Pre-definition

If there is a function $\lambda$, from a integral domain $\R$, to a non-negative sets $\{0,1,2,3,4...\}$, such that if $a,b\in \R, b\neq 0$, there exists $c,d\in\R$ with $a=cb+d$ and either $d=0$ or $\lambda(d) < \lambda(b)$. Then we termed $\R$ as Euclidean Domain.

As we discussed before, $\Z$ and $K[X]$​ are Euclidean domains.

For elements $a_1,a_2...a_n\in\R$, the set $(a_1,a_2...a_n)=Ra_1+Ra_2+...+Ra_n=\{\sum_{i=1}^{n} r_i a_i|r_i\in\R\}$is an ideal. If $I=(a)$ for $a\in I$, $I$ is called principal ideal.

$\R$ as a principal ideal domain (PID) if every ideal in $\R$ is principal.

Euclidean Domain

Proposition 1

If $\R$ is a Euclidean Domain and $I \sube \R$ is an ideal, then there exists an element $a\in \R$ such that $I = \R a= \{ra|r\in\R\}$.

Proof:

Consider a non-negative set $S = \{\lambda(b) | b \in I, b\neq 0 \}$. For all $b\in I,b\neq0$ there exists a least element $a\in I, a\neq0$, such that $\lambda(a)\leq\lambda(b)$. Here we can call a is the smallest norm. Thus clearly, $I=\R a$ and $\R a \sube I$.

Moreover, we can have $b = ca+d$ for $c,d\in\R$, and either $d=0$ or $\lambda(d) < \lambda(a)$. Now we have $ca \in I$. So $d=b-ca\in\R$. It implies that $d=0$ because a is our smallest norm, so $b=ca \in \R a$, $I\sube \R a$.

Therefore, every Euclidean domain is a PID.

As we defined before, elements $p\in\R$ are irreducible if $u|p$ or $up|p$ ($u$ is the unit element that $u|1$). And non-unit $p\in\R$ if prime if $p\neq0, p|ab$ that $p|a$ or $p|b$. In Euclidean Domain, irreducible and prime do coincide.

That is $$a|b \iff (b)\sube(a)$$

$$u\in\R \iff (u) =R$$

$$a = bu \iff (a) = (b)$$

$$p\text{ is prime } \iff ab\in(p) \text{ either } a\in (p) \text{ or } b\in(p)$$

Proposition 2

Define $d\in R$ is a GCD of $a,b\in\R$ if $d|a$ and $d|b$, with any $d'|a$ and $d'|b$, we have $d'|d$. Clearly, $d'$ is associate with $d$ as GCDs.

Let $\R$ be a PID and $a,b\in\R$. Then $a,b$ have a GCD d and $(a,b)=(d)$

Proof:

By definition, if $R$ is a $PID$, then there exists d such that $(a,b)=(d)$. Since, $(a)\sube(d)$ and $(b)\sube(d)$, it follows that $d|a$ and $d|b$. Assume some $d'|a$ and $d'|b$, we have $(a)\sube(d')$ and $(b)\sube(d')$. Thus $(d)=(a,b)\sube(d')$, implying $d'|d$. Proved.

Corollary 1

If $\R$ is a PID, and $a,b\in\R$ are relatively prime, then $(a,b)=\R$.

Proof:

If $a,b$ are relatively prime, meaning the only common divisor are units. So $(a,b)=(u)=R$

Corollary 2

If $R$ is a PID and $p\in\R$ is irreducible, then p is prime.

Proof:

Assume $p|ab$ and $p\nmid a$. By corollary 1, $(a,p)=\R$, they are relatively prime. It follows that $(ab,pb)=(b)$. Since $ab\in (p)$ and $pb\in(p)$ , we deduce $(b)\sube(p)$, hence $p|b$.

Lemma 1

Given an ascending chain of ideals: $(a_1)\sube(a_2)\sube(a_3)\sube ...$. Then there will be a break off at $(a_k)=(a_{m})$ for $k\geq m$.

Proof:

Considering $I=\bigcup_{i=1}^\infin (a_i)$ as an ideal of $\R$. Since $\R$ is PID, so $I$ is principle ideal. Implying $I=(a)=aR$ for some a $a \in \R$. Consequently, $a\in (a_k) = I$ for some k. Indicating $(a_k)=I$. Thus, for any $n\geq m$, $(a_n)=(a_m)$. Proved.

Proposition 3

Every non-zero non-unit of $\R$ can be expressed as a product of irreducible elements (primes) .

Proof:

Assume $a\in\R, a\neq0, a\neq u$ and $a=pq$ where $p,q$ are non-units. We can decompose $p$ continuously to form an ascending chain: $(p_1)\sube(p_2)\sube(p_3)\sube ...$. As we proved before, there will be a break off at $(p_k)$. So $a$ is divisible a irreducible elements.

Now, let $p_1$ be irreducible such that $p_1|a$. Now we have $a=p_1q$ and q is not a unit(or we are done). We can decompose $q$ to form an ascending chain. As we shown before, the chain cannot go on indefinitely. So in the end we will have $a=p_1p_2p_3..p_ku$. All of the elements are irreducible, so proved.

Lemma 2

Now we can start to define $\text{ord}$ function. Let $p$ is prime and $a\neq0$. There will be an unique integer $n$ such that $p^n|a$ however $p^{n+1}\nmid a$.

Proof:

Assume any integer $m > 0$ and $p^m|a$. We have $a=p^m*b_m = p^{m+1}b_{m+1}$ , so that $b_m=b_{m+1}*p$​. Then the ascending chain can goes forever which is a contradiction.

So we can define, there is a unique n that allows $p^n|a$ however $p^{n+1}\nmid a$, denoted $n=\text{ord}_p a$.

Lemma 3

Keep the same route. If $a,b\in R$ and $a,b\neq0$. $\text{ord}_pab=\text{ord}_pa+\text{ord}_pb$.

Proof:

Let $\alpha$ that $a=p^\alpha c$ and $\beta$ that $b=p^\beta d$, where $p\nmid c$ and $p\nmid d$. We have $ab=p^{\alpha+\beta}cd$. Obviously, $p\nmid cd$, so $\text{ord}_pab=\text{ord}_pa+\text{ord}_pb$.

Theorem 3

First define a prime set $S$ to allows:

• All primes in $\R$ associate to primes in $S$.
• No two primes in $S$ are associated.

For example, that gives all positive primes in $\Z$ and all monic irreducible polynomials in $K[X]$.

For such $\R$ is PID and $S$ is given prime set. For any $a\in R$ and $a\neq0$, we have: $$a=u\prod_pp^{e(p)}$$ where $u$ is unit and $p\in S$. $e(p) = \text{ord}_p a$.

Proof:

The proof is no different than previous posts, since we have all Proposition 3 to prove the existence. And by Lemma 3, we can apply $\text{ord}_q$ on both side: $$\text{ord}_p a= \text{ord}_q u + \sum_pe(p)\text{ord}_qp$$ Since $\text{ord}_qu=0$ and $\text{ord}_qp=1$ only when $q=p$, otherwise $\text{ord}_qp=0$. So $\text{ord}_qa=e(q)$.