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My question is how would I go about proving this?

Prove that $R$ is a local ring if and only if all elements of R that are not units form an ideal.

I understand that I need to prove both directions so:

$(\Rightarrow)$ Local ring means has a unique maximal ideal, so I want to show that this implies the elements are not units.

$(\Leftarrow)$ non unit elements of $R$ form an ideal, so if I show this is unique maximal ideal I can then conclude local ring?

Any hints would be appreciated.

pjs36
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Robert Thompson
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    Are you familiar with the fact that any non-unit is contained in a maximal ideal? – Tobias Kildetoft Feb 22 '16 at 14:05
  • I'm not, so i think that is what i need to be able to prove in my answer? – Robert Thompson Feb 22 '16 at 14:06
  • Using the axiom of choice, you can show that every non-unit lies in a maximal ideal. Now, this exercize is a direct consequence of this fact. – Crostul Feb 22 '16 at 14:34
  • From the context it looks like you are assuming a commutative ring. Actually, the statement still holds for noncommutative rings (although the solutions given so far all seem to rely on a fact that holds for commutative rings and not all noncommutative rings.) – rschwieb Feb 22 '16 at 18:32

1 Answers1

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The book "Introduction to Commutative Algebra" by Atiyah-Macdonald has in Corollary 1.5:

Every non-unit of $R$ is contained in a maximal ideal.

So
Let $R$ be local. If $m$ is the only maximal of $R$, then $m$ will be the set of non-units.
Conversely:
Let $n$ be the set of non-units of $R$. Let $m$ be a proper ideal s.t. $n\subseteq m$. Since $m$ is proper, no element of $m$ would be unit and so $m\subseteq n$. So $n$ is maximal.

Update:
This is also lemma 3.13 of the book "Steps in Commutative Algebra" by "Sharp" (in that terminology quasi-local means your "local".

user 1
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