local ring

Definition

A local ring is a commutative ring $R$ with a unique maximal ideal $\mathfrak{m}\subset R$.

The field $R/\mathfrak{m}$ is called the residue field of $R$.

Examples

Any field

In a field, every non-zero element is a unit. An ideal that is a proper subset cannot contain a unit, so the maximal ideal is the trivial ideal $\{0\}$. Therefore, the residue field is $k/\{0\}=k$.

Formal power series

Let $k$ be a field. The ring of formal power series with coefficients in $k$ (denoted $k[[x]]$) is the ring with elements that are infinite formal sums

$$\sum _{n=0}^{\infty }{a}_n{x}^n,a_i\in k$$

and are added an multiplied in the familiar way. (Since these sums are formal, without concern for convergence, there is no danger in multiplying them.)

Claim: $k[[x]]$ is a local ring.

Let ${\sum }_{n=0}^{\infty }{a}_n{x}^n$ be a unit. Then

$$\left(\sum _{n=0}^{\infty }{a}_n{x}^n\right)\left(\sum _{n=0}^{\infty }{b}_n{x}^n\right)=\sum _{n=0}^{\infty }\left(\sum _{j=1}^n{a}_j{b}_{n-j}\right)x^n=1$$

Hence, $a_0{b}_0=1$ so since $k$ is a field, $a_0\ne 0$.

Next, let $a_0\ne 0$. We want to build a multiplicative inverse to the infinite formal sum. To fix the constant term, let $b_0=a_0^{-1}$. Then for $n\ge 1$, let

$$b_n=a_0^{-1}\left(-\sum _{j=1}^n{a}_j{b}_{n-j}\right)$$

This is chosen such that

$$\begin{array}{rl}{b}_n& =a_0^{-1}\left(-\sum _{j=1}^n{a}_j{b}_{n-j}\right)\\ a_0{b}_n& =\left(-\sum _{j=1}^n{a}_j{b}_{n-j}\right)\\ 0& =a_0{b}_n+\sum _{j=1}^n{a}_j{b}_{n-j}\end{array}$$

Therefore,

$$\left(\sum _{n=0}^{\infty }{a}_n{x}^n\right)\left(\sum _{n=0}^{\infty }{b}_n{x}^n\right)=a_0{a}_0^{-1}+\sum _{n=1}^{\infty }0x^n=1$$

Thus, ${\sum }^{\infty }{a}_n{x}^n$ is a unit. That means that the only elements that could be in a maximal ideal are those without constant terms, that is

$$\mathfrak{m}=\sum _{n=1}^{\infty }{a}_n{x}^n$$

Note this is an ideal generated by $x$, and it is maximal since if there were any other element it would have a non-zero constant term which would force the ideal to be the entire ring. Any other proper ideal is contained in $\mathfrak{m}$.