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?

  • 5
    $\begingroup$ 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. $\endgroup$
    – YCor
    Sep 9, 2022 at 23:55
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    $\begingroup$ 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 $\endgroup$ Sep 10, 2022 at 3:19
  • $\begingroup$ @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? $\endgroup$ Sep 10, 2022 at 18:05
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    $\begingroup$ 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. $\endgroup$ Sep 10, 2022 at 18:08
  • $\begingroup$ @KevinBuzzard Ah, ok, I see. $\endgroup$ Sep 10, 2022 at 18:10

1 Answer 1


Nice question ! I believe the homework exercise implies AC.

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 $R=0$.


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