supsetneq V^{\oplus (n-1)}\supsetneq \cdots\supsetneq V^{\oplus 1}\supsetneq V \supsetneq 0 \qquad\text{ and} \\ V^{\oplus m}\supsetneq V^{\oplus (m-1)}\supsetneq \cdots\supsetneq V^{\oplus 1}\supsetneq V \supsetneq 0.$$ By the Jordan Hölder theorem, since $V^{\oplus n}\cong V^{\oplus m}$, we have that $$n = \text{length}(V^{\oplus n}) = \text{length}(V^{\oplus m}) = m.$$

Can anyone find the mistake in the argument?

You cannot conclude that $R^{\oplus n} \otimes_R V \cong R^{\oplus m} \otimes_R V$. The definition of these tensor products uses the right $R$-module structure on $R^{\oplus n}$ and $R^{\oplus m}$, whereas you've only assumed that the isomorphism is an isomorphism of left $R$-modules.

Your proof correctly shows the following two true statements:

• If $R$ is a nonzero ring and $R^{\oplus n} \cong R^{\oplus m}$ as $(R, R)$-bimodules, then $n = m$.
• If $R$ is a nonzero commutative ring and $R^{\oplus n} \cong R^{\oplus m}$ as $R$-modules, then $n = m$.

A generalization of the second statement which is also true is that if a ring admits a morphism to a ring with IBN (e.g. a division ring), then it has IBN. But not every ring admits such a morphism.

The nonzero requirement here is necessary; the zero ring has no maximal left ideals and no simple modules.

• I think it can be considered the tensor product $V\otimes_R R^{\oplus_n}$ but then we only have an isomorphism at the level of abelian groups. I'm right? Sep 27, 2017 at 22:17
• @Hero: no, we don't have an isomorphism at all, for exactly the reasons described in the second sentence of my answer. Sep 27, 2017 at 22:36