p-adic Valuation and its Connection to Algebraic Geometry and Complex Analysis

Initial idea and first obervation

I would like to find the connection between $p-\mathrm{a}\mathrm{d}\mathrm{i}\mathrm{c}$$p-\rm adic$ number and algebraic geometry and complex analysis.

The basic idea of this essay comes from:

Math Essays: Coprime-Orthogonal, Module and p-adic valuation (wuyulanliulongblog.blogspot.com)

Math Essays: A contravariant functor from sub cat of Top to Algebra, a kind of algebra-geometry duality (Typo, in the lattice isomorphism, S should be the closed set, so we do not have the Boolean Homomorphism) (wuyulanliulongblog.blogspot.com)

Math Essays: From Calculus to Algebraic Geometry (wuyulanliulongblog.blogspot.com)

If you read them before, this blog will be easy to follow.

Let $\mathcal{O}(\mathrm{\Omega })$$\mathcal O(\Omega)$ be the holomorphic function ring defined on a domain $\mathrm{\Omega }$$\Omega$.

That is a functor.

$$\begin{array}{}\text{(1)}& \mathrm{Hom}(-,\mathbb{C})\end{array}$$

We could induce the evaluation map at one point $a\in \mathrm{\Omega }$$a\in\Omega$.

$$\begin{array}{}\text{(2)}& a(f):=f(a)\end{array}$$

Then, all the functions vanish at $a$ will from a maximal ideal since $\mathrm{im}a\cong \mathbb{C}$$\mathrm{im}\,a\cong\mathbb C$.

It is not hard to see that the evaluation map is equivalence to the quotient of the maximal ideal $m_a$$m_a$.

Conversely, for any commutative ring, consider the Spectrum functor.

$$\begin{array}{}\text{(3)}& \mathrm{Spec}:\mathrm{CRing}⟶\mathrm{Top}\end{array}$$

That is, define Zariski Topology on $\mathrm{Spec}(R)$$\mathrm{Spec}(R)$. Then, each maximal ideal is the closed point in $\mathrm{Spec}(R)$$\mathrm{Spec}(R)$.

Then, we could view $R$$R$ as the function ring over $\mathrm{Spec}(R)$$\mathrm{Spec}(R)$.

Each $r$$r$ is a function

$$\begin{array}{}\text{(4)}& r:\mathrm{Spec}(R)⟶\coprod _{m_x\in \mathrm{MSpec}(R)}\frac{R}{m_x}\end{array}$$

The value of a function $r\in R$$r\in R$ at one point $m_a$$m_a$ is defined by

$$\begin{array}{}\text{(1)}& r(a):=r\mathrm{mod}{m}_a\end{array}$$

We will deal with this example, $\mathbb{Z}$$\mathbb Z$ and $\mathrm{Spec}(\mathbb{Z})$$\mathrm{Spec}(\mathbb Z)$ next.

How to describe an integral number $n$$n$ (as a function)?

You might consider the value of $n$$n$ at each point.

That is,

$$\begin{array}{}\text{(2)}& n\mathrm{mod}{p}_i,i\in {\mathbb{N}}^{+}\end{array}$$

That is,

$$\begin{array}{}\text{(3)}& a=b⟺a\equiv b\mathrm{mod}{p}_i\end{array}$$

But this condition needs to check infinitely many values for a function. It needs to be better.

In fact, we only need to consider the zero point of each function.

In other words, the local information shows us the global information!

That is, let $\{p_i,i\in I\}$$\{p_i,i\in I\}$ be the zero point of a function $m$$m$, then

$$\begin{array}{}\text{(4)}& a=\pm b⟺{\mathrm{ord}}_{p_i}(a)={\mathrm{ord}}_{p_i}(b)\end{array}$$

Where ${\mathrm{ord}}_{p_i}$$\mathrm{ord}_{p_i}$ is the p-adic valuation.

