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Unique factorization domain

In abstract algebra, a unique factorization domain (UFD) is an integral domain R in which every nonzero non-unit element x \in R admits a factorization into irreducible elements, and any two such factorizations of x have the same length and are identical up to reordering and multiplication by units (i.e., associates).^[1] This property generalizes the fundamental theorem of arithmetic from the integers to more abstract algebraic structures, ensuring that factorization behaves predictably without ambiguity beyond trivial equivalences.^[2] Unique factorization domains play a central role in commutative algebra, as they provide a framework for studying divisibility, primes, and ideals in rings beyond the familiar case of the integers \mathbb{[Z](/page/Z)}, which is itself a prototypical UFD.^[3] In a UFD, every irreducible element is prime, meaning that if an irreducible divides a product, it divides one of the factors; this equivalence distinguishes UFDs from more general atomic domains where factorization exists but uniqueness may fail.^[1] Moreover, UFDs satisfy the ascending chain condition on principal ideals, preventing infinite ascending chains of ideals generated by single elements, which facilitates proofs of finiteness in factorization processes.^[1] Classic examples of UFDs include the ring of integers \mathbb{Z}, where unique factorization into primes holds by Gauss's theorem, and polynomial rings R over any UFD R, such as \mathbb{Z} or k for a field k.^[4] Principal ideal domains (PIDs), a stricter class where every ideal is principal, are always UFDs—e.g., Euclidean domains like \mathbb{Z} and k—but the converse does not hold, as there exist UFDs that are not PIDs, such as \mathbb{Z}.^[5] Non-examples, like the ring \mathbb{Z}[\sqrt{-5}], illustrate the subtlety of the condition, as they admit factorization into irreducibles but lack uniqueness.^[2]

Core Concepts

Definition

An integral domain is a commutative ring with unity (where $1 \neq 0) that has no zero divisors, meaning that if ab = 0 for elements a, b in the ring, then either a = 0 or b = 0.^[6] A unique factorization domain (UFD) is an integral domain R in which every nonzero nonunit element can be expressed as a product of irreducible elements, and this factorization is unique up to the order of the factors and association by units.^[6] The existence part of this definition guarantees that such a factorization into irreducibles always exists and is finite for any qualifying element.^[6] The uniqueness theorem states that any two such factorizations of the same element differ only in the ordering of the irreducible factors and by multiplication of those factors by units from R.^[6] Two elements a, b \in R are associates, denoted a \sim b, if there exists a unit u \in R such that a = u b.^[6] For an element x \in R, the notation \pi(x) denotes the number of irreducible factors in its factorization (counting multiplicity).^[6]

Primes, Irreducibles, and Units

In an integral domain R, a unit is a nonzero element u \in R such that there exists v \in R with uv = 1; the set of units forms a multiplicative group.^[7] The units play a crucial role in factorization by allowing elements to be scaled without altering their essential structure. An irreducible element in an integral domain R is a nonzero non-unit p \in R such that if p = ab for some a, b \in R, then either a or b is a unit.^[7] Irreducibles serve as the "atoms" in factorizations, representing elements that cannot be further decomposed nontrivially. A prime element in an integral domain R is a nonzero non-unit p \in R such that if p divides ab for some a, b \in R, then p divides a or p divides b.^[7] In any integral domain, every prime element is irreducible: if p is prime and p = ab, then p divides ab, so p divides a or b; assuming without loss of generality that p divides a, say a = pc, then p = pcb, implying c = 1 up to units, so b is a unit (and symmetrically).^[7] However, the converse does not hold in general; for instance, in the ring \mathbb{Z}[\sqrt{-3}], the element $1 + \sqrt{-3} is irreducible (as its norm is 4, a prime power in \mathbb{Z}, preventing nontrivial factorization) but not prime, since it divides $4 = 2 \cdot 2 yet divides neither factor.^[7] However, in a UFD, every irreducible element is prime.^[1] Two elements a, b \in R are associates if a = ub for some unit u \in R; this defines an equivalence relation, and factorizations into irreducibles are considered up to associates and ordering.^[7] In unique factorization domains, every nonzero non-unit element factors uniquely into irreducibles up to such associates.

