Localization

局部化和取商是交换代数里面最常用的技巧. 取商能够给出包含理想的理想, 而局部化则能给出素理想内部的理想. 可以通过saturation 来一般地刻画 $S^{-1}R$ 中的理想. 而对于module 而言, $S^{-1}(-)$ 这一个函子的正合性十分重要, 这给出了很多局部化module 的商的结果. 同样的, 也可以通过saturation 来刻画 $S^{-1}M$ 的子模. 局部环在几何上是在一点处的连续函数germ 空间, 这为我们研究局部环提供了理由.

Localization of Ring

First we give the universal property of localization.

Theorem 1. Let $R$ be a ring, $S$ is a multiplicative subset. For any homomorphism $\psi:R\to R'$ , $\psi(S)\subset R'^{\times}$ if and only if there exists a ring homomorphism $\rho: S^{-1}R\to R'$ s.t. $\rho\varphi_{S}=\psi$ , where $\varphi_{S}$ is the canonical homomorphism: $\varphi_{S}:R\to S^{-1}R$ .

If so , $\rho$ is unique and $\mathrm{Ker}\rho=\mathrm{Ker}\psi S^{-1}R$ .

Proof

Proof. If $\rho$ exists, for $s\in S$ , $\psi(s)=\rho(s /1)$ so $\psi(s)\rho(1 /s)=\rho(1)=1$ . $\psi(S)\subset R'^{\times }$ .

Conversely, if $\psi(S)\subset R'^{\times }$ , take $\rho(x /s)=\psi(x)\psi(s)^{-1}$ . It’s easy to check $\rho$ is well-defined homomorphism. And $\forall x\in R$ , $\rho\varphi_{S}(x)=\rho(x /1)=\psi(x)\psi(1)^{-1}=\psi(x)$ .

Lastly, $\mathrm{Ker}\rho\supset \mathrm{Ker}\psi S^{-1}R$ is trivial. Conversely, $\forall x/s\in\mathrm{Ker}\rho$ , $\psi(x)\psi(s)^{-1}=0\Rightarrow \psi(x)=0$ . So $x /s\in\mathrm{Ker}\psi S^{-1}R$ . ◻

Corollary 1. Let $R$ be a ring, $S$ be a multiplicative subset. The canonical homomorphism $\varphi_{S}:R\to S^{-1}R$ is an isomorphism if and only if $S$ consists of units.

Proof

Proof. Omitted. ◻

We can localize a ring at any element $f\in R$ .

Definition 1. Let $R$ be a ring, $S_{f}=\{f^{n}|n\geq 0\}$ . Then we call $S_{f}^{-1}R$ the localization of $R$ at $f$ , denoted by $R_{f}$ .

Localization at an element can be easily charactorized.

Theorem 2. Let $f\in R$ , $X$ is an indeterminate, then $R_{f}\cong R[X] /(1-fX)$ .

Proof

Proof. We use the universal property to prove it. Let $x\in R[X] /(1-fX)$ be the image of $X$ . Then $1-\varphi(f)X=0$ . So $\varphi(f)$ is a unit. $\varphi(S_{f})\subset (R[X] /(1-fX))^{\times }$ . Let any $\psi:R\to R'$ with $\psi(f)$ a unit in $R'$ . Define $\theta:R[X]\to R'$ by $\theta|_{R}=\psi$ and $\theta(X)=\psi(f)^{-1}$ . Then $\theta(1-fX)=0$ , $(1-fX)\subset \mathrm{Ker}\theta$ , we have unique homomorphism $\rho:R[X] /(1-fX)\to R'$ s.t. $\psi=\rho\varphi$ . Thus $R[X] /(1-fX)$ satisfies the universal property, i.e. $R[X] /(1-fX)\cong R_{f}$ . ◻

Proposition 1. Let $R$ be a ring, $S$ multiplicative subset, $\mathfrak{a}$ an ideal of $R$ .

$\mathfrak{a}S^{-1}R=\{a /s|a\in \mathfrak{a}, s\in S\}$ .
$\mathfrak{a}\cap S=\varnothing$ if and only if $\mathfrak{a}S^{-1}R=S^{-1}R$ if and only if $\varphi_{S}^{-1}(\mathfrak{a}S^{-1}R)=R$ .

