local rings

DEFINITION 1.44   A commutative ring $A$ is said to be a local ring if it has only one maximal ideal.

EXAMPLE 1.45   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 1.46  

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 1.47   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 $\mathfrak{p}A_\mathfrak{p}$ .

PROPOSITION 1.48   Let $A$ be a commutative ring. Let $\mathfrak{p}\in \operatorname{Spec}(A)$ then the stalk $\mathcal{O}_\mathfrak{p}$ of $\mathcal{O}$ on $\mathfrak{p}$ is isomorphic to $A_\mathfrak{p}$ .

DEFINITION 1.49   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 local if

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

EXAMPLE 1.50 (of NOT being a local homomorphism)  

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

is not a local homomorphism.