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 semilocal.
Example:

Any field \(F\) is local and \(F\) is its own residue field.

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:
+
+ 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 \(\{a/ba,b\in \mathbb{Z}, 2\nmid b, 3 \nmid b\}\) is semilocal. We can continue in this fashion: by taking the first \(n\) primes we can construct a semilocal 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 \(R = \langle M, \{x\} \rangle = \{a x +ma\in R, m\in M\}\). 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.