Using Linear Algebra to prove that C[x] is PID (Via proving Educlidean domain is PID and for all K, K[x] is Euclidean domain)

In the previous essay, we used the Chinese Remainder Theorem over $\mathrm{P}\mathrm{I}\mathrm{D}$$\rm PID$, but why $\mathbb{C}[x]$$\mathbb C[x]$ is a $\mathrm{P}\mathrm{I}\mathrm{D}$$\rm PID$?

In fact, any Euclidean domain is $\mathrm{P}\mathrm{I}\mathrm{D}$$\rm PID$ , and for any field $K,K[x]$$K,K[x]$ is Euclidean domain.

So what is Euclidean domain?

Definition 1. An integral domain R is called a Euclidean domain if it admits a map $\delta :R-\{0\}\to \mathbb{N}$$δ : R−\{0\}\to \mathbb N$ making possible Euclidean division in R in the following sense:

Given elements $f,g\in R,g\ne 0$$f, g &#8712; R, g \ne 0$, there are elements $q,r\in R$$q, r &#8712; R$ such that $f=qg+r$$f = qg + r$, where $\delta (r)<\delta (g)$$δ(r) < δ(g)$ or $r=0$$r = 0$. The map $\delta$$δ$ is referred to as a Euclidean function of the Euclidean domain $R$$R$

So, Let us prove

Proposition. Every Euclidean domain is $\mathrm{P}\mathrm{I}\mathrm{D}$$\rm PID$

$\mathrm{P}\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{f}.$$\rm Proof.$

Let $R$$R$ be a Euclidean domain, and $\mathfrak{a}\in R$$\mathfrak{a}\in R$ be a non-zero ideal; choose $a\in \mathfrak{a},a\ne 0$$a\in\mathfrak{a},a\ne 0$, such that $\delta (a)$$δ(a)$ is minimal with respect to the Euclidean function $\delta$$δ$ considered on $R$$R$. We claim that $(a)=\mathfrak{a}$$(a) = \mathfrak{a}$.

Let $f\in \mathfrak{a}$$f\in \mathfrak a$, and consider $f=sa+r,\delta (r)<\delta (a)$$f=sa+r,\delta(r)<\delta(a)$, or $r=0$$r=0$ and $r=f-sa$$r=f-sa$, thus $r\in \mathfrak{a}$$r\in\mathfrak{a}$, since $\delta (a)$$\delta(a)$ is minimal, $r=0$$r=0$

Thus $\mathrm{\forall }f\in \mathfrak{a},\mathrm{\exists }s,f=sa$$\forall f\in\mathfrak{a},\exists s,f=sa$, then $\mathfrak{a}=aR$$\mathfrak{a}=aR$, and $0$$0$ is principle ideal. Thus $R$$R$ is $\mathrm{P}\mathrm{I}\mathrm{D}$$\rm PID$. $◼$$\blacksquare$

Then we just need proof for any field $K,K[x]$$K,K[x]$ is Euclidean domain. We will use linear algebra to prove that.

$\mathrm{P}\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{f}.$$\rm Proof.$

Let $\delta :=\mathrm{deg}:K[x]\to \mathbb{N}$$\delta:=\deg:K[x]\to\mathbb N$

For a $f,g\in K[x]$$f,g\in K[x]$, assume $\mathrm{deg}(f)=m,\mathrm{deg}(g)=n$$\deg(f)=m,\deg(g)=n$, assume that $m\ge n$$m\ge n$

We need to prove that there exists unique $s,r,f=sg+r$$s,r,f=sg+r$

View $K[x]$$K[x]$ is a linear space over $K$$K$, consider the linear map $T_g(s,r)=sg+r$$T_g(s,r)=sg+r$

The linear map is injective, since $\mathrm{Ker}{T}_g=0$$\mathrm{Ker}T_g=0$, thus $f=sg+r$$f=sg+r$ is unique.

To see the existence, we need to prove that $T_g$$T_g$ is surjective.

And since $\mathrm{deg}(f)=\mathrm{deg}(sg)=\mathrm{deg}(s)+\mathrm{deg}(g)$$\deg(f)=\deg(sg)=\deg(s)+\deg(g)$,

$T_g:P_{m-n}\times P_{n-1}\to P_m$$T_g:P_{m-n}\times P_{n-1}\to P_m$, ($P_k$$P_k$) is the polynomial space with $\mathrm{deg}(p)\le k$$\deg(p)\le k$

And $\dim P_{m-n}\times \dim P_{n-1}=(m-n+1)+(n-1+1)=m+1=\dim P_m$$\dim P_{m-n}\times\dim P_{n-1}=(m-n+1)+(n-1+1)=m+1=\dim P_m$

And by $T_g$$T_g$ is injective, that implies $T_g$$T_g$ is surjective. Thus we prove that for $\mathrm{\forall }f\in K[x],\mathrm{\exists }f=sg+r$$\forall f\in K[x],\exists f=sg+r$

Thus $K[x]$$K[x]$ is Euclidean domain $⟹\mathbb{C}[x]$$\Longrightarrow\C[x]$ is Euclidean domain $⟹$$\Longrightarrow$ $\mathbb{C}[x]$$\C[x]$ is $\mathrm{P}\mathrm{I}\mathrm{D}$$\rm PID$

$Q.E.D.$$Q.E.D.$