Proof

Proof.

Let $N=\{a /s|a\in \mathfrak{a}, s\in S\}$ . Obviously, $N\subset \mathfrak{a}S^{-1}R$ . Conversely, for $a, b\in \mathfrak{a}$ , $x /s, y /t\in S^{-1}R$ . Then $\varphi(a)x /s+\varphi(b)y /t=ax /s+by /t=(axt+bys) /st$ , so $\mathfrak{a}S^{-1}R\subset N$ .
If $s\in\mathfrak{a}\cap S$ , then $1_{S^{-1}R}=s /s\in\mathfrak{a}S^{-1}R$ , so $\mathfrak{a}S^{-1}R=S^{-1}R$ . $\varphi_{S}^{-1}(\mathfrak{a}S^{-1}R)=R$ .
Conversely, if $\varphi_{S}^{-1}(\mathfrak{a}S^{-1}R)=R$ , then $1\in \mathfrak{a}S^{-1}R$ . So $\exists a\in\mathfrak{a}$ , $s\in S$ s.t. $a /s=1$ , so $\exists t\in S$ with $at=st$ . Since $at\in\mathfrak{a}$ , $st\in S$ , $\mathfrak{a}\cap S=\varnothing$ . ◻

To describe the ideals of $S^{-1}R$ more precisely, we use the saturation: $\mathfrak{a}^{S}=\{a\in R|\exists s\in S \text{ s.t. }as\in\mathfrak{a}\}$ , i.e. $\mathfrak{a}^{S}=\cap_{s\in S}(\mathfrak{a}:s)$ . And it’s easy to show $\mathrm{Ker}\varphi_{S}=(0)^{S}$ and $\mathfrak{a}\subset \mathfrak{a}^{S}$ . $\mathfrak{a}^{S}$ is an ideal.

Proposition 2. Let $R$ be a ring, $S$ a multiplicative subset,

Let $\mathfrak{b}$ be an ideal of $S^{-1}R$ , then $\varphi_{S}^{-1}(\mathfrak{b})=(\varphi_{S}^{-1}(\mathfrak{b}))^{S}$ , $\mathfrak{b}=\varphi_{S}^{-1}(\mathfrak{b})S^{-1}R$ .
Let $\mathfrak{a}$ be an ideal of $R$ , then $\mathfrak{a}S^{-1}R=\mathfrak{a}^{S}S^{-1}R$ , $\varphi_{S}^{-1}(\mathfrak{a}S^{-1}R)=\mathfrak{a}^{S}$ .
Let $\mathfrak{p}$ be a prime ideal of $R$ , $\mathfrak{p}\cap S=\varnothing$ , then $\mathfrak{p}=\mathfrak{p}^{S}$ and $\mathfrak{p}S^{-1}R$ is prime in $S^{-1}R$ .

Proof

Proof.

