Unit 7: Integral Domains, Euclidean Domains, PIDs & UFDs

1. Zero Divisors and Integral Domains

In standard high-school arithmetic, if $a \cdot b = 0$, we conclude that $a = 0$ or $b = 0$. In general abstract rings, this property can fail dramatically (e.g., in $\mathbb{Z}_6$, $2 \cdot 3 \equiv 0 \pmod 6$ even though neither $2$ nor $3$ is zero).

Definition 7.1 (Zero Divisor): Let $R$ be a ring. A non-zero element $a \in R \setminus \{0\}$ is called a left zero divisor (resp. right zero divisor) if there exists a non-zero element $b \in R \setminus \{0\}$ such that $a \cdot b = 0$ (resp. $b \cdot a = 0$).

Definition 7.2 (Integral Domain): A commutative ring $R$ with unity $1_R \ne 0$ is called an integral domain if it contains no zero divisors:

Examples of integral domains: $\mathbb{Z}$, any field $\mathbb{F}$ (such as $\mathbb{Q}, \mathbb{R}, \mathbb{C}, \mathbb{Z}_p$), polynomial rings over domains $R[x]$, and the Gaussian integers $\mathbb{Z}[i]$. Non-examples: $\mathbb{Z}_n$ for composite $n$, $M_n(\mathbb{R})$ for $n \ge 2$, and direct products $R_1 \times R_2$.

2. The Cancellation Law

Theorem 7.1 (Equivalence of Domain Property and Cancellation): A commutative ring $R$ with $1 \ne 0$ is an integral domain if and only if the cancellation law holds:

Proof:

• $(\implies)$ Let $R$ be an integral domain. If $ab = ac$, then $ab - ac = 0 \implies a(b - c) = 0$.

Since $a \ne 0$ and $R$ has no zero divisors, $b - c = 0 \implies b = c$.

• $(\impliedby)$ Suppose cancellation holds. If $ab = 0$, and $a \ne 0$, we write $ab = a \cdot 0$.

By cancellation of $a$, $b = 0$. Thus $R$ contains no zero divisors. $\blacksquare$

3. Finite Integral Domains are Fields

A remarkable bridge connecting finiteness, algebra, and number theory:

Theorem 7.2 (Finite Integral Domains are Fields): Every finite integral domain is a field.

Complete Proof:

Let $D$ be a finite integral domain with unity $1 \ne 0$, and let $n = |D| < \infty$. Let $a \in D$ be any non-zero element ($a \ne 0$). We must show that $a$ has a multiplicative inverse $a^{-1} \in D$.

Consider the left-multiplication mapping:

1. Injectivity of $\lambda_a$:

Suppose $\lambda_a(x) = \lambda_a(y)$. Then $ax = ay$. Since $a \ne 0$ and $D$ is an integral domain, the cancellation law (Theorem 7.1) implies $x = y$. Thus $\lambda_a$ is an injective mapping.

2. Surjectivity by Pigeonhole Principle:

Since $D$ is a finite set, any injective map from $D$ to itself is automatically surjective! Therefore, $\lambda_a(D) = D$.

3. Existence of Inverse:

Since $\lambda_a$ is surjective and $1 \in D$, there must exist some element $b \in D$ such that:

Since $D$ is commutative, $b \cdot a = a \cdot b = 1$. Thus $b = a^{-1}$ is the multiplicative inverse of $a$. Since every non-zero element of $D$ is invertible, $D$ is a field. $\blacksquare$

1. Generalizing $\mathbb{Z} \hookrightarrow \mathbb{Q}$

Just as the rational numbers $\mathbb{Q}$ are formed from the integers $\mathbb{Z}$ by introducing formal fractions $a/b$ ($b \ne 0$), any integral domain $R$ can be canonically embedded into a minimal field called its field of fractions (or quotient field), denoted $\text{Frac}(R)$ or $Q(R)$.

2. Formal Rigorous Construction

Let $R$ be an integral domain. Let $S = R \setminus \{0\}$. Consider the Cartesian product $R \times S = \{ (a, b) : a \in R, b \in R, b \ne 0 \}$.

