A Partial Order Isomorphism on PID and its application

A Partial Order Isomorphism on PID

Consider a PID $R$$R$. Let $(I,\subseteq )$$(I, \subseteq)$ denote the set of all ideals in $R$$R$ with the relation $\subseteq$$\subseteq$. It is easy to see that this forms a partial order set.

Now, consider a preorder set $(R,|)$$(R, |)$ where $a|b$$a|b$ denotes $b=ac$$b=ac$.

We define an equivalence relationship on $R$$R$ as $a\sim b$$a\sim b$ if and only if $a|b$$a|b$ and $b|a$$b|a$. The equivalence class of $a$$a$ is denoted by $[a]$$[a]$.

Proposition 1: $a\sim b$$a\sim b$ if and only if $(a)=(b)$$(a)=(b)$.

Proof:

$(\Rightarrow )$$(\Rightarrow)$ If $a\sim b$$a\sim b$, then $a|b$$a|b$ and $b|a$$b|a$.

Thus, $(b=ac,(b)\subseteq (a))$$(b=ac,(b)\subseteq(a))$ and $(a=bd,(a)\subseteq (b))$$(a=bd,(a)\subseteq(b))$, which implies $(a)=(b)$$(a)=(b)$.

$(\Leftarrow )$$(\Leftarrow)$ If $(a)=(b)$$(a)=(b)$, then $a=bc$$a=bc$ and $b=ad$$b=ad$, which implies $a|b$$a|b$ and $b|a$$b|a$.

Therefore, $a\sim b$$a\sim b$. $◻$$\square$

Proposition 2: $(R/\sim ,|)\cong (I,\supseteq )$$(R/\sim,|)\cong (I,\supseteq)$.

That is, $[a]|[b]$$[a]|[b]$ if and only if $(a)\supseteq (b)$$(a)\supseteq(b)$.

Proof:

$(\Rightarrow )$$(\Rightarrow)$ If $[a]|[b]$$[a]|[b]$, then $a\sim b$$a\sim b$. Since $b=ac$$b=ac$, we have $b\in (a)$$b\in(a)$.

Thus, $(a)\supseteq (b)$$(a)\supseteq(b)$.

$(\Leftarrow )$$(\Leftarrow)$ If $(a)\supseteq (b)$$(a)\supseteq(b)$, then $[a]|[b]$$[a]|[b]$ obviously holds. $◻$$\square$

Application: Using this Isomorphism to Prove that Every Nonzero Nonunit $a\in R$$a ∈ R$ is a Product of Prime Elements

In order to prove that a PID is a UFD, we typically rely on the Fundamental Theorem of Arithmetic.

It is assumed that readers have some knowledge about the Fundamental Theorem of Arithmetic, which states that every subset of $\mathbb{N}$$\mathbb{N}$ has a least element.

However, it is hard to say that in a PID, every chain of $(R/\sim ,|)$$(R/\sim,|)$ has a least element.

But we have the following lemma:

Lemma 1: Every principal ideal domain $R$$R$ is Noetherian, i.e., every ascending chain of ideals ${\mathfrak{a}}_{\mathfrak{1}}\subseteq {\mathfrak{a}}_{\mathfrak{2}}\subseteq \dots \subseteq R$$\mathfrak{a_1} \subseteq \mathfrak{a_2} \subseteq \ldots \subseteq R$ becomes stationary in the sense that there is some $n\in \mathbb{N}$$n \in \mathbb{N}$ such that ${\mathfrak{a}}_{\mathfrak{i}}={\mathfrak{a}}_{\mathfrak{n}}$$\mathfrak{a_i} = \mathfrak{a_n}$ for all $i\ge n$$i \geq n$.

Proof:

Since $\mathfrak{a}=\bigcup _{i\ge 1}{\mathfrak{a}}_{\mathfrak{i}}$$\mathfrak{a} = \bigcup_{i\geq 1}\mathfrak{a_i}$ is an ideal, $\mathfrak{a}=(a)$$\mathfrak{a}=(a)$ for some $a\in {\mathfrak{a}}_{\mathfrak{n}}$$a\in\mathfrak{a_n}$, where $(a)\subseteq {\mathfrak{a}}_{\mathfrak{n}}\subseteq (a)$$(a)\subseteq\mathfrak{a_n}\subseteq(a)$.

Proposition 3: Let $R$$R$ be a principal ideal domain. Then every nonzero nonunit $a\in R$$a \in R$ is a product of prime elements.

Proof:

Let $S$$S$ be the set of all principal ideals in $R$$R$ admitting a generator $a\in R-(R^{\ast }\cup \{0\})$$a \in R-(R^* \cup \{0\})$ such that $a$$a$ does not allow a finite factorization into irreducible elements. We have to show that $S=\mathrm{\varnothing }$$S=\emptyset$. Assume $S\ne \mathrm{\varnothing }$$S\neq \emptyset$.

By Lemma 1 and Zorn's Lemma, there exists a maximal element $(a)\in S$$(a)\in S$, corresponding to a least element $[a]\in (R/\sim ,|)$$[a]\in(R/\sim,|)$.

Since $a$$a$ is reducible, $a=bc$$a=bc$ where $b,c$$b,c$ are not units. Thus, $(b),(c)\ne R$$(b),(c)\neq R$.

Also, $(a)\subseteq (b)$$(a)\subseteq(b)$ and $(a)\subseteq (c)$$(a)\subseteq(c)$, but $(b),(c)\notin S$$(b),(c) \notin S$. This implies that $b$$b$ and $c$$c$ are products of prime elements. Therefore, $a$$a$ is a product of prime elements.

This leads to a contradiction, thus $S=\mathrm{\varnothing }$$S=\emptyset$. $◻$$\square$