Local Ring and Localization.

 

Local ringLocalization.

 

Local ring

Definition. A ring is called local if it contains precisely one maximal ideal $\mathfrak{m}$$\mathfrak m$. The field $R/\mathfrak{m}$$R/\mathfrak m$ is called the residue field of the local ring $R$$R$​.

Example.

Arithmetic function ring is a local ring. $R[[X]]$$R[[X]]$ is local ring as well. The stalk of $C^{\mathrm{\infty }}(M)$$C^{\infty}(M)$ at a point $p$$p$​ is a local ring.

Proposition.

Let $R$$R$ be a ring and $\mathfrak{m}$$\mathfrak m$ be a proper ideal, the following conditions on $\mathfrak{m}$$\mathfrak m$ is equivalent:

1. $R$$R$ is a local ring with maximal ideal $\mathfrak{m}$$\mathfrak m$.

2. Every element $R-\mathfrak{m}$$R-\mathfrak m$ is unit in $R$$R$.

3. $\mathfrak{m}$$\mathfrak m$ is a maximal ideal and every element of type $1+m$$1+m$ with $m\in \mathfrak{m}$$m\in\mathfrak m$ is a unit in $R$$R$.

Proof.

$1⟹2$$1\implies 2$. Let $a$$a$ be a non unit element. Then there exists a maximal ideal $(a)\subseteq \mathfrak{n}$$(a)\subseteq\mathfrak n$ (see last essay, prime and maximal ideal). Hence $a\in \mathfrak{n}$$a\in\mathfrak n$ and since $R$$R$ is a local ring, $\mathfrak{n}=\mathfrak{m}$$\mathfrak n=\mathfrak m$. Therefore every non unit element belongs to $\mathfrak{m}$$\mathfrak m$.

$2⟹1.$$2\implies 1.$ Since every proper ideal can not contain unit, and $\mathfrak{m}$$\mathfrak m$ is the set contain all the non unit elements, every ideal are contained in $\mathfrak{m}$$\mathfrak m$.

$2⟹3.$$2\implies 3.$ By $1⟺2$$1\iff2$ we know that $\mathfrak{m}$$\mathfrak m$ is a maximal ideal. Since $1+m\notin \mathfrak{m}$$1+m\notin\mathfrak m$ (Otherwise $1+m-m=1\in \mathfrak{m}$$1+m-m=1\in\mathfrak m$), $1+m$$1+m$ is a unit in $R$$R$.

$3⟹2.$$3\implies 2.$ Let $x\notin \mathfrak{m}$$x\notin\mathfrak m$, then $(x)+\mathfrak{m}=(1)$$(x)+\mathfrak m=(1)$ since $\mathfrak{m}$$\mathfrak m$ is a maximal ideal and for a lattice, $b\le a\vee b,a\vee b=b⟺a\le b$$b\le a\vee b,a\vee b=b\iff a\le b$.

Hence there exist a equation $1=ax-m$$1=ax-m$ for an element $a\in R$$a\in R$. Hence $ax=1+m$$ax=1+m$ is a unit. But $ax$$ax$ is unit iff $a$$a$ is unit and $x$$x$ is unit. Hence $x\notin \mathfrak{m}⟹x$$x\notin\mathfrak m\implies x$ is unit. $◻$$\square$

One important example of local ring in number theory is

$$\begin{array}{}\text{(1)}& {\mathbb{Z}}_{(p)}=\{\frac{m}{n}\in \mathbb{Q}|v_p\left(\frac{m}{n}\right)\ge 0\}\subset \mathbb{Q}\end{array}$$

Then ${\mathbb{Z}}_{(p)}$$\mathbb Z_{(p)}$ is an integral domain, we claim that ${\mathbb{Z}}_{(p)}$$\mathbb Z_{(p)}$ is a PID.

To show that every ideal $\mathfrak{a}\subset {\mathbb{Z}}_{(p)}$$\mathfrak a\subset\mathbb Z_{(p)}$ is principle, consider ${\mathfrak{a}}^{\prime }:=\mathfrak{a}\cap \mathbb{Z}=(a)$$\mathfrak a':=\mathfrak a\cap\mathbb Z=(a)$, which is a principle ideal. Now we claim that $\mathfrak{a}=a{\mathbb{Z}}_{(p)}$$\mathfrak a=a\mathbb Z_{(p)}$ since ${\mathfrak{a}}^{\prime }$$\mathfrak a'$ is the numerator of $\begin{array}{c}\frac{m}{n},v_p\left(\frac{m}{n}\right)\ge 0\end{array}$$\frac{m}{n},v_{p}\left(\frac{m}{n}\right)\ge 0\\$.