First, $\forall a\in (\varphi_{S}^{-1}(\mathfrak{b}))^{S}$ , $\exists s\in S$ s.t. $as\in \varphi_{S}(\mathfrak{b})$ . Then $as /1\in\mathfrak{b}$ , so $a /1\in\mathfrak{b}$ . Thus $\mathfrak{a}=\varphi_{S}^{-1}(a /1)\in \varphi_{S}^{-1}(\mathfrak{b})\Rightarrow \varphi_{S}^{-1}(\mathfrak{b})=\varphi_{S}^{-1}(\mathfrak{b})$ .
Second, $\forall a /s\in \mathfrak{b}$ , $a /1\in\mathfrak{b}$ . So $a\in \varphi_{S}^{-1}(\mathfrak{b})\Rightarrow a /s\in(\varphi_{S}^{-1}(\mathfrak{b}))S^{-1}R$ . The converse is obvious.
First, $\forall a\in \mathfrak{a}^{S}$ , $\exists s\in S$ , $as\in \mathfrak{a}$ so $a /1=as /1\cdot 1 /s\in\mathfrak{a}S^{-1}R$ . Thus $\mathfrak{a}^{S}S^{-1}R\subset \mathfrak{a}S^{-1}R$ . So $\mathfrak{a}^{S}S^{-1}R=\mathfrak{a}S^{-1}R$ .
Second, $\varphi_{S}^{-1}(\mathfrak{a}S^{-1}R)=\varphi_{S}^{-1}(\mathfrak{a}^{S}S^{-1}R)\supset \mathfrak{a}^{S}$ . Conversely, $\forall x\in \varphi_{S}^{-1}(\mathfrak{a}^{S}S^{-1}R)$ , $x /1\in\mathfrak{a}^{S}S^{-1}R=\mathfrak{a}S^{-1}R$ . So $x /1=a /s$ for $a\in \mathfrak{a}$ , $s\in S$ . Thus $(xs-a)t=0$ for some $t\in S\Rightarrow x\in \mathfrak{a}^{S}$ .
First, if $sa\in \mathfrak{p}$ for $s\in S$ , $s\notin \mathfrak{p}\Rightarrow a\in \mathfrak{p}$ . So $\mathfrak{p}^{S}=\mathfrak{p}$ .
Second, for any $a /s\cdot b /t\in \mathfrak{p}S^{-1}R$ , we have $ab\in \varphi_{S}^{-1}(\mathfrak{p}S^{-1}R)=\mathfrak{p}^{S}=\mathfrak{p}$ . So $a\in \mathfrak{p}$ or $b\in\mathfrak{p}$ . Then $a /s\in S^{-1}R$ or $b /t\in \mathfrak{p}S^{-1}R$ . Hence $\mathfrak{p}S^{-1}R$ is prime. ◻

From the above theorem, we can conclude that $\varphi_{S}$ gives a bijection between ideals $\mathfrak{a}$ of $R$ with $\mathfrak{a}=\mathfrak{a}^{S}$ and ideals of $S^{-1}R$ .

For the prime ideals $\mathfrak{p}\cap S=\varnothing$ , we have $\mathfrak{p}=\mathfrak{p}^{S}$ so there is bijection between prime ideal $\mathfrak{p}\cap S=\varnothing$ and prime ideals of $S^{-1}R$ .

Definition 2. Let $R$ be a ring, $\mathfrak{p}$ a prime ideal of $R$ . Let $S_{\mathfrak{p}}=R-\mathfrak{p}$ . We call the ring $S_{\mathfrak{p}}^{-1}R$ the localization of $R$ at $\mathfrak{p}$ , and denoted by $R_{\mathfrak{p}}$ .

$R_{\mathfrak{p}}$ is the local ring with the maximal ideal $\mathfrak{p}R_{\mathfrak{p}}$ .

Theorem 3. Let $R$ be a ring, $S$ multiplicative subset, Let $T'$ be a multiplicative subset of $S^{-1}R$ , $T=\varphi_{S}^{-1}(T')$ . Assume $S\subset T$ , then $(T')^{-1}(S^{-1}R)=T^{-1}R$ .

Proof

Proof. Idea: use the universal property. Omitted. ◻

Corollary 2. Let $R$ be a ring, $\mathfrak{p}\subset \mathfrak{q}$ prime ideals. Then $R_{\mathfrak{p}}$ is the localization of $R_{\mathfrak{q}}$ at the prime ideal $\mathfrak{p}R_{\mathfrak{q}}$ .

Proof

Proof. Let $S=R-\mathfrak{q}$ , $T'=R_{\mathfrak{q}}-\mathfrak{p}R_{\mathfrak{q}}$ . Set $T=\varphi_{S}^{-1}(T')$ . Then $T=R-\mathfrak{p}$ . By above theorem, we have $R_{\mathfrak{p}}$ the localization of $R_{\mathfrak{q}}$ at $\mathfrak{p}R_{\mathfrak{q}}$ . ◻

Localization of Module

Theorem 4. Let $R$ be a ring, $S$ multiplicative subset. Then $M$ has a compatible $S^{-1}R$ -module structure if and only if $\forall s\in S$ , the homomorphism $\mu_{s}:m\mapsto sm$ is an isomorphism. If so , the $S^{-1}R$ -module structure is unique.

