📂Abstract Algebra

Principal Ideal Domain

Definition ¹

Assume that $D$ of the domain is not a unit.

PID

If all ideals of $D$ are principal ideals, then $D$ is called a Principal Ideal Domain.

Subsidiary Definitions

1. Let the commutative ring $R$ have a unity $1$. If there exists an $c \in R$ that satisfies $b=ac$ for a given $a,b \in R$, then $a$ is said to divide $b$ or $a$ is a factor of $b$, denoted as $a \mid b$.
2. If $a \mid b$ and $b \mid a$, then $a,b$ are called associates.
3. If for $\forall a,b \in D$ and $p=ab$, one of $a$ or $b$ is a unit, then $p$ is called an irreducible element.
4. If $\forall a,b \in D$ and $p \mid ab$, then $p \mid a$ or $p \mid b$ is called a prime element.

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• Unity is the identity element $1$ for multiplication, and unit refers to elements that have an inverse element for multiplication.

Theorem ²

Assume $D$ is a Principal Ideal Domain.

• [1]: $D$ is a Noetherian ring.
• [2]: A non-unit $d \in D$ in $D$ is expressed as a product of irreducible elements.
• [3]: If $\left< p \right>$ is a maximal ideal in $D$, $p$ is an irreducible element of $D$.
• [4]: The irreducible elements of $D$ are prime elements.

Explanation

The term ‘Principal Ideal Domain’ is often abbreviated as PID due to its length.

Note that the term associative can refer to both the noun meaning of associates as well as the associative property, which has the same spelling but different meaning. Associative elements $-3,3 \in \mathbb{Z}$ can be represented with units .

Examples

Integer Ring $\mathbb{Z}$

The integer ring $\mathbb{Z}$ has all ideals expressed as principal ideals like $n \mathbb{Z} = \left< n \right>$.

Any Field $\mathbb{F}$

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

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

Proof

[1]

Definition of Noetherian Ring: Let $N$ be a ring.

1. When the ideals of $N$ satisfy $S_{1} \le S_{2} \le \cdots$, it is called an ascending chain.
2. If there exists $n \in \mathbf{n}$ that satisfies $S_{n} = S_{n+1} = \cdots$ for an ascending chain $\left\{ S_{i} \right\}_{i \in \mathbb{N} }$, it is called stationary. In a stationary ascending chain, the ideal does not grow indefinitely from a certain point.
3. A ring where every ascending chain is stationary is called a Noetherian ring.

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Consider an ascending chain of ideals $N_{1} \le N_{2} \le \cdots$ in $D$ and their union $\displaystyle N := \bigcup_{k=1}^{ \infty } N_{k}$. For some $i, j \in \mathbb{N}$, we have $$a \in N_{i} \\ b \in N_{j} \\ N_{i} \le N_{j}$$ Since $( N_{j} , + , \cdot )$ is clearly defined as an ideal, it is a subring, and the additive inverse $(-b) \in N_{j}$ for $b$ must exist. Furthermore, $ab \in N_{j}$ implies $(a-b), ab \in N$, and by the subring validation theorem, $N$ becomes a subring of $D$. More than that, since $N_{i}$ is an ideal, for all $d \in D$, we have $d a = a d$, and since $da \in N$, $N$ becomes an ideal of $D$.

Since $D$ is a PID, all ideals are principal ideals, representable as $N = \left< c \right>$ for some $c \in N$. As $\displaystyle N = \bigcup_{k=1}^{ \infty } N_{k}$, if $c \in N$, there must exist a natural number $r \in \mathbb{N}$ that satisfies $c \in N_{r}$. This means there exists a principal ideal with generator $c$ among the ideals smaller than $N_{r}$. In symbolic terms, $$\left< c \right> \le N_{r} \le N_{r+1} \le \cdots \le N = \left< c \right>$$ Therefore, $N_{r} = N_{r+1} = \cdots$. Hence, $D$ is a Noetherian ring.

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[2]

If $d$ is an irreducible element, there is nothing to prove. For a non-unit $d_{1}, c_{1} \in D$, assume it is expressed 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 obtain 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 a terminal $a_{r}$ in this sequence, which becomes an irreducible factor of $a$. Denote the irreducible dividing $d$ by $p_{1}$, and for a non-unit $f_{1}$, assume $d = p_{1} f_{1}$. Then, by continually defining $f_{i} := p_{i+1} f_{i+1}$, we obtain another ascending chain $$\left< d \right> \le \left< f_{1} \right> \le \left< f_{2} \right> \le \cdots$$ According to theorem [1] again, there must exist a terminal $f_{s}$ in this chain which becomes an irreducible factor of $f_{i}$.

By repeating this procedure finitely, $d$ can be expressed as a product of irreducible elements.

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[3]

$( \implies )$

Assume $p$ of the maximal ideal $\left< p \right>$ of $D$ can be expressed as $p=ab$ for a non-unit $a,b$.

Then $\left< p \right> \le \left< a \right>$, but if $\left< p \right> = \left< a \right>$, since $b$ must be a unit, actually $\left< p \right> \lneq \left< a \right>$ is obtained. Since $\left< p \right>$ is a maximal ideal, $\left< a \right> = D = \left< 1 \right>$, and hence $a$ and $1$ are associates. Summarizing:

• If $\left< p \right> \ne \left< a \right>$, $a$ is a unit.
• If $\left< p \right> = \left< a \right>$, $b$ is a unit.

Hence, $p$ must be an irreducible element.

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$( \impliedby )$

Assuming $\left< p \right> \le \left< a \right>$ for an irreducible element $p=ab$.

If $a$ is a unit, then $\left< a \right> = D$, which poses no issue, but if $a$ is not a unit, then $b$ must be a unit.

The fact that $b$ is a unit implies $bu =1$ for some $u \in D$, $$pu = abu = a$$ Therefore, $\left< p \right> \ge \left< a \right>$, thus $\left< p \right> = \left< a \right>$. Summarizing:

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

Therefore, $\left< p \right>$ becomes a maximal ideal.

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[4]

If $p$ is an irreducible element, then by theorem [3] $\left< p \right>$ is a maximal ideal and thus $1 \in D$ implies a prime ideal.

If $p$ divides $ab$, then $(ab) \in \left< p \right>$, and since $\left< p \right>$ is a prime ideal, $a \in \left< p \right>$ or $b \in \left< p \right>$. Rewriting this in another form, if $p \mid ab$, then $p \mid a$ or $p \mid b$, meaning $p$ is a prime element.

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Related Reading

• Euclidean Domain $\implies$ Principal Ideal Domain $\implies$ Unique Factorization Domain $\implies$ Domain
• Euclidean Domain $\implies$ Principal Ideal Domain $\implies$ Noetherian Ring