Definition 7.3 (Fraction Equivalence Relation): Define a binary relation $\sim$ on $R \times S$ by:

Lemma 7.1: $\sim$ is an equivalence relation on $R \times S$.

Proof of Transitivity (where domain property is essential):

Suppose $(a, b) \sim (c, d)$ and $(c, d) \sim (e, f)$. Then $ad = bc$ and $cf = de$. Multiply the first equation by $f$ and the second by $b$:

Equating them yields:

Since $d \in S$, $d \ne 0$. Because $R$ is an integral domain (no zero divisors), $(af - be)d = 0$ forces:

We denote the equivalence class of $(a, b)$ by $\frac{a}{b}$ or $[a, b]$, and the set of classes by $Q(R) = (R \times S) / \sim$.

3. Field Operations on $Q(R)$

Define addition and multiplication on $Q(R)$ by:

Notice $bd \ne 0$ because $R$ has no zero divisors, so $bd \in S$. Both operations are easily verified to be well-defined.

Theorem 7.3 (Properties of $Q(R)$):

1. $(Q(R), +, \cdot)$ is a field, with zero element $0_{Q(R)} = \frac{0}{1}$ and unity $1_{Q(R)} = \frac{1}{1}$.
2. The inverse of any non-zero element $\frac{a}{b} \ne \frac{0}{1}$ ($a \ne 0$) is $\left(\frac{a}{b}\right)^{-1} = \frac{b}{a}$.
3. The canonical map $\iota: R \to Q(R)$ defined by $\iota(r) = \frac{r}{1}$ is an injective ring homomorphism (embedding).
4. Universal Property: If $K$ is any field and $\phi: R \hookrightarrow K$ is an injective ring homomorphism, there exists a unique field homomorphism $\widetilde{\phi}: Q(R) \hookrightarrow K$ such that $\widetilde{\phi} \circ \iota = \phi$, given by $\widetilde{\phi}\left(\frac{a}{b}\right) = \phi(a)\phi(b)^{-1}$.

1. Divisibility, Units and Associates

Let $R$ be an integral domain with unity $1$.

Definition 7.4 (Divisibility, Units, Associates):

1. Divisibility: For $a, b \in R$, we say $a$ divides $b$ (written $a \mid b$) if there exists $c \in R$ such that $b = ac$. Equivalently, $\langle b \rangle \subseteq \langle a \rangle$.
2. Unit: An element $u \in R$ is a unit if $u \mid 1$, meaning $\exists v \in R$ such that $uv = 1$. The set of all units forms a multiplicative group $R^\times = U(R)$.
3. Associates: Elements $a, b \in R$ are called associates (written $a \sim b$) if $a = ub$ for some unit $u \in R^\times$. Equivalently, $a \mid b$ and $b \mid a \iff \langle a \rangle = \langle b \rangle$.

2. Irreducible Elements vs. Prime Elements

In elementary arithmetic in $\mathbb{Z}$, the terms "prime" and "irreducible" are used interchangeably. In general rings, however, they represent two fundamentally distinct concepts!

Definition 7.5 (Irreducible Element): A non-zero element $p \in R \setminus \{0\}$ that is not a unit ($p \notin R^\times$) is called irreducible if whenever $p = ab$ for $a, b \in R$, then either $a$ is a unit or $b$ is a unit. (Irreducible elements cannot be factored non-trivially.)

Definition 7.6 (Prime Element): A non-zero element $p \in R \setminus \{0\}$ that is not a unit ($p \notin R^\times$) is called prime if whenever $p \mid ab$ for $a, b \in R$, then $p \mid a$ or $p \mid b$. (Prime elements preserve the Euclid's Lemma divisibility property; equivalently, the principal ideal $\langle p \rangle$ is a prime ideal.)