Now we prove that ${\mathbb{Z}}_{(p)}$$\mathbb Z_{(p)}$ is a local ring. Firstly, notice that $p$$p$ is not invertible in ${\mathbb{Z}}_{(p)}$$\mathbb Z_{(p)}$ since $\frac{1}{p}\notin {\mathbb{Z}}_{(p)}$$\frac{1}{p}\notin\mathbb Z_{(p)}$.

Hence $\begin{array}{c}p{\mathbb{Z}}_{(p)}=\{\frac{m}{n}\in \mathbb{Q}:p|m\}\end{array}$$p\mathbb Z_{(p)}=\left\{\frac{m}{n}\in\mathbb Q:p|m\right\}\\$ is a proper ideal. Notice that every $\frac{k}{n}\in {\mathbb{Z}}_{(p)}-p{\mathbb{Z}}_{(p)}$$\frac{k}{n}\in \mathbb Z_{(p)}-p\mathbb Z_{(p)}$ is a unit since $(\frac{k}{n}{)}^{-1}=\frac{n}{k}\in {\mathbb{Z}}_{(p)}$$(\frac{k}{n})^{-1}=\frac{n}{k}\in\mathbb Z_{(p)}$.

Geometrically speaking, consider $\mathrm{Spec}(\mathbb{Z})$$\mathrm{Spec}(\mathbb Z)$ with zariski topology and view integral number as functions over $\mathrm{Spec}(\mathbb{Z})$$\mathrm{Spec}(\mathbb Z)$,

then ${\mathbb{Z}}_{(p)}$$\mathbb Z_{(p)}$ is the ring of function that without singular point at ${\mathfrak{p}}_x$$\mathfrak p_x$ and $\begin{array}{c}p{\mathbb{Z}}_{(p)}=\{\frac{m}{n}\in \mathbb{Q}|p|m\}\end{array}$$p\mathbb Z_{(p)}=\left\{\frac{m}{n}\in\mathbb Q|p|m\right\}\\$ is the set of function that vanishing at ${\mathfrak{p}}_x$$\mathfrak p_x$.

An proper analogy is to consider the local ring of a point $P\in {\mathbb{A}}_K^n$$P\in\mathbb A_K^n$. i.e.

$$\begin{array}{}\text{(2)}& {\mathcal{O}}_P:=\{\frac{f}{g}|f,g\in K[T_1,...,T_n],g(P)\ne 0\}=\{\frac{f}{g}|f,g\in K(T_1,...,T_n),{\mathrm{ord}}_P(g)\ge 0\}\end{array}$$

Section 3 of my paper: (PDF) ODE: An Algebraic Approach (researchgate.net) provide a way to solve ODE via local ring.

Now let us consider a special kind of local ring.

Let $R$$R$ be a ring, the following conditions is equivalent.

$(a)$$(a)$ $R$$R$ has exactly one prime ideal $\mathfrak{p}$$\mathfrak p$.

$(b)$$(b)$ Every elements of $R$$R$ is either unit or nilpotent.

$(c)$$(c)$ $\sqrt{(0)}$$\sqrt{(0)}$​ is a maximal ideal.

Proof.

$a⟹b$$a\implies b$

$(a)$$(a)$ shows us that $R$$R$ is a local ring, hence $R-\mathfrak{p}$$R-\mathfrak p$ contains all the unit. Now we need to prove that all the element in $\mathfrak{p}$$\mathfrak p$ are nilpotent. Since there exists only one prime ideal, we have $\sqrt{(0)}=\mathfrak{p}$$\sqrt{(0)}=\mathfrak p$, hence every elements in $\mathfrak{p}$$\mathfrak p$ are nilpotent.

$b⟹c$$b\implies c$

Suppose $\sqrt{0}$$\sqrt{0}$ is not a maximal ideal, let $I$$I$ be an ideal such that $\sqrt{(0)}\subset I$$\sqrt{(0)}\subset I$, and $a\in I-\sqrt{(0)}$$a\in I-\sqrt{(0)}$, then $a$$a$ is a unit, hence $I=R$$I=R$​.

$c⟹a$$c\implies a$

Notice that ${\displaystyle \sqrt{(0)}=\bigcap _{\mathfrak{p}\subseteq R}\mathfrak{p}}$$\displaystyle\sqrt{(0)}=\bigcap_{\mathfrak p\subseteq R}\mathfrak p$. Suppose that $R$$R$ has more than one prime ideal, then $\sqrt{(0)}$$\sqrt{(0)}$ is not maximal ideal anymore, contradiction. $◻$$\square$

Example. $R[T]/(T^n),\mathbb{Z}/p^n\mathbb{Z}$$R[T]/(T^n),\mathbb Z/p^n\mathbb Z$.

Localization.

Definition.

