local rings

DEFINITION 02.10   A commutative ring $A$ is said to be a local ring if it has only one maximal ideal.

EXAMPLE 02.11   We give examples of local rings here.

• Any field is a local ring.
• For any commutative ring $A$ and for any prime ideal $\mathfrak{p}\in \operatorname{Spec}(A)$ , the localization $A_\mathfrak{p}$ is a local ring with the maximal ideal $\mathfrak{p}A_\mathfrak{p}$ .

LEMMA 02.12  

1. Let $A$ be a local ring. Then the maximal ideal of $A$ coincides with $A\setminus A^\times$ .
2. A commutative ring $A$ is a local ring if and only if the set $A\setminus A^\times$ of non-units of $A$ forms an ideal of $A$ .

PROOF.. (1) Assume $A$ is a local ring with the maximal ideal $\mathfrak{m}$ . Then for any element $f \in A\setminus A^\times$ , an ideal $I=f A +\mathfrak{m}$ is an ideal of $A$ . By Zorn's lemma, we know that $I$ is contained in a maximal ideal of $A$ . From the assumption, the maximal ideal should be $\mathfrak{m}$ . Therefore, we have

$$f A \subset \mathfrak{m}$$

which shows that

$$A\setminus A^\times \subset \mathfrak{m}.$$

The converse inclusion being obvious (why?), we have

$$A\setminus A^\times =\mathfrak{m}.$$

(2) The ``only if'' part is an easy corollary of (1). The ``if'' part is also easy.

$\qedsymbol$

COROLLARY 02.13   Let $A$ be a commutative ring. Let $\mathfrak{p}$ its prime ideal. Then $A_\mathfrak{p}$ is a local ring with the only maximal ideal $\mathfrak{p}A_\mathfrak{p}$ .

DEFINITION 02.14   Let $A,B$ be local rings with maximal ideals $\mathfrak{m}_A, \mathfrak{m}_B$ respectively. A local homomorphism $\varphi:A \to B$ is a homomorphism which preserves maximal ideals. That means, a homomorphism $\varphi$ is said to be loc al if

$$\varphi^{-1}(\mathfrak{m}_B) =\mathfrak{m}_A$$

EXAMPLE 02.15 (of NOT being a local homomorphism)  

$$\mathbb{Z}_{(p)}\to \mathbb{Q}$$

is not a local homomorphism.

In the argument above, we have used the following lemma.

LEMMA 02.16 (Zorn's lemma)   Let $\mathcal S$ be a partially ordered set. Assume that every totally ordered subset of $\mathcal S$ has an upper bound in $\mathcal S$ . Then $\mathcal S$ has at least one maximal element.

We prove here another consequence of the lemma.

PROPOSITION 02.17   Let $A$ be a commutative ring. let $I$ be an ideal of $A$ such that $A\neq I$ . Then there exists a maximal ideal $\mathfrak{m}$ of $A$ which contains $I$ .