For example

$$\begin{array}{}\text{(5)}& {\mathrm{ord}}_2(18)=1,{\mathrm{ord}}_3(18)=2\end{array}$$

That tells us $2$$2$ is the first order zero of $18$$18$, and $3$$3$ is the second order of $18$$18$.

The order of each zero determines the function over $\mathrm{Spec}\mathbb{Z}$$\mathrm{Spec}\mathbb Z$ up to a unit. Since $-18$$-18$ has the same order at the same zero.

In fact, it's just the fundamental theorem of arithmetic!

It is not hard to generalize $\mathbb{Z}$$\mathbb Z$ to $\mathbb{Q}$$\mathbb Q$. See Math Essays: Coprime-Orthogonal, Module and p-adic valuation (wuyulanliulongblog.blogspot.com).

It is not hard to see the similarities between $\mathbb{Z}$$\mathbb Z$ and $\mathbb{C}[X]$$\mathbb C[X]$. For example, there are both Euclidean domains.

For $\mathbb{Z}$$\mathbb Z$, we have the fundamental theorem of arithmetic, for $\mathbb{C}[X]$$\mathbb C[X]$, we have the fundamental theorem of algebra.

The order of each zero determines the function up to a unit.

For example

$$\begin{array}{}\text{(6)}& p(x)=\lambda (z-2)(z-3{)}^2,{\mathrm{ord}}_2(p)=1,{\mathrm{ord}}_3(p)=2\end{array}$$

For any $q(x)\in \mathbb{C}[X]$$q(x)\in\mathbb C[X]$If you want to know the ${\mathrm{ord}}_a(q)$$\mathrm{ord}_a(q)$

You might need to consider the Taylor series of $q(x)$$q(x)$ at $a$.

That is

$$\begin{array}{}\text{(7)}& q(x)=c_0+c_1(x-a)+c_2(x-a{)}^2+...+c_n(x-a{)}^n\end{array}$$

Then the ${\mathrm{ord}}_a(q)$$\mathrm{ord}_a(q)$ is the least $k,c_k\ne 0$$k,c_k\ne 0$.

For example

$$\begin{array}{}\text{(8)}& p(x)=(z-2)(z-3{)}^2=(z-2)(z^2-6z+9)=(x-2)-2(x-2{)}^2+(x-2{)}^3\end{array}$$

i.e.

$$\begin{array}{}\text{(9)}& {\mathrm{ord}}_2(p(x))=1\end{array}$$

Similarly, if you want to know the ${\mathrm{ord}}_p(n)$$\mathrm{ord}_p(n)$, you might expand the function $n$$n$ at $p$$p$!

$$\begin{array}{}\text{(10)}& n=c_0+c_{1}p+c_2{p}^2+...+c_n{p}^n\end{array}$$

That is why we define the p-adic norm of $n$$n$ as $p^{-{\mathrm{ord}}_p(n)}$$p^{-\mathrm{ord}_p(n)}$.

${\mathrm{ord}}_p$$\mathrm{ord}_p$ show us the rate of convergence to $0$$0$ at $(p)$$(p)$ of $n$$n$. Thus $p^{-{\mathrm{ord}}_p(n)}$$p^{-\mathrm{ord}_p(n)}$ show us the distance between $n$$n$ and $0$$0$ function at $(p)$$(p)$.

Then we could do the localization here.

