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

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?

• Without the axiom of choice, you can have a vector space with two bases of different cardinalities. I don't know ring theory, but I think that's an example of a commutative ring without the invariant basis number property. It's probably weaker than full AC though. Sep 21, 2016 at 16:08
• @MitchellSpector Without AC, you can have bases with different infinite cardinalities, but the IBN property is that you specifically can't have two finite bases of different cardinalities. Sep 21, 2016 at 17:04
• Never mind then :) Sep 21, 2016 at 17:06
• The fact about maximal ideals and Zorn's lemma being equivalent is true. Sep 21, 2016 at 17:14

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.

• Could you explain why $\phi$ has a right-inverse in his proof of Theorem 1? Feb 27, 2017 at 23:57
• @Jack: That's true for arbitrary, even noncommutative rings. If a linear map $A^n \to A^m$ is surjective, we can find for any of the canonical basis vectors $e_i \in A^m$ a preimage $v_i \in A^n$. The map obtained by linearly extending $e_i \mapsto v_i$ is then a right inverse $A^m \to A^n$. A fancy way to express this fact is to say that $A^m$ is a projective module. Finite free modules are projective. The axiom of choice implies that all free modules (possibly of infinite rank) are projective. Feb 28, 2017 at 20:17
• @Jack: But personally, I'd prove the statement that the existence of a linear surjection $A^n \to A^m$ with $n < m$ implies $1 = 0 \in A$ in a different way. Using a device called "internal language of a topos", one can pretend that $A$ is a local ring. In this case a short argument using elementary row and column transformations shows the claim. This device is quite useful, also for verifying other statements in algebra. Rough notes of mine contain an exposition of these ideas, but they're directed at algebraic geometers. Feb 28, 2017 at 20:26

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 $F\otimes_RK$.

• What form of choice does the existence of a prime ideal for commutative rings require? Sep 21, 2016 at 18:06
• @JackM I'm no expert, but there's a fairly detailed discussion here: mathoverflow.net/questions/98549/… Sep 21, 2016 at 18:16
• It's true that one needs some amount of choice to deduce the existence of prime ideals in nontrivial rings. However, one doesn't need any prime ideals to verify that nonzero commutative rings enjoy the IBN property -- see my answer below. Feb 25, 2017 at 21:51

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.