Proof

Proof. Assume $M$ has a compatible $S^{-1}R$ -module structure, take $s\in S$ . Then $\mu_{s}=\mu_{s /1}$ . So $\mu_{s}\cdot \mu_{1 /s}=1$ . Similarly, $\mu_{1 /s}\cdot \mu_{s}=1$ . $\mu_{s}$ is bijective.

Conversely,if $\mu_{s}$ is bijective for all $s\in S$ . Then $\mu:R\to \mathrm{End}(M)$ given by $\mu:s\mapsto \mu_{s}$ sends $S$ to units of $\mathrm{End}(M)$ . Hence there’s unique $M_{S^{-1}R}:S^{-1}R\to \mathrm{End}(M)$ , there’s unique compatible $S^{-1}R$ -module structure on $M$ . ◻

Now we give the universal property of localized modules.

Theorem 5. Let $R$ be a ring. $S$ a multiplicative subset, $M$ an $R$ -module. Then for any $S^{-1}R$ -module $N$ with $R$ -module homomorphism $\psi:M\to N$ , there’s a $S^{-1}R$ -module homomorphism: $\rho:S^{-1}M\to N$ s.t. $\rho\varphi_{S}=\psi$ .

Proof

Proof. Similar as the analogy in ring. ◻

We can show that $S^{-1}(-)$ is actually a functor. Moreover, it’s the left adjoint to the functor of restriction of scalars, i.e. for $R$ -module $M$ and $S^{-1}R$ -module $N$ , we can regard $N$ as an $R$ -module. Then $\mathrm{Hom}_{S^{-1}R}(S^{-1}M,N)\cong \mathrm{Hom}_{R}(M,N)$ by universal property.

Corollary 3. Let $R$ be a ring, $S$ be a multiplicative subset. Then $S^{-1}(-)$ and $S^{-1}R\otimes_{R}-$ are naturally isomorphic.

Similar as the localization of rings, we use the saturation of a submodule to charactorize $\varphi_{S}^{-1}$ . For $N\subset M$ be $R$ -modules, the saturation $N^{S}=\{m\in M|\exists s \text{ s.t. }ms\in N\}$ .

Proposition 3. Let $R$ be a ring, $M$ be an $R$ -module, $N$ , $P$ submodules. Let $S,T$ be multiplicative subsets, $K$ an $S^{-1}R$ -submodule of $S^{-1}M$ . Then

$\varphi_{S}^{-1}(K)=(\varphi_{S}^{-1}(K))^{S}$ and $K=S^{-1}(\varphi_{S}^{-1}(K))$ .
$\varphi_{S}^{-1}(S^{-1}N)=N^{S}$ , $S^{-1}N=S^{-1}N^{S}$ , so $\mathrm{Ker}\varphi_{S}=0^{S}$ .
$(N^{S})^{T}=N^{ST}$ , $S^{-1}(S^{-1}N)=S^{-1}N$ .
Assume $N\subset P$ , then $N^{S}\subset P^{S}$ and $S^{-1}N\subset S^{-1}P$ .
$(N\cap P)^{S}=N^{S}\cap P^{S}$ and $S^{-1}(N\cap P)=S^{-1}N\cap S^{-1}P$ .
$(N+P)^{S}\supset N^{S}+P^{S}$ and $S^{-1}(N+P)=S^{-1}N+S^{-1}P$ .
Assume $S\subset T$ , then $N^{S}\subset N^{T}$ .

Proof

Proof.

