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.

Ben Lynn blynn@cs.stanford.edu 💡