Consider $\mathbb{Z}-(p)$$\mathbb Z-(p)$, that is, all the functions do not vanish at $(p)$$(p)$, make them invertible.

i.e.

$$\begin{array}{}\text{(11)}& {\mathbb{Z}}_{(p)}=\{\frac{m}{n}\in \mathbb{Q}|m,n\in \mathbb{Z},n\notin (p)\}\end{array}$$

if you expand $\begin{array}{c}\frac{m}{n}\end{array}$$\frac{m}{n}\\$ at $(p)$$(p)$,

i.e.

$$\begin{array}{}\text{(12)}& \frac{m}{n}=\frac{c_0+c_{1}p+c_2{p}^2+...+c_m{p}^m}{c_0+c_{1}p+c_2{p}^2+...+c_n{p}^n}=a_0+a_{1}p+a_2{p}^2+...+a_k{p}^k+...\end{array}$$

Then consider the completion under the p-adic metric

$$\begin{array}{}\text{(13)}& d_p(a,b)=p^{-{\mathrm{ord}}_p(a-b)}\end{array}$$

We get ${\mathbb{Z}}_{(p)}$$\mathbb Z_{(p)}$.

To see

$$\begin{array}{}\text{(14)}& {\mathrm{ord}}_p(a-b)\end{array}$$

Let $a=a_0+a_{1}p+a_2{p}^2+...+a_k{p}^k+...,b=b_0+b_{1}p+b_2{p}^2+...+b_k{p}^k+...$$a=a_0+a_1p+a_2p^2+...+a_kp^k+...,b=b_0+b_1p+b_2p^2+...+b_kp^k+...$

Then take the difference

$$\begin{array}{}\text{(15)}& a-b=(a_0-b_0)+(a_1-b_1)p+(a_2-b_2)p^2+...+(a_k-b_k)p^k+...\end{array}$$

Then...

If we consider the completion of ${\mathbb{Z}}_{(p)}$$\mathbb Z_{(p)}$ under the p-adic metric, we get the p-adic integral number ${\mathbb{Z}}_p$$\mathbb Z_p$.

That will correspond $\mathbb{C}[[X]]$$\mathbb C[[X]]$, in other words, holomorphic function.

Using the language of Algebraic Geometry

If you are farmilar with Zariski Topology, it is not hard to see that the closed set in $\mathrm{Spec}\mathbb{Z}$$\mathrm{Spec}\mathbb Z$ is a finite union of prime ideals.

Thus the open set is $\mathrm{Spec}\mathbb Z-\{(p_{i_1}),...,(p_{{i_n}})\}$. That corresponds to the domain where $\begin{array}{c}\frac{b}{p_{i_1}^{j_1}...p_{i_n}^{j_n}}\end{array}$$\frac{b}{p^{j_1}_{i_1}...p^{j_n}_{i_n}}\\$ holomorphic!

Indeed, what we should do here is consider the concept scheme.

Scheme is a locally ringed space, thus manifold is a kind of scheme as well.

For any open set $X=\mathrm{Spec}\mathbb{Z}-\{(p_{i_1}),...,(p_{i_n})\}\subseteq \mathrm{Spec}(\mathbb{Z})$$X=\mathrm{Spec}\mathbb Z-\{(p_{i_1}),...,(p_{{i_n}})\}\subseteq\mathrm{Spec}(\mathbb Z)$, We could consider the ring $\mathcal{F}(X)$$\mathcal F(X)$ associate with $X$$X$.

It may not be the most natural way, but I want to let $\mathcal{F}(X)$$\mathcal F(X)$ be the ring of rational function over $X$$X$.

i.e.

$$\begin{array}{}\text{(16)}& \{\frac{b}{p_{i_1}^{j_1}...p_{i_n}^{j_n}},(j_1,...,j_n)\in {\mathbb{N}}^n,b\in \mathbb{Z}\}\end{array}$$

Then we could consider the stalk

$$\begin{array}{}\text{(17)}& {\mathcal{F}}_{(p)}:=\underset{X\ni (p)}{\underrightarrow{\lim }}\mathcal{F}(X)={\mathbb{Z}}_{(p)}=\{\frac{m}{n}\in \mathbb{Q}|m,n\in \mathbb{Z},n\notin (p)\}\end{array}$$

Just like what we do in differential geometry, we can define $\begin{array}{c}{\mathrm{m}}_p/{{\mathrm{m}}_{\mathrm{p}}}^2\end{array}$${\mathrm {m}_p}/{\mathrm{m_p}^2}\\$ be the cotangent space, similarly $(p)/(p{)}^2$$(p)/(p)^2$ .

