Principal Ideal Domain 📂Abstract Algebra

Principal Ideal Domain

Definition ¹

Let’s assume that the $p \ne 0$ of the integral domain $D$ is not a field.

PID

An integral domain $D$ is called a Principal Ideal Domain (PID) if every ideal in $D$ is a principal ideal.

Consequent Definitions

Let’s say a commutative ring $R$ has a unity $1$. If for $a,b \in R$ there exists $c \in R$ that satisfies $b=ac$, then $a$ divides $b$ or $a$ is a factor of $b$, denoted as $a \mid b$.
If $a \mid b$ and $b \mid a$, then $a,b$ are called associates.
For $\forall a,b \in D$ and $p=ab$, if one of $a$ or $b$ is a unit, then $p$ is called an irreducible element.
For $\forall a,b \in D$, if $p \mid ab$, then $p$ is called a prime element, which is either $p \mid a$ or $p \mid b$.

A unity is the multiplicative identity $1$, a unit is an element with a multiplicative inverse.

Theorem ²

Let’s assume $D$ is a principal ideal domain.

[1]: $D$ is a Noetherian ring.
[2]: Every element in $D$ that is neither $0$ nor a unit can be expressed as a product of irreducible elements of $D$.
[3]: If $\left$ is a maximal ideal of $D$ and $p$ is an irreducible element in $D$.
[4]: Every irreducible element in $D$ is a prime element.

Explanation

The term “Principal Ideal Domain” is often abbreviated as PID due to its length.

Note that associative in “associates” shares the same spelling with associative in “associative property” but is a noun, indicating a relation where two elements can be expressed as products of a unit.

Examples

Integer Ring $\mathbb{Z}$

In the integer ring $\mathbb{Z}$, all ideals can be expressed as principal ideals like $n \mathbb{Z} = \left< n \right>$.

All Fields $\mathbb{F}$

Gaussian Integer Ring $\mathbb{Z} [i]$ and Eisenstein Integer Ring $\mathbb{Z} [\omega]$

The Gaussian integer ring and the Eisenstein integer ring are rings formed by adding the imaginary unit $i := \sqrt{-1}$ or $\omega := (-1)^{1/3}$ to the integer ring $\mathbb{Z}$, respectively.

Proof

[1]

Definition of Noetherian Ring: Let’s consider a ring $N$.
An ascending chain $\left\{ S_{i} \right\}_{i \in \mathbb{N} }$ of ideals in $N$ that satisfies $S_{1} \le S_{2} \le \cdots$ is called an Ascending Chain.
If for an ascending chain $\left\{ S_{i} \right\}_{i \in \mathbb{N} }$, there exists $n \in \mathbb{n}$ satisfying $S_{n} = S_{n+1} = \cdots$, it is called Stationary. Meaning, at some point, the ideals in the chain no longer increase.
A ring in which every ascending chain of ideals is stationary is called a Noetherian Ring.

Consider an ascending chain of ideals $N_{1} \le N_{2} \le \cdots$ in $D$ and its union $\displaystyle N := \bigcup_{k=1}^{ \infty } N_{k}$. For some $i, j \in \mathbb{N}$, if we assume $$ a \in N_{i} \\ b \in N_{j} \\ N_{i} \le N_{j} $$, then $( N_{j} , + , \cdot )$, defined as an ideal, is a subring, and there exists an additive inverse $(-b) \in N_{j}$ for $b$. Since $ab \in N_{j}$, we have $(a-b), ab \in N$, and by the subring test, $N$ is a subring of $D$. Moreover, since $N_{i}$ is an ideal, for all $d \in D$, $d a = a d$ holds, and because of $da \in N$, $N$ is an ideal of $D$.

Since $D$ is a PID, all ideals are principal ideals, and for some $c \in N$, it can be expressed as $N = \left< c \right>$. Here, since $\displaystyle N = \bigcup_{k=1}^{ \infty } N_{k}$, if $c \in N$, then a natural number $r \in \mathbb{N}$ exists satisfying $c \in N_{r}$. $c \in N_{r}$ implies the existence of a principal ideal smaller than $c$ with $c$ as its generator. Mathematically, $$ \left< c \right> \le N_{r} \le N_{r+1} \le \cdots \le N = \left< c \right> $$, leading to $N_{r} = N_{r+1} = \cdots$. Hence, $D$ is a Noetherian ring.

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[2]

If $d$ is an irreducible element, there is nothing to prove. Assuming a non-unit element $d_{1}, c_{1} \in D$, let’s express it as $d = d_{1} c_{1}$.

Then $\left< d \right> \le \left< d_{1} \right>$, and by continually defining $d_{i} := d_{i+1} c_{i+1}$, we get an ascending chain $$ \left< d \right> \le \left< d_{1} \right> \le \left< d_{2} \right> \le \cdots $$. However, according to theorem [1], there must exist an end $a_{r}$ to this chain, and $a_{r}$ simultaneously becomes an irreducible factor of $a$. By setting the irreducible element dividing $d$ as $p_{1}$ and considering a non-unit element $f_{1}$ such that $d = p_{1} f_{1}$, we have $\left< d \right> \le \left< f_{1} \right>$, and by continually defining $f_{i} := p_{i+1} f_{i+1}$, we get an ascending chain $$ \left< d \right> \le \left< f_{1} \right> \le \left< f_{2} \right> \le \cdots $$. This chain must also end at some point $f_{s}$ according to theorem [1], making $f_{s}$ an irreducible factor of $f_{i}$.

This process, repeated a finite number of times, confirms that $d$ can be expressed as a product of irreducible elements.

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[3]

$( \implies )$

Assuming that $p$, an irreducible element, is expressed as $p=ab$ with respect to a maximal ideal $\left$ of $D$ and a non-unit element $a,b$.

Then $\left \le \left< a \right>$, but if $\left = \left< a \right>$, then $b$ must be a unit, leading to the actuality of $\left \lneq \left< a \right>$. Since $\left$ is a maximal ideal, it must be $\left< a \right> = D = \left< 1 \right>$, making $a$ and $1$ associates. In summary:

If $\left \ne \left< a \right>$, then $a$ is a unit, and
If $\left = \left< a \right>$, then $b$ is a unit, making

$p$ an irreducible element by necessity.

$( \impliedby )$

Assuming a prime element $p=ab$ and $\left \le \left< a \right>$,

If $a$ is a unit, then $\left< a \right> = D$, causing no issues, but if $a$ is not a unit, then $b$ must necessarily be a unit.

If $b$ is a unit, it means for some $u \in D$, $bu =1$ holds, and since $$ pu = abu = a $$, we get $\left \ge \left< a \right>$, meaning $\left = \left< a \right>$ must hold. In summary:

Either $\left< a \right> = D$ must hold or
$\left< a \right> = \left$ must hold, making

$\left$ a maximal ideal.

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[4]

If $p$ is an irreducible element, then $\left$, by theorem [3], is a maximal ideal and hence a prime ideal.

If $p$ divides $ab$, then $(ab) \in \left$ holds, and since $\left$ is a prime ideal, either $a \in \left$ or $b \in \left$ holds. Rewritten differently, if $p \mid ab$, then either $p \mid a$ or $p \mid b$, making $p$ a prime element.

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