Let $R$$R$ be a ring and $S$$S$ be a multiplicative set, i.e. a submonoid of $(R,\cdot )$$(R,\cdot)$​.

Example. Let $f\in R$$f\in R$ and consider $1,f,f^2,...,f^n...$$1,f,f^2,...,f^n...$ this form a multiplicative set. Also, if $\mathfrak{p}$$\mathfrak p$ is a prime ideal, then $R-\mathfrak{p}$$R-\mathfrak p$ is a multiplicative set.

Let us consider the category $S^{-1}(R)$$S^{-1}(R)$ defined as follows:

Objects in this category are ring homomorphism $f:R\to R^{\prime }$$f:R\to R'$ such that $f(S)\subseteq U(R^{\prime })$$f(S)\subseteq U(R')$.

For $f:R\to R^{\prime },R\to R^{″}$$f:R\to R',R\to R''$, the morphism between $f,g$$f,g$ is a ring homommorphism $h:R^{\prime }\to R^{″}$$h:R'\to R''$

such that $h_{\ast }(f)=h\circ f=g$$h_*(f)=h\circ f=g$.

Does the initial object of $S^{-1}(R)$$S^{-1}(R)$ exists? In other words, if we have a ring homomorphism $f:R\to R^{\prime }$$f:R\to R'$ such that $f(S)\subseteq U(R^{\prime })$$f(S)\subseteq U(R')$, does $f$$f$ uniquely passing through a ring $S^{-1}R$$S^{-1}R$​?

The answer is yes.

Let us construct the initial object $\varphi :R\to S^{-1}R$$\phi:R\rightarrow S^{-1}R$ in category of $S^{-1}(R)$$S^{-1}(R)$​​.

If $R$$R$ is an integral domain, then $S^{-1}R$$S^{-1}R$ is a subring of $F(R)$$F(R)$. But what if $R$$R$ is not integral domain?

For elements in $S^{-1}R$$S^{-1}R$, we denote it as ${\displaystyle \frac{a}{s}},a\in R,s\in S$$\dfrac{a}{s},a\in R,s\in S$, and $\varphi :a\mapsto {\displaystyle \frac{a}{1}}$$\phi:a\mapsto\dfrac{a}{1}$.

Proposition.

$(1).$$(1).$The set $I=\{a\in R:\mathrm{\exists }s\in S,as=0\}$$I=\{a\in R:\exists s\in S,as=0\}$ is an ideal of $R$$R$.

$(2)$$(2)$ For all $f$$f$ such that $f(S)\subseteq U(R^{\prime })$$f(S)\subseteq U(R')$, we have $I\subseteq \mathrm{Ker}f$$I\subseteq\mathrm{Ker} f$.

Proof.

$(1).$$(1).$ If $as=0,a^{\prime }{s}^{\prime }=0$$as=0,a's'=0$, then $(a+a^{\prime })s{s}^{\prime }=(as)s^{\prime }+(a^{\prime }{s}^{\prime })s=0+0$$(a+a')ss'=(as)s'+(a's')s=0+0$, $a{a}^{\prime }(s{s}^{\prime })=(as)(a^{\prime }{s}^{\prime })=0\cdot 0=0$$aa'(ss')=(as)(a's')=0\cdot0=0$.

$(2).$$(2).$ $f(as)=0⟹f(a)f(s)=0$$f(as)=0\implies f(a)f(s)=0$, but $f(s)$$f(s)$ is invertible in $R^{\prime }$$R'$, hence $f(a)=0$$f(a)=0$. $◻$$\square$​​​

Let us define a relationship on $R\times S$$R\times S$ such that $(a,s)\sim (a^{\prime },s^{\prime })⟺{\displaystyle \frac{f(a)}{f(s)}}={\displaystyle \frac{f(a^{\prime })}{f(s^{\prime })}}$$(a,s)\sim (a',s')\iff \dfrac{f(a)}{f(s)}=\dfrac{f(a')}{f(s')}$ for all $f:R\to R^{\prime }$$f:R\to R'$ such that $f(S)\subseteq U(R^{\prime })$$f(S)\subseteq U(R')$. Easy to see that the relationship $\sim$$\sim$ is an equivalence relation.

Now we define a ring sturcture on $S^{-1}R:=(R\times S)/\sim$$S^{-1}R:=(R\times S)/\sim$ as follows.

Firstly we denote the pair $(a,s)$$(a,s)$ as ${\displaystyle \frac{a}{s}}$$\dfrac{a}{s}$, and

$$\begin{array}{}\text{(3)}& {\displaystyle \frac{a}{s}}+{\displaystyle \frac{b}{t}}={\displaystyle \frac{at+bs}{st}},{\displaystyle \frac{a}{s}}\cdot {\displaystyle \frac{b}{t}}={\displaystyle \frac{ab}{st}}\end{array}$$

This is well defined.

