Local ring

Last updated January 27, 2022

Let $R$ be a commutative ring with identity.

Definition. $R$ is called local if there exists just one maximal ideal $M\subset R$. Conversely, if there exists an ideal $M$ such that all elements outside of $M$ are invertible, then $R$ is local and $M$ is its maximal ideal.

Definition. The field $A/M$ is called the residue field of $A$.

Definition. A ring with only a finite number of maximal ideals is called semi-local.

Properties

1. Every element outside of $M$ is invertible. Maximal ideal
2. (Moved)
3. Let $\mathfrak{m}\neq (1)$ be an ideal of $A$ such that every $x\in A-\mathfrak{m}$ is a unit in $A$. Then $A$ is a local ring and $\mathfrak{m}$ is its maximal ideal
4. Let $A$ be a ring and $\mathfrak{m}$ a maximal ideal of $A$ such that every element of $1+\mathfrak{m}$ is a unit in $A$. Then $A$ is a local ring.

Proofs

3

This is a corollary of (1): every ideal $\neq (1)$ contains only non-units, thus is contained in $\mathfrak{m}$.

4

Let $x\in A-\mathfrak{m}$. Since $\mathfrak{m}$ is maximal, the ideal generated by $x$ and $\mathfrak{m}$ is $(1)$. So there exists $y\in A$ and $t\in\mathfrak{m}$ such that $xy+\mathfrak{m}y=xy+t=1$. Thus $xy=1-t$ is in $1+\mathfrak{m}$ hence is invertible by assumption. Then by (3) we are done.