Pre-Definitions

Consider the ring $K[X]$ of polynomials with coefficients in a field k. $$K[X]=\\{ p_nx^n+...+p_2x^2+p_1x+p_0|p_i\in Z,i=0,1,2,...n\\}$$

If $f,g\in K[X]$ and there is a $h\in K[X]$, allows $g=fh$, then $f|g$.

Denote $\deg{f}$ as the degree of $f$, $\deg{fg}=\deg{f}+\deg{g}$.

Iff $f$ is a nonzero constant, $\deg{f}=0$. So iff $f=cg$ and c is nonzero constant, $f|g,g|f$​.

Prime in K[X]: Irreducible Polynomial

A nonconstant polynomial p is said to be irreducible if $q|p$ that q is either a constant or a constant times p.

Lemma 1

Every nonconstant polynomial is the product of irreducible polynomial.

Proof:

First, all polynomial of degree 1 are irreducible. Assume a lowest degree polynomials f that have $f=gh$, which $\deg{f}=n$ and $1 \geq \deg{g},\deg{h} < n$. So it will be a contradiction.

Theorem 2

Define f is a monic polynomial, if its leading coefficient is 1. So any nonzero polynomial will be a constant times of monic polynomial.

Now we can define $\text{ord}_pf$ similar in $\Z$. Let $p$ is a monic polynomial. $\text{ord}_pf$ is the integer a that $p^a|f$ and $p^{a+1}\nmid f$​.

Let $f\in K[X]$, then we have: $$f = c \prod_pp^{a(p)}$$

As in last posts, the constant c and the $a(p)$ are unique. $a(p)=\text{ord}_pf$.

Lemma 2

Let $f,g \in K[X]$ with $g\neq 0$, there exist unique $q,r\in K[X]$ such that $f=hg+r$ where $0 \leq \deg{r} < \deg{g}$.

Proof:

If $g|f$, we will have $h=f/g$ and $r=0$. Otherwise, consider a set $S=\\{f -lg | l \in K[X]\\}$ and $r \in S$ as the least degree polynomial. If $\deg{r} \geq \deg{g}$, let the leading term of $r$ be $ax^d$ and leading term of $g$ be $bx^m$. Then $f - (h + ab^{-1}x^{d-m})g = r - ab^{-1}x^{d-m}g$ which degree is smaller than r. Contradiction.

Lemma 3

Define the ideal of ring $K[X]$, if $f_1,f_2,...,f_n \in K[X]$, then we have $(f_1,f_2,...,f_n)$ for all polynomials of the form $f_1h_1,f_2h_2,...,f_nh_n$ where $h_1,h_2,...,h_n \in K[X]$.

Given $f,g \in K[X]$, there will be a $d \in K[X]$ such that $(f,g)=(d)$.

Proof:

Let d be the smallest degree polynomial that $d\in(f,g)$. Let $c \in (f,g)$, if $d\nmid c$, we shall have $c = hd+r$ and $\deg{r} < \deg{d}$. Since $c,d \in (f,g)$, so $r \in (f,g)$​ which is a contradiction that r has smaller degree than d.

Lemma 4

Now we can have a polynomial version of GCD:

Let $f,g\in K[X]$, if there is a $d\in K[X]$ and $d$ divides $f$ and $g$ and any common divisor of $f,g$ divides $d$, we call $d$ as a GCD. Also, we can let the uniquely monic $d$ be the GCD.

Let $f,g \in K[X]$, if there is $d\in K[X]$ and $(f,g)=(d)$, then $d$ is a GCD of $f,g$.

Proof:

By definition, $f\in(d), g\in(d)$, we know $d|f,d|g$. Assume a $h$ allows $h|f,h|g$, so with any $l,m \in K[X]$, $h$ can divide every polynomial in form of $fl+gm$. Therefore, $h|d$.

Proposition 1

As same as in $\Z$, we can define two polynomial $f,g$ are relatively prime if $(f,g)=(1)$​.

If $f,g$ are relatively prime, and $f|gh$, then $f|h$.

Proof:

Construct a polynomial $lf+mg =1$ where $l,m \in K[X]$, since $(f,g)=(1)$. Then we have $lfh+mgh=h$. The left side $f|lfh$ and $f|mgh$, so $f|h$

Corollary 1

If p is irreducible and $p|fg$, then $p|f$ or $p|g$.

Proof:

Since p is irreducible, $(p,f)=1$ or $(p,f)=p$. In the first case, by last proposition, $p|g$. In the second case, $p|f$.

Corollary 2

If p is a monic irreducible polynomial and $f,g\in K[X]$, $\text{ord}_p fg=\text{ord}_p f+\text{ord}_p g$.

Proof:

Let $\alpha$ that $f=p^\alpha c$ and $\beta$ that $g=p^\beta d$ where $p\nmid c$ and $p\nmid d$. Now we have, $fg=p^{\alpha+\beta}cd$ and by Corollary 1. So $\text{ord}_p fg=\alpha+\beta=\text{ord}_p f+\text{ord}_p g$. This is exactly as same as in $\Z$,

Proof of Unique Factorization in K[X]:

By corollary 2: $$\text{ord}_q f=\text{ord}_qc + \sum_p a(p)\text{ord}_q(p)$$ Obviously, $\text{ord}_qc=0$ and $\text{ord}q(p)=0$ when $p\neq q$. Only when $p=q$, $\text{ord}_q(p)=1$.

So: $$\text{ord}_qf=a(q)$$ Proved.