Examples and Counterexamples

Principal Examples

The ring of integers \mathbb{Z} is a prototypical example of a unique factorization domain, where every nonzero non-unit element factors uniquely into a product of prime numbers, up to ordering and association by units \pm 1. For instance, the integer 6 factors as $6 = 2 \cdot 3, and this representation is unique modulo units and permutation of factors. This property follows from the fundamental theorem of arithmetic, which guarantees both existence and uniqueness of such factorizations for all integers greater than 1. Polynomial rings in one variable over a field k, denoted k, are also unique factorization domains, with irreducible elements being the irreducible polynomials and units consisting of the nonzero constants in k. In this setting, every nonconstant polynomial factors uniquely into irreducible polynomials, up to ordering and multiplication by units; for example, x^2 - 1 = (x-1)(x+1) in \mathbb{Q}, and no alternative irreducible factorization exists. This result relies on Gauss's lemma, which preserves primitivity and irreducibility when extending from the field to the polynomial ring.^[8] The Gaussian integers \mathbb{Z} = \{a + bi \mid a, b \in \mathbb{Z}\}, where i = \sqrt{-1}, form a unique factorization domain equipped with the norm function N(a + bi) = a^2 + b^2, which is multiplicative and takes nonnegative integer values. Factorizations in \mathbb{Z} are unique up to units \{\pm 1, \pm i\}; for example, $5 = (1 + 2i)(1 - 2i), and since N(1 + 2i) = 5 is prime in \mathbb{Z}, this is an irreducible factorization with no distinct alternative. The norm enables a Euclidean algorithm in \mathbb{Z}, ensuring the unique factorization property.^[9] Polynomial rings in multiple variables over a field k, such as k[x_1, \dots, x_n] for n \geq 2, are unique factorization domains, extending the one-variable case by induction. Specifically, k[x_1, \dots, x_n] = k[x_1, \dots, x_{n-1}][x_n], and since the base ring is a UFD, the extension preserves unique factorization into irreducible polynomials. Hilbert's basis theorem underpins this by confirming that such rings are Noetherian, allowing finite generation of ideals and supporting the iterative UFD structure without introducing non-unique factorizations.^[10]

Non-Examples

A classic example of an integral domain that fails to be a unique factorization domain (UFD) arises in quadratic integer rings, particularly \mathbb{Z}[\sqrt{-5}]. In this ring, the element 6 admits two distinct factorizations into irreducibles: $6 = 2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5}).^[2] The norm function N(a + b\sqrt{-5}) = a^2 + 5b^2 plays a key role in establishing irreducibility; for instance, N(2) = 4, N(3) = 9, and N(1 \pm \sqrt{-5}) = 6, all of which are prime in \mathbb{Z}, implying that 2, 3, and $1 \pm \sqrt{-5} cannot be factored nontrivially in \mathbb{Z}[\sqrt{-5}].^[2] However, these irreducibles are not prime elements, as 2 divides (1 + \sqrt{-5})(1 - \sqrt{-5}) but divides neither factor, demonstrating the failure of unique factorization.^[2] Polynomial rings and their subrings provide further non-examples. Consider the subring R = k[x^2, x^3] of the polynomial ring k over a field k; here, x^2 and x^3 are irreducible elements, yet x^6 = (x^2)^3 = (x^3)^2 yields two distinct factorizations into irreducibles, so R is not a UFD.^[11] Integral domains that are atomic—meaning every nonzero nonunit factors into irreducibles—but not UFDs highlight the specific failure of uniqueness. The ring \mathbb{Z}[\sqrt{-5}] is atomic, as all elements factor into irreducibles despite the non-unique factorization of elements like 6, but the irreducibles are not prime, allowing distinct factorizations.^[12] Similarly, k[x^2, x^3] is atomic, with every nonzero nonunit expressible as a product of irreducibles like powers of x^2 and x^3, yet uniqueness fails as shown.^[11] For failures of existence rather than uniqueness, non-atomic domains serve as counterexamples, where some nonzero nonunits do not factor into irreducibles at all due to infinite descending chains of divisors. A prominent algebraic example is the ring \overline{\mathbb{Z}}_\mathbb{C} of all algebraic integers in \mathbb{C}, which is an integral domain but non-atomic, as its multiplicative structure admits elements without finite factorizations into irreducibles.^[13] Such domains underscore that UFDs require both existence and uniqueness of factorizations.

