Is every commutative ring having the invariant basis number property equivalent to AC?

The statement that nonzero commutative rings enjoy the invariant basis number property is true without any form of choice. In fact, it even holds in constructive mathematics, so without using the law of excluded middle.

A proof is contained in Fred Richman's three-page jewel Nontrivial uses of trivial rings. More specifically, he shows that:

Let $A$ be a commutative ring.

• If there is a linear injection $A^n \to A^m$ with $n > m$, then $1 = 0$ in $A$.
• If there is a linear surjection $A^n \to A^m$ with $n < m$, then $1 = 0$ in $A$.

The proof is fully explicit, showing how one can derive the equation $1 = 0$ from the (conditional) equations expressing the assumptions.

It's not equivalent to the axiom of choice, although it may well need some weak form of choice.

It is well-known that the statement that every commutative ring has a prime ideal is strictly weaker than AC.

If $\mathfrak{p}$ is a prime ideal of $R$, and $K$ is the field of fractions of the integral domain $R/\mathfrak{p}$, then the rank of a free $R$-module $F$ is the $K$-dimension of $F\otimes_RK$.