TITLE: Is every commutative ring having the invariant basis number property equivalent to AC?
QUESTION [10 upvotes]: 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?

REPLY [3 votes]: We do not need axiom of choice to show that commutative rings have invariant basis number property. We can prove it with the help of exterior algebra $\wedge(V)$. Let $A$ be a commutative ring. First of all, it is obvious that if $m>n$, the $A$-module $\wedge^m(A^n)=0$ by considering a basis $p_1,p_2,...p_n$ and writing each element of $A^n$ in terms of the basis. Now we will show that $\wedge^n(A^n)$ is not $0$, by providing a surjective homomorphism onto $A$. Take a labelled basis $p_1,p_2,...p_n$ of $A^n$. Now for any permutation $\sigma$ in $S_n$ and elements $a_1,a_2,...a_n$ in $A$, send $a_1*p_{\sigma(1)} \wedge a_2*p_{\sigma(2)} .... \wedge a_n*p_{\sigma(n)}$ to $s(\sigma)*a_1a_2....a_n$ where $s(\sigma)$ is the sign of the permutation. Check that this uniquely extends to a surjective homomorphism $\wedge^n(A^n) \rightarrow A$ by bilinearity and the $v \wedge v=0$ property of the exterior algebra. Thus if $A$ is commutative and not trivial, $\wedge^n(A^n)$ isn't trivial. Now if $A^m$ and $A^n$ are isomorphic for some $m \neq n$, if we take $m>n$ WLOG, we see that there is a surjective homomorphism from $0=\wedge^m(A^n)=\wedge^m(A^m)$ to $A$, contradiction as $A$ is non-trivial. Thus non-trivial commutative rings satisfy IBN property.