Local Rings

We call a ring $R$ local if $R$ has exactly one maximal ideal $M$. In this case, we call $A/M$ the residue field of $R$. A ring with only finitely many maximal ideals is called semi-local.


  1. Any field $F$ is local and $F$ is its own residue field.

  2. Let $R$ be any (possibly noncommutative) ring and let $G$ be any group. Then the group ring $R[G]$ is defined to be the set of formal linear combinations $\sum_{g\in G}\alpha_g g$ where every $\alpha_g \in R$, and only finitely many $\alpha_g$ are nonzero, with componentwise addition:

    \[ \sum{\alpha_g}g + \sum{\beta_g}g = \sum{(\alpha_g+\beta_g)}g \]

    and convolution product:

    \[ {(\sum_{g\in G}\alpha_g g)}{(\sum_{h\in G}\beta_h h)} = \sum_{k\in G}{(\sum_{g h = k} \alpha_g \beta_h)}k \]

    Take the cyclic group of order 2 $C_2 = \{x, x^2 = 1\}$. Then $\mathbb{Z}_2 [C_2] = \{1,x,0,1+x\}$. This is a local ring with maximal ideal $\{0, 1+x\}$ and its residue field is isomorphic to $\mathbb{Z}_2$.

    Let $F$ be a field of characteristic $p$, that is $sum_{i=1}^p 1 = 0$, and let $G$ be any abelian $p$-group, that is, the order of every element of $G$ is a power of $p$. Then $F[G]$ is local with unique maximal ideal $M = \{\sum \alpha_g g|\sum \alpha_g = 0\}$ with residue field isomorphic to $F$. 3. The ring $\{a/b | a,b \in \mathbb{Z}, 2\nmid b\}$ is local with residue field $\mathbb{Z}_2$. The ring $\{a/b|a,b\in \mathbb{Z}, 2\nmid b, 3 \nmid b\}$ is semi-local. We can continue in this fashion: by taking the first $n$ primes we can construct a semi-local ring with exactly $n$ maximal ideals. 4. The ring $\{p(x)/q(x)|p(x),q(x)\in\mathbb{R}[x],q(0)\ne 0\}$ is local with residue field $\mathbb{R}$. 5. The ring $\mathbb{R}[[x]]$ is local with residue field $\mathbb{R}$.

Proposition: Let $R$ be a ring and $M \ne R$ be an ideal such that every element of $R\setminus M$ is a unit. Then $R$ is local and $M$ is maximal.

Proof: No proper ideal can contain a unit, thus $M$ contains every proper ideal of $R$.

Proposition: Let $R$ be a ring and $M$ be a maximal ideal such that every element of $1 + M$ is a unit. Then $R$ is local.

Proof: Since $M$ is maximal, for any $x \in R\setminus M$ we have $R = \langle M, \{x\} \rangle = \{a x +m|a\in R, m\in M\}$. In particular, $1 = a x + m$ for some $a \in R, m\in M$ so $a x = 1 - m \in 1 + M$. If $a x$ is a unit then $x$ must also be a unit, and hence all of $R \setminus M$ are units, showing that $R$ is local.