$A^n\cong A^m$ implies $n=m$ without using tensor products.

Question

Let $A$ be a nonzero commutative ring with 1. Then we can show that $A^m\cong A^n$ implies $m=n$ by picking a maximal ideal $I\subset A$ and tensoring both sides of $$ A^m \cong A^m $$ with $k=A/I$. Is it possible to prove the statement without using tensor products?

Answer 1

A ring $R$ (not necessarily commutative) is called stably finite if for each $n \in \Bbb N$, the matrix ring $M_n(R)$ has the property that $AB=1$ implies $BA=1$ for $A,B \in M_n(R)$.

Claim Any commutative ring $R$ is stably finite.

Proof: Suppose $AB=1$ for $A,B \in M_n(R)$, then $\det(A)\det(B)=\det(B)\det(A)=1$, so $\det(A) \in R^\times$, this implies that $A$ is invertible by Cramer's rule. Thus we get $AB=1=A A^{-1}$, so $B=A^{-1}$ after multiplying with $A^{-1}$ from the left.

Claim If $R$ is a non-zero stably finite ring, then $R^n\cong R^m$ implies $n=m$

Proof: Assume $n \leq m$ and let $k=m-n$, then we have by assumption $R^n \cong R^m = R^n \oplus R^k$. Take this isomorphism and compose it with the projection $R^n \oplus R^k \twoheadrightarrow R^n$ onto the first summand, to get a surjective homomorphism $f:R^n \to R^n$. Because $R^n$ is projective, the sequence $$0 \to \operatorname{ker}(f) \to R^n \xrightarrow{f} R^n \to 0$$

splits, so we get a homomorphism $g: R^n \to R^n$ such that $f \circ g = \operatorname{id}$. If we regard $f$ and $g$ as matrices, then we get $g \circ f=\operatorname{id}$ since we assume that $R$ is stably finite. This shows that $f$ is injective. But since $f$ is the composition of an isomorphism with the projection $p:R^n \oplus R^k \twoheadrightarrow R^n$, $p$ must also be injective. But $\operatorname{ker}(p)=R^k$, so $R^k=0$, thus $k=0$ and $n=m$.

All of the above and more on rings that satisfy "$R^n\cong R^m \Rightarrow n=m$" can be found in the first section of the book "Lectures on Modules and Rings" by T.Y. Lam (though some of the details I worked out are left to the reader).