# A Krull-like Theorem and its possible equivalence to AC

A well known equivalent of the Axiom of Choice is Krull's Maximal Ideal Theorem (1929): if $$I$$ is a proper ideal of a ring $$R$$ (with unity), then $$R$$ has a maximal ideal containing $$I$$. The proof is easy with Zorn's Lemma. The converse, Krull implies Zorn, is due to Hodges (1979).

A ring with a unique maximal ideal is said to be local. A standard homework problem is this: Show that if $$R$$ is a local ring with maximal ideal $$M$$, then every element outside of $$M$$ is a unit. The "usual" solution is as follows: Let $$x$$ be a nonunit and let $$I$$ be the ideal generated by $$x$$. By Krull, $$I$$ is contained in a maximal ideal which, by locality, must be $$M$$.

Since the solution relies on AC, a natural question arises:

Is the homework problem itself equivalent to AC? More precisely, assume that in every local ring, every element outside of the unique maximal ideal is a unit. Does this imply AC?

If the answer is no, is there a proof of the homework problem which does not rely on AC?

• Of course the morale of the answer to this question is that this is not the right ZF definition of a local (commutative) ring. A right definition is maybe: a nonzero ring in which the sum of any two non-units is a non-unit.
– YCor
Sep 9, 2022 at 23:55
• There's some discussion of this issue on the nLab. The nLab uses the definition that if $a + b = 1$ then either $a$ or $b$ is a unit. Johnstone apparently calls YCor's proposal a "weak local ring": ncatlab.org/nlab/show/local+ring#in_weak_foundations Sep 10, 2022 at 3:19
• @YCor I know very little about foundations; could you explain what you mean by "right ZF definition"? Do you just mean a definition which is not equivalent to one of the ZF axioms? If so, what's wrong with that? Sep 10, 2022 at 18:05
• If you drop the axiom of choice then suddenly definitions which used to be equivalent are no longer equivalent (e.g. there are two definitions of closure of a subset in a metric space, two definitions of a local ring, ...) so now you have to decide which one is the "right" one. This is not really a mathematical question any more but mathematicians still have opinions. Sep 10, 2022 at 18:08
• @KevinBuzzard Ah, ok, I see. Sep 10, 2022 at 18:10

Indeed, assume its conclusion holds, and let $$R$$ be a ring with no maximal ideal. I'm going to prove that $$R$$ is zero, thus proving Krull's theorem (apply this to $$R/I$$ for a proper ideal $$I$$).
Let $$k$$ be your favourite field. Then $$k\times R$$ is local : $$0\times R$$ is a maximal ideal, and any ideal of a product is of the form $$I\times J$$, so because $$R$$ has no maximal ideal, $$0\times R$$ is the only maximal ideal.
Thus, by the homework problem, anything outside is invertible - but $$(1,0)$$ is outside. This proves that $$0$$ is invertible in $$R$$, hence $$R=0$$.