[Math] Is every commutative ring having the invariant basis number property equivalent to AC

Question

The proof I know that every commutative ring has the invariant basis number property involves quotienting by a maximum ideal to get a field, and so reducing to the case where the commutative ring is a field, for which the result is already proven. I wondered if there might be a "direct" proof, and tried to formalize it like this.

The fact that every commutative ring has a maximal ideal is proven with Zorn's lemma, and I suspect is in fact equivalent to Zorn's lemma. If we work in a set theory without Zorn's lemma/AC, is it still true that every commutative ring has the invariant basis number?

Answer 1

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.