Valuation Function on Polynomial Ring, From the AG point of View.

Let $R$$R$ be a UFD, then for any prime element $p$$p$, we could define the valuation function $v_p(-):R^{\ast }\to \mathbb{N}$$v_p(-):R^*\to\mathbb N$.

$$\begin{array}{}\text{(1)}& a=u\prod _{p\in P}{p}^{v_p(a)}\end{array}$$

It is not hard to see that $v_p(-):R^{\ast }\to \mathbb{N}$$v_{p}(-):R^*\to\mathbb N$ is a monoid homomorphism.

$$\begin{array}{}\text{(2)}& v_p(mn)=v_p(m)+v_p(n)\end{array}$$

From the Algebraic Geometry point of view, the valuation function tells us the order of zero at the point $(p)\in \mathrm{Spec}(R)$$(p)\in\mathrm{Spec}(R)$.

If you can not understand what I am saying, this blog will be helpful.

It can be extended on $Q(R)$$Q(R)$ naturally. ${\displaystyle v_p\left(\frac{a}{b}\right)=v_p(a)-v_p(b)}$$\displaystyle v_p\left(\frac ab\right)=v_{p}(a)-v_p(b)$. $v_p(-):Q(R{)}^{\ast }\to \mathbb{Z}$$v_p(-):Q(R)^*\to\mathbb Z$ will be a group homomorphism.

I see it has an extension to $R[X]$$R[X]$. Let $f=\sum a_i{X}^i\in R[X],v_p(f):=min{v}_p(a_i)$$f=\sum a_iX^i\in R[X],v_{p}(f):=\min v_p(a_i)$.

But why do we define $v_p(f)$$v_p(f)$ like this?

 

Let us consider an analogy from Differential Geometry. The vector field over a manifold $M$$M$ is a $C^{\mathrm{\infty }}(M)$$C^{\infty}(M)$-Module.

For example, let $M={\mathbb{R}}^3$$M=\mathbb R^3$, then the vector field has the form $\mathrm{\Phi }=f\frac{\partial }{\partial x}+g\frac{\partial }{\partial y}+h\frac{\partial }{\partial z},f,g,h\in C^{\mathrm{\infty }}({\mathbb{R}}^3)$$\Phi=f\frac{\partial}{\partial x}+g\frac{\partial}{\partial y}+h\frac{\partial}{\partial z},f,g,h\in C^{\infty}(\mathbb R^3)$.

Given a point $x\in {\mathbb{R}}^3$$x\in\mathbb R^3$, the evaluation map $x(\mathrm{\Phi }):=\mathrm{\Phi }(x)$$x(\Phi):=\Phi(x)$ give a surjection from vector field to $T_x({\mathbb{R}}^3)$$T_x(\mathbb R^3)$.

The evaluation map trans $C^{\mathrm{\infty }}({\mathbb{R}}^3)$$C^{\infty}(\mathbb R^3)$ to the residue field of $C^{\mathrm{\infty }}({\mathbb{R}}^3)$$C^{\infty}(\mathbb R^3)$ at the maximal ideal ${\mathfrak{m}}_x$$\mathfrak m_x$.

Hence for $\mathrm{\forall }x\in M$$\forall x\in M$, we give an $C^{\mathrm{\infty }}({\mathbb{R}}^3)/{\mathfrak{m}}_x$$C^{\infty}(\mathbb R^3)/\mathfrak m_x$-Module structure.

Similarly, view $R[X]$$R[X]$ as an $R$$R$-Module over $\mathrm{Spec}(R)$$\mathrm{Spec}(R)$, consider a point $\mathfrak{p}\in \mathrm{Spec}(R)$$\mathfrak p\in \mathrm{Spec}(R)$, the evaluation map give a surjection from $R[X]$$R[X]$ to $\begin{array}{c}\frac{R}{\mathfrak{p}}[X]\end{array}$$\frac{R}{\mathfrak p}[X]\\$.

Define the map $\begin{array}{c}{\mathrm{\Phi }}_p:R[X]⟶\frac{R}{\mathfrak{p}}[X]\end{array}$$\Phi_p:R[X]\longrightarrow \frac{R}{\frak p}[X]\\$ for each $\mathfrak{p}\in \mathrm{Spec}(R)$$\mathfrak p\in\mathrm{Spec}(R)$.

The kernel of ${\mathrm{\Phi }}_p$$\Phi_p$ is $\{f\in R[X]|v_p(f)>0\}$$\{f\in R[X]|v_p(f)>0\}$.

Then the value of $f\in R[X]$$f\in R[X]$ at $p\in \mathrm{Spec}(R)$$p\in\mathrm{Spec}(R)$ is ${\mathrm{\Phi }}_p(f)$$\Phi_{p}(f)$, in other words, mod $\mathfrak{p}[X]$$\mathfrak p[X]$.

Moreover, we have the following diagram commute.

$$\begin{array}{}\text{(3)}& \begin{array}{c}\begin{array}{ccc}R& \stackrel{\iota }{\to }& R[X]\\ {\mathrm{\Phi }}_p\downarrow & & {\mathrm{\Phi }}_p\downarrow & \\ \frac{R}{\mathfrak{p}}& \stackrel{\iota }{\to }& \frac{R}{\mathfrak{p}}[X]\end{array}\end{array}\end{array}$$

Then $v_p(f)$$v_{p}(f)$ show us the order of $f$$f$ at $\begin{array}{c}\mathfrak{p}\end{array}$$\frak p\\$ as well.

The primitive polynomial is the non-vanishing ''vector field'' on $\mathrm{Spec}(R)$$\mathrm{Spec}(R)$.

Since by definition, primitive polynomial is $f(X)=\sum a_i{X}^i,gcd(a_1,...,a_n)=1$$f(X)=\sum a_iX^i,\gcd(a_1,...,a_n)=1$. Hence $v_p(f)\ne 0,\mathrm{\forall }p\in P$$v_p(f)\ne 0,\forall p\in P$.

The set $v_p(f)\ge 0$$v_p(f)\ge 0$ is the discrete valuation ring.