If we have

$$\begin{array}{}\text{(4)}& {\displaystyle \frac{a}{s}}={\displaystyle \frac{a^{\prime }}{s^{\prime }}},{\displaystyle \frac{b}{t}}={\displaystyle \frac{b^{\prime }}{t^{\prime }}}\end{array}$$

then

$$\begin{array}{}\text{(5)}& {\displaystyle \frac{a^{\prime }}{s^{\prime }}}+{\displaystyle \frac{b^{\prime }}{t^{\prime }}}={\displaystyle \frac{a^{\prime }{t}^{\prime }+b^{\prime }{s}^{\prime }}{s^{\prime }{t}^{\prime }}}\end{array}$$

and

$$\begin{array}{}\text{(6)}& {\displaystyle \frac{f(a^{\prime }{t}^{\prime }+b^{\prime }{s}^{\prime })}{f(s^{\prime }{t}^{\prime })}}={\displaystyle \frac{f(a^{\prime })f(t^{\prime })+f(b^{\prime })f(s^{\prime })}{f(s^{\prime })f(t^{\prime })}}={\displaystyle \frac{f(a)f(t)+f(b)f(s)}{f(s)f(t)}}={\displaystyle \frac{f(at+bs)}{f(st)}}\end{array}$$

Similarly for ${\displaystyle \frac{ab}{st}}={\displaystyle \frac{a^{\prime }{b}^{\prime }}{s^{\prime }{t}^{\prime }}}$$\dfrac{ab}{st}=\dfrac{a'b'}{s't'}$​.

The universal map $\varphi :R\to S^{-1}R$$\phi:R\to S^{-1}R$ is defined as $a⟼{\displaystyle \frac{a}{1}}$$a\longmapsto\dfrac{a}{1}$.

Now we need to check the universal property of $S^{-1}R$$S^{-1}R$.

Let $\psi :R\to A$$\psi:R\to A$ be a ring homomorphism such that $\psi (S)\subseteq U(A)$$\psi(S)\subseteq U(A)$​.

We need to find the universal map such that ${\psi }^{\prime }\circ \varphi =\psi$$\psi'\circ\phi=\psi$.

Then we define ${\psi }^{\prime }:S^{-1}R:A$$\psi':S^{-1}R:A$ such that ${\psi }^{\prime }\left({\displaystyle \frac{a}{s}}\right)={\displaystyle \frac{\psi (a)}{\psi (s)}}$$\psi'\left(\dfrac{a}{s}\right)=\dfrac{\psi(a)}{\psi(s)}$. Firstly, we need to check that ${\psi }^{\prime }$$\psi'$ is a ring homomorphism.

$$\begin{array}{}\text{(7)}& {\psi }^{\prime }\left({\displaystyle \frac{at+bs}{st}}\right)={\displaystyle \frac{\psi (at+bs)}{\psi (st)}}={\displaystyle \frac{\psi (a)\psi (t)+\psi (b)\psi (s)}{\psi (s)\psi (t)}}={\displaystyle \frac{\psi (a)}{\psi (s)}}+{\displaystyle \frac{\psi (b)}{\psi (t)}}={\psi }^{\prime }\left({\displaystyle \frac{a}{s}}\right)+{\psi }^{\prime }\left({\displaystyle \frac{b}{t}}\right)\end{array}$$

And $\psi ={\psi }^{\prime }\circ \varphi$$\psi=\psi'\circ\phi$:

$$\begin{array}{}\text{(8)}& \psi (a)={\psi }^{\prime }\left({\displaystyle \frac{a}{1}}\right)={\displaystyle \frac{\psi (a)}{\psi (1)}}=\psi (a)\end{array}$$

Suppose there exists another ${\psi }^{″}$$\psi''$ such that $\psi ={\psi }^{″}\circ \varphi$$\psi=\psi''\circ\phi$.

Then

$$\begin{array}{}\text{(9)}& {\psi }^{″}\left({\displaystyle \frac{a}{1}}\right)=\psi (a)={\psi }^{\prime }\left({\displaystyle \frac{a}{1}}\right)\end{array}$$

Hence ${\psi }^{″}(s^{-1})={\psi }^{″}{\left({\displaystyle \frac{s}{1}}\right)}^{-1}=\psi (s{)}^{-1}={\psi }^{\prime }(s^{-1})$$\psi''(s^{-1})=\psi''\left(\dfrac{s}{1}\right)^{-1}=\psi(s)^{-1}=\psi'(s^{-1})$, this implies that

$$\begin{array}{}\text{(10)}& {\psi }^{″}\left(\frac{a}{s}\right)={\psi }^{\prime }\left(\frac{a}{s}\right)\end{array}$$