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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.

some1fromhell
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    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. http://math.stackexchange.com/q/54194/ – Alex Wertheim Jan 24 '17 at 05:37
  • 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 05:38
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    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 05:41
  • @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 05:48
  • @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 05:50
  • Related: http://math.stackexchange.com/questions/1225428 – Watson Jan 25 '17 at 15:37

1 Answers1

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As noted in comments, Zorn's Lemma is known to be equivalent to the statement that every (nonzero, unital) commutative ring has a maximal ideal.

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.

Jeremy Rickard
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