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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:

- Does the rank of homology and cohomology groups always coincide?
- A ring with few invertible elements
- Galois group of the extension $E:= \mathbb{Q}(i, \sqrt{2}, \sqrt{3}, \sqrt{2})$
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- $A^m\hookrightarrow A^n$ implies $m\leq n$ for a ring $A\neq 0$
- Isomorphism of $\mathbb{Z}/n \mathbb{Z}$ and $\mathbb{Z}_n$

$(\Rightarrow)$ 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.

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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”.

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