Invariant Basis Number

Question

How can we show that $R$ has IBN if either of the following conditions hold:

1. All finitely generated subrings $S$ of $R$ have IBN, or

2. $R$ has a subring $S$ with IBN such that $R$ is a finitely generated free $S$-module.

Thanks.

Answer 1

Assume 1. holds. If $R^n \cong R^m$ is an isomorphism of $R$-modules, then it is given by an invertible $m \times n$-matrix with entries in $R$. These entries, as well as the entries of the inverse matrix, generate a finitely generated subring $S$ of $R$. We get an isomorphism of $S$-modules $S^n \cong S^m$, which implies $n=m$.

Assume 2. holds. Let $d \in \mathbb{N}$ such that $R \cong S^d$ as $S$-modules. Since $R \neq 0$ (otherwise $S=0$ has not IBN), we have $d>0$. Then $R^n \cong R^m$ as $R$-modules implies $S^{nd} \cong S^{md}$ as $S$-modules, hence $nd=md$, and therefore $n=m$ (since $d>0$).

---

By the way, 1. offers a nice proof that commutative rings $R \neq 0$ have IBN (without reducing to the field case, which will be a special case): We may assume that $R$ is finitely generated. Then it is known that every residue field is a finite field. Therefore we may even assume that $R$ is a finite ring, say with $e>1$ elements. But then $R^n \cong R^m$ implies $e^n=e^m$, hence $n=m$.