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?



A Krull-like Theorem and its possible equivalence to AC
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 $R=0$.


