# Can the following be done without zorn's lemma

Prove that $R$ is local ring with maximal ideal $M$ if and only if every element of $R\backslash M$ is a unit.

I understand that this is a easy consequence of zorn's lemma. But I'm wondering if this can be done without zorn's lemma.

Any help or insight is deeply appreciated.

• Certainly, one direction does not require Zorn's lemma: if every element of $R\setminus M$ is a unit, then $M$ is maximal, since any ideal strictly containing $M$ must contain an element of $R\setminus M$, and hence is all of $R$. Further, any ideal which is not contained in $M$ also contains an element of $R \setminus M$, and so must be all of $R$, so $M$ is the unique maximal ideal of $R$: any proper ideal of $R$ is contained in $M$. I don't believe that the ''forward'' direction can be done without Zorn's lemma: see here for more details. math.stackexchange.com/q/54194 – Alex Wertheim Jan 24 '17 at 5:37
• Specifically, I don't see a way to get around using the fact that every nonunit must be contained in a maximal ideal of $R$, which is equivalent to Zorn's lemma, as noted in the linked question. – Alex Wertheim Jan 24 '17 at 5:38
• You must intend to assume that $M$ is an ideal, since there are subsets of a ring $R$ (with identity) whose complement consists only of units, but which are not ideals (much less maximal ones). You must further intend to assume $M$ is proper, ie. the complement $R\setminus M$ is nonempty, since otherwise $M=R$ is not a maximal ideal but vacuously the complement contains nothing but units. – hardmath Jan 24 '17 at 5:41
• @AlexWertheim: We don't need to "get around" the fact that every nonunit is contained in a maximal ideal of $R$ is the setup includes the assumption that $M$ is an ideal that contains all the nonunits. For this is precisely the stated assumption that $R\setminus M$ consists only of units. – hardmath Jan 24 '17 at 5:48
• @hardmath: yes, we're in agreement. This is the direction whose proof I present above. I am referring to the direction whose hypothesis is ''$R$ is a local ring with maximal ideal $M$'' and whose conclusion is "every element of $R \setminus M$ is a unit. Perhaps I've misunderstood your comment though. – Alex Wertheim Jan 24 '17 at 5:50

Suppose $S$ is a commutative ring without a maximal ideal, and $k$ a field. Then $R=k\times S$ has a unique maximal ideal $M=0\times S$, but not every element of $R\setminus M$ is a unit.
So the full strength of Zorn's Lemma is needed to prove that if a commutative ring $R$ has a unique maximal ideal $M$, then every element of $R\setminus M$ is a unit.