Then do the completion for the stalk we get the ${\mathbb{Z}}_p$$\mathbb Z_p$, the holomorphic function around $(p)$$(p)$.

Another interesting thing is ${\mathbb{Z}}_p$$\mathbb Z_p$ is the unit ball in the ${\mathbb{Q}}_p$$\mathbb Q_p$ .

One fact we know that ${\mathbb{Z}}_p$$\mathbb Z_p$ Corresponding to holomorphic function over the unit ball.

We know that ${\mathbb{Q}}_p-{\mathbb{Z}}_p$$\mathbb Q_p-\mathbb Z_p$ corresponding to Laurent series on $1<|z|$$1<|z|$.

i.e.

$$\begin{array}{}\text{(18)}& \begin{array}{c}{\mathbb{Z}}_p⟷\mathbb{D}⟷\mathcal{O}(\mathbb{D})\\ {\mathbb{Q}}_p⟷\mathbb{C}-\mathbb{D}⟷\mathcal{O}(\mathbb{C}-\mathbb{D})\end{array}\end{array}$$

Further

In general, Let $R$$R$ be a $\mathrm{U}\mathrm{F}\mathrm{D}$$\rm UFD$, we could consider the spectrum functor as well.

$$\begin{array}{}\text{(19)}& \mathrm{Spec}:\mathrm{CRing}⟶\mathrm{Top}\end{array}$$

Since in $\mathrm{U}\mathrm{F}\mathrm{D}$$\rm UFD$, $p$$p$ is prime $\Rightarrow (p)$$\Rightarrow (p)$ is a maximal ideal, all the $(p)\in \mathrm{Spec}(R)$$(p)\in\mathrm{Spec}(R)$ are closed points.

Then similarly we could view $r\in R$$r\in R$ as a function over $\mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}(\mathrm{R})$$\rm Spec(R)$.

Then similarly, you could consider the p-adic valuation, if $r=p^{n}b$$r=p^nb$, then ${\mathrm{ord}}_p(r)=n.$$\mathrm{ord}_p(r)=n.$

The order of each zero determines the function up to a unit as well.

But the issue here is $p\notin \mathbb{R}$$p\notin\mathbb R$, thus we could not define the p-adic norm as $\parallel a{\parallel }_p=p^{-{\mathrm{ord}}_{\mathrm{p}}(a)}$$\|a\|_p=p^{-\mathrm{ord_p}(a)}$.

However, we could define the norm as

$$\begin{array}{}\text{(20)}& \parallel a{\parallel }_p=2^{-{\mathrm{ord}}_p(a)}\end{array}$$

One important example I would like to consider is related to ODE an Algebraic Approach.

Math Essays: ODE An Algebraic Approach (2) (wuyulanliulongblog.blogspot.com)

It is not hard to see that $\mathbb{R}[D]$$\mathbb R[D]$, where $\begin{array}{c}D=\frac{d}{dx}\end{array}$$D=\frac{d}{dx}\\$ is an $\mathrm{E}\mathrm{D}\Rightarrow \mathrm{P}\mathrm{I}\mathrm{D}\Rightarrow \mathrm{U}\mathrm{F}\mathrm{D}$$\rm{ED}\Rightarrow{PID}\Rightarrow UFD$.

Indeed,

$$\begin{array}{}\text{(21)}& \mathbb{C}[\frac{\partial }{\partial x_1},\frac{\partial }{\partial x_2},...,\frac{\partial }{\partial x_n}]\end{array}$$

is $\mathrm{U}\mathrm{F}\mathrm{D}$$\rm UFD$ as well.

When I try to use the formal power series inverse of a differential opreator polynomial to solve ode, what I do, in fact is consider the localization.

I will back to it.