3. The Prime $\implies$ Irreducible Theorem

Theorem 7.4 (Primes are Always Irreducible in a Domain): In any integral domain $R$, every prime element is irreducible:

Complete Proof:

Let $p \in R$ be a prime element. Suppose $p = ab$ for some $a, b \in R$. Since $p = ab$, $p \mid ab$. Because $p$ is prime, by Definition 7.6, $p \mid a$ or $p \mid b$. Without loss of generality, assume $p \mid a$. Then there exists $c \in R$ such that $a = pc$. Substituting this into $p = ab$:

Using the cancellation law in the domain $R$ (since $p \ne 0$):

This equation proves that $b$ is a unit in $R$ (with inverse $c$)! Similarly, if $p \mid b$, then $a$ is a unit. Thus any factorization of $p$ involves a unit factor. Therefore, $p$ is irreducible. $\blacksquare$

Warning (The Converse Fails in General!): An irreducible element is NOT necessarily prime in an arbitrary integral domain! In $\mathbb{Z}[\sqrt{-5}]$, the element $2$ is irreducible, but $2 \mid (1+\sqrt{-5})(1-\sqrt{-5}) = 6$, while $2 \nmid (1+\sqrt{-5})$ and $2 \nmid (1-\sqrt{-5})$. Thus $2$ is irreducible but NOT prime!

1. Euclidean Domains (ED)

A Euclidean Domain is an integral domain equipped with a notion of "size" that allows division with remainder.

Definition 7.7 (Euclidean Domain): An integral domain $R$ is called a Euclidean Domain (ED) if there exists a function $d: R \setminus \{0\} \to \mathbb{Z}_{\ge 0}$ (called a Euclidean norm or degree function) such that:

1. For all $a, b \in R \setminus \{0\}$, $d(a) \le d(ab)$.
2. (Division Algorithm): For any $a \in R$ and $b \in R \setminus \{0\}$, there exist $q, r \in R$ such that:

Classic Examples:

• $\mathbb{Z}$ with $d(n) = |n|$.
• Polynomial ring $F[x]$ over any field $F$ with $d(f) = \deg(f)$.
• Gaussian integers $\mathbb{Z}[i]$ with $d(a+bi) = a^2 + b^2$.

2. Principal Ideal Domains (PID)

Definition 7.8 (Principal Ideal Domain): An integral domain $R$ is called a Principal Ideal Domain (PID) if every ideal of $R$ is principal (generated by a single element):

Examples: $\mathbb{Z}, F[x], \mathbb{Z}[i]$. Non-example: $\mathbb{Z}[x]$ (the ideal $\langle 2, x \rangle$ is not principal).

3. Unique Factorization Domains (UFD)

Definition 7.9 (Unique Factorization Domain): An integral domain $R$ is called a Unique Factorization Domain (UFD) if:

1. (Factorization Existence): Every non-zero non-unit $a \in R \setminus (R^\times \cup \{0\})$ can be written as a product of irreducible elements:

1. (Uniqueness up to Associates): If $a = p_1 p_2 \cdots p_r = q_1 q_2 \cdots q_s$ are two factorizations into irreducibles, then $r = s$, and after reordering, $p_i$ is associate to $q_i$ for all $i$ ($p_i = u_i q_i$ for some $u_i \in R^\times$).

1. The Master Hierarchy Chain

The fundamental relationships among commutative domains form a strict linear hierarchy:

2. Proof 1: Every Euclidean Domain is a PID ($\text{ED} \implies \text{PID}$)

Theorem 7.5: If $R$ is a Euclidean Domain, then $R$ is a Principal Ideal Domain.

Complete Proof:

Let $I \trianglelefteq R$ be any ideal.

• If $I = \{0\}$, then $I = \langle 0 \rangle$, which is principal.
• If $I \ne \{0\}$, consider the set of norms of non-zero elements:

Since $I \ne \{0\}$, $S$ is a non-empty subset of non-negative integers. By the Well-Ordering Principle of $\mathbb{Z}$, $S$ contains a minimum element. Choose $b \in I \setminus \{0\}$ such that $d(b) = \min(S)$.

We claim $I = \langle b \rangle$:

• Since $b \in I$ and $I$ is an ideal, $rb \in I$ for all $r \in R$, so $\langle b \rangle \subseteq I$.
• Now let $a \in I$ be any element.