Fundamental Properties

Key Structural Properties

A unique factorization domain (UFD) is an atomic domain, meaning that every non-zero non-unit element admits a factorization into irreducible elements.^[14] This property follows directly from the existence part of the UFD definition, ensuring that factorization into irreducibles is always possible. In a UFD, every irreducible element is prime. To see this, suppose \pi is an irreducible that is not prime, so there exist a, b \in R such that \pi \mid ab but \pi \nmid a and \pi \nmid b. Since R is a UFD, a and b factor uniquely into irreducibles, leading to a contradiction with the uniqueness of factorization if \pi divides ab without dividing either factor. This equivalence between irreducibles and primes distinguishes UFDs from more general atomic domains.^[15] The uniqueness of factorization in a UFD implies that all irreducible factorizations of a given non-zero non-unit element have the same length, counting multiplicities up to units. This uniformity in factorization length is a direct consequence of the unique decomposition into primes (or irreducibles, since they coincide). Every UFD is a Schreier domain, an integrally closed integral domain in which every non-zero element is primal—meaning that for any factorization a = uv, the irreducible factors of u and v can be uniquely distributed into the irreducible factors of a. Since UFDs are GCD domains, they satisfy this primal condition, ensuring weak divisibility properties across factorizations.^[16] While the Noetherian property (every ideal is finitely generated) is not required for a domain to be a UFD—for instance, the polynomial ring in infinitely many variables over a field is a non-Noetherian UFD—many important UFDs are Noetherian. The Hilbert basis theorem guarantees that if R is a Noetherian ring, then the polynomial ring R is also Noetherian, preserving the UFD structure in finite extensions.^[17]^[18]

Divisibility and Factorization Behavior

In a unique factorization domain R, any two nonzero elements a, b \in R possess a greatest common divisor, defined up to multiplication by units. This GCD, denoted \gcd(a, b), is constructed by taking the product of the irreducible factors common to both a and b, each raised to the minimum of their exponents in the unique factorizations of a and b.^[8]^[2] The least common multiple (LCM) of a and b also exists in R, again up to units, and satisfies the relation \operatorname{lcm}(a, b) \cdot \gcd(a, b) \sim a \cdot b, where \sim denotes association (i.e., differing by a unit factor). When a and b are coprime (i.e., \gcd(a, b) is a unit), this simplifies to \operatorname{lcm}(a, b) \sim a \cdot b. This product formula generalizes to finite sets of elements, leveraging the unique factorization to determine maximal exponents for each irreducible.^[19] In the polynomial ring R[x_1, \dots, x_n] where R is a UFD (such as the polynomial ring over a field k), the content of a polynomial f = \sum c_I x^I with coefficients c_I \in R is defined as the GCD of its nonzero coefficients, \operatorname{cont}(f) = \gcd(\{c_I \mid c_I \neq 0\}), which exists by the GCD property of R. A polynomial is primitive if its content is a unit in R. Gauss's lemma states that the content is multiplicative: for any f, g \in R[x_1, \dots, x_n], \operatorname{cont}(f g) = \operatorname{cont}(f) \cdot \operatorname{cont}(g) up to units. Moreover, the product of two primitive polynomials is primitive, ensuring that factorization into irreducibles in the quotient field ring K[x_1, \dots, x_n] (where K is the fraction field of R) lifts to primitive factors in R[x_1, \dots, x_n], preserving unique factorization.^[20]^[21] Algorithmically, GCDs in a UFD can be computed by obtaining the unique factorizations of the elements and selecting the minimal exponents for common irreducibles, though this requires an effective factorization procedure. In specific UFDs admitting a Euclidean function—such as the integers \mathbb{Z} (with the absolute value norm) or polynomial rings k over a field k (with the degree norm)—the Euclidean algorithm extends naturally to compute GCDs efficiently via repeated division with remainder, yielding the same result as the factorization method.^[2]^[22]

