local rings

DEFINITION 07.44   A commutative ring $A$ is said to be a local ring if it has only one maximal ideal.

EXAMPLE 07.45   We give examples of local rings here.

LEMMA 07.46  
  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

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

which shows that

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

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

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

(2) The “only if” part is an easy corollary of (1). The “if” part is also easy.

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COROLLARY 07.47   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 07.48   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 07.49   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

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

EXAMPLE 07.50 (of NOT being a local homomorphism)  

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

is not a local homomorphism.