Commutative Algebra

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.

Example:

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:

   $$(\sum _{g\in G}{\alpha }_{g}g)(\sum _{h\in G}{\beta }_{h}h)=\sum _{k\in G}(\sum _{gh=k}{\alpha }_g{\beta }_h)k$$

   Take the cyclic group of order 2 $C_2=\{x,x^2=1\}$. Then ${\mathbb{Z}}_2[C_2]=\{1,x\mathrm{,0,1}+x\}$. 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 ${\mathrm{sum}}_{i=1}^{p}1=0$, 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 $M=\{\sum {\alpha }_{g}g\mid \sum {\alpha }_g=0\}$ with residue field isomorphic to $F$. 3. The ring $\{a/b\mid a,b\in \mathbb{Z},2\nmid b\}$ is local with residue field ${\mathbb{Z}}_2$. The ring $\{a/b\mid a,b\in \mathbb{Z},2\nmid b,3\nmid b\}$ 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 $\{p(x)/q(x)\mid p(x),q(x)\in ℝ[x],q(0)\ne 0\}$ is local with residue field $ℝ$. 5. The ring $ℝ[[x]]$ is local with residue field $ℝ$.

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 $R=⟨M,\{x\}⟩=\{ax+m\mid a\in R,m\in M\}$. In particular, $1=ax+m$ for some $a\in R,m\in M$ so $ax=1-m\in 1+M$. If $ax$ is a unit then $x$ must also be a unit, and hence all of $R\setminus M$ are units, showing that $R$ is local.