Characterizations and Relations

Equivalent Conditions

A fundamental equivalent condition for an integral domain R to be a unique factorization domain is that every nonzero non-unit element of R admits a factorization into irreducibles and that any two such factorizations of the same element are equal up to multiplication by units and reordering of factors.^[1] This formulation emphasizes both the existence of irreducible factorizations and their uniqueness modulo associates.^[23] Another set of equivalent conditions is that R satisfies the ascending chain condition on principal ideals (ACCP)—meaning every ascending chain of principal ideals stabilizes—and that every irreducible element of R is prime.^[24] The ACCP guarantees the existence of factorizations into irreducibles by preventing infinite strictly ascending chains of divisors for any element, while the primality of irreducibles ensures uniqueness via the property that irreducibles divide products only if they divide one factor, leading to the standard cancellation in factorizations.^[25] In a UFD, uniqueness of factorization implies the ACCP, as equal-length factorizations into irreducibles force stabilization in divisor chains.^[26] For Noetherian integral domains, R is a UFD if and only if every prime ideal of height one is principal.^[27] This characterization leverages the Noetherian structure to relate ideal-theoretic properties to element factorization; in such domains, height-one primes arise minimally over principal ideals generated by irreducibles, and principality ensures controlled generation.^[23] An additional equivalent condition is that every nonzero prime ideal of R contains a nonzero principal prime ideal.^[23] In a UFD, irreducible elements generate such principal prime ideals, and every prime ideal contains one generated by an irreducible factor. Conversely, this condition implies that irreducibles are prime and factorizations exist and are unique, as principal primes enforce the necessary divisibility control.^[28]

Connections to Other Integral Domains

Unique factorization domains (UFDs) exhibit strong connections to other classes of integral domains, particularly in terms of divisibility and factorization properties. A key implication is that every UFD is a greatest common divisor (GCD) domain, meaning that for any two elements a, b in the domain, there exists a greatest common divisor d such that d divides both a and b, and any other common divisor divides d.^[29] This follows from the unique factorization property, which allows the GCD to be constructed by taking the minimum exponents in the irreducible factorizations of a and b.^[30] Principal ideal domains (PIDs) form a subclass of UFDs, as every PID admits unique factorization into irreducibles.^[31] In a PID, every ideal is principal, generated by a single element, which strengthens the divisibility structure and ensures the existence of unique factorizations. Euclidean domains, such as the ring of integers \mathbb{Z} or the polynomial ring k over a field k, are PIDs and thus UFDs, where the Euclidean algorithm facilitates division and GCD computations.^[32] However, the converse does not hold: there exist UFDs that are not PIDs. A classic example is the polynomial ring k[x, y] in two variables over a field k, which is a UFD by Gauss's lemma applied iteratively but not a PID, as the ideal (x, y) requires two generators.^[33] All UFDs are atomic domains, where every nonzero nonunit element factors into a finite product of irreducible (atomic) elements, due to the well-founded nature of the factorization process.^[34] The converse fails, however: atomic domains need not have unique factorizations. For instance, the ring of integers \mathbb{Z}[\sqrt{-5}] in the quadratic field \mathbb{Q}(\sqrt{-5}) is atomic (as it is Noetherian, satisfying the ascending chain condition on principal ideals) but not a UFD, since $6 = 2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5}) provides two distinct irreducible factorizations.^[35] UFDs can be generalized to weaker structures that retain some factorization control but relax uniqueness. Half-factorial domains (HFDs) are atomic integral domains where all irreducible factorizations of a