First for $a\in(\varphi_{S}^{-1}(K))^{S}$ , $\exists s\in S$ , $sa /1\in K$ . Then $a /1=1/s\cdot sa /1\in K$ . So $a\in \varphi_{S}^{-1}(K)$ .
Second, $K\subset S^{-1}(\varphi_{S}^{-1}(K))$ is obvious. For any $a /s\in S^{-1}(\varphi_{S}^{-1}(K))$ , $a /1\in K\Rightarrow a /s=1 /s\cdot a /1\in K$ . So $S^{-1}(\varphi_{S}^{-1}K)\subset K$ .
First, for any $a\in N^{S}$ , $\exists s\in S s.t. sa \in N$ . So $sa /1\in S^{-1}N$ , $a /1=1 /s\cdot sa /1\in S^{-1}N$ . Thus $a\in \varphi_{S}^{-1}(S^{-1}N)$ .
Conversely, $\forall a\in \varphi_{S}^{-1}S^{-1}N$ , $a /1\in S^{-1}N$ , so $t(sa-b)=0$ for some $b\in N$ , $s,t\in S$ . Thus $sta=tb\in N$ , $a\in N^{S}$ . Immediately, we have $\mathrm{Ker}\varphi_{S}=0^{S}$ .
Second, $S^{-1}N\subset S^{-1}N^{S}$ is obvious. $\forall a /t\in S^{-1}N^{S}$ , $sa\in N$ for some $s\in S$ . Then $a /t=sa /ts\in S^{-1}N$ .
First just note $a\in (N^{S})^{T}$ if and only if $\exists t\in t$ , $s\in S$ s.t. $s(tn)=(st)n\in N$ .
Second one is obvious.
Obvious.
First, it’s easy to see $(N\cap P)^{S}\subset N^{S}\cap P^{S}$ . Converselym if $a\in N^{S}\cap P^{S}$ , $\exists s,t\in S$ , $sa\in N$ , $ta\in P$ , so $sta\in N\cap P$ . Thus $a\in (N\cap P)^{S}$ .
Second, $N\cap P\subset N, P$ , so $S^{-1}(N\cap P)=\subset S^{-1}N\cap S^{-1}P$ . Conversely, if $a /s=b/ t\in S^{-1}N\cap S^{-1}P$ with $a\in N$ , $b\in P$ , $s,t\in S$ .Then $\exists w\in S$ s.t. $wta=wsb\in N\cap P$ . Thus $wta /wts=a /s\in S^{-1}(N\cap P)$ .
First, for $a\in N^{S}$ , $b\in P^{S}$ , $\exists s,t\in S$ s.t. $sa\in N$ , $tb\in P$ , so $st(a+b)\in N+P$ . $a+b\in (N+P)^{S}$ .
Second, $N,P\subset N+P$ , so $S^{-1}(N+P)\supset S^{-1}N+S^{-1}P$ . The converse is obvious.
Obvious. ◻

Theorem 6. The functor $S^{-1}(-)$ is exact.

Proof

Proof. For $M'\xrightarrow{f} M\xrightarrow{g}M''$ exact at $M$ , it’s easy to see $S^{-1}g\circ S^{-1}f=0$ . So $\mathrm{Im}S^{-1}f\subset \mathrm{Ker}S^{-1}g$ .

Conversely, for $a /s\in \mathrm{Ker}S^{-1}g$ , $g(a) /s=0$ so there is $t\in S$ s.t. $tg(a)=g(ta)=0$ . So $ta\in \mathrm{Ker}g=\mathrm{Im}f$ . $ta=f(b)$ for some $b\in M'$ . Hence $a /s=S^{-1}f(b /st)\in \mathrm{Im}S^{-1}f$ . Thus $\mathrm{Ker}S^{-1}g=\mathrm{Im}S^{-1}f$ . ◻

Corollary 4. $S^{-1}R$ is flat over $R$ .

Proof

Proof. Notice that $S^{-1}R\otimes_{R}M\cong S^{-1}M$ . ◻

Corollary 5. Let $M$ be an $R$ -module. Then $S^{-1}(M /\mathfrak{a}M)\cong S^{-1}M /S^{-1}(\mathfrak{a}M)\cong S^{-1}M /\mathfrak{a}S^{-1}M$ .

Proof

Proof. It’s easy to see $S^{-1}(\mathfrak{a}M)=\mathfrak{a}S^{-1}M$ . To show the first equation, just apply $S^{-1}(-)$ to $0\to \mathfrak{a}M\to M\to M/\mathfrak{a}M$ . ◻