By the Division Algorithm in the Euclidean Domain $R$, there exist $q, r \in R$ such that:

Rearranging gives:

Since $a \in I$ and $b \in I$, and $I$ is an ideal, $r \in I$. If $r \ne 0$, then $d(r) < d(b)$, which contradicts the minimality of $d(b)$ among all non-zero elements in $I$! Therefore, $r$ must be $0$. This implies $a = qb \in \langle b \rangle$. Thus $I \subseteq \langle b \rangle$. Combining both inclusions gives $I = \langle b \rangle$. Hence every ideal of $R$ is principal, proving that $R$ is a PID. $\blacksquare$

3. Proof 2: In a PID, Irreducible Elements are Prime

Lemma 7.2: If $R$ is a PID and $p \in R$ is irreducible, then $p$ is prime.

Complete Proof:

Let $p \in R$ be irreducible, and suppose $p \mid ab$ for $a, b \in R$. Consider the ideal $I = \langle p, a \rangle$. Since $R$ is a PID, $I$ is principal: $\langle p, a \rangle = \langle d \rangle$ for some $d \in R$. Since $p \in \langle d \rangle$, $d \mid p$. Because $p$ is irreducible, its only divisors are units and associates of $p$.

• Case 1: $d$ is associate to $p$.

Then $\langle d \rangle = \langle p \rangle$, which means $\langle p, a \rangle = \langle p \rangle$. This implies $a \in \langle p \rangle \implies p \mid a$.

• Case 2: $d$ is a unit.

Then $\langle d \rangle = R$, so $\langle p, a \rangle = R = \langle 1 \rangle$. By Bézout's identity in a PID, there exist $x, y \in R$ such that:

Multiplying both sides by $b$:

Since $p \mid p$ and $p \mid ab$, $p$ divides the entire left-hand side:

Thus either $p \mid a$ or $p \mid b$. Hence $p$ is prime. $\blacksquare$

4. Proof 3: Every PID is a UFD ($\text{PID} \implies \text{UFD}$)

Theorem 7.6: If $R$ is a Principal Ideal Domain, then $R$ is a Unique Factorization Domain.

Outline of Complete Proof:

1. Factorization Existence (Ascending Chain Condition on Principal Ideals - ACCP):

In a PID, any ascending chain of ideals $\langle a_1 \rangle \subseteq \langle a_2 \rangle \subseteq \langle a_3 \rangle \subseteq \dots$ must stabilize. (Proof: The union $J = \bigcup \langle a_i \rangle$ is an ideal, hence $J = \langle c \rangle$ for some $c \in R$. Then $c \in \langle a_N \rangle$ for some $N$, so the chain stabilizes at $N$.) This ACCP property guarantees that the process of factoring into irreducibles cannot continue indefinitely, ensuring the existence of irreducible factorizations.

2. Uniqueness:

By Lemma 7.2, every irreducible element in a PID is prime. Using Euclid's Lemma repeatedly on $p_1 \cdots p_r = q_1 \cdots q_s$, $p_1 \mid q_j$ for some $j$. Since $q_j$ is irreducible, $p_1$ and $q_j$ are associates. Cancelling $p_1$ and proceeding by induction establishes uniqueness up to associates. $\blacksquare$

5. Counterexamples Delineating the Strict Inclusions

| Strict Inclusion | Counterexample | Why the Inversion Fails | | :--- | :--- | :--- | | $\text{ED} \subsetneq \text{PID}$ | $\mathbb{Z}\left[\frac{1 + \sqrt{-19}}{2}\right]$ | Is a PID (every ideal is principal), but admits NO Euclidean norm function. | | $\text{PID} \subsetneq \text{UFD}$ | $\mathbb{Z}[x]$ | Is a UFD (by Gauss's Lemma), but $\langle 2, x \rangle$ is not principal, so NOT a PID. | | $\text{UFD} \subsetneq \text{ID}$ | $\mathbb{Z}[\sqrt{-5}]$ | Is an integral domain, but $6 = 2 \cdot 3 = (1+\sqrt{-5})(1-\sqrt{-5})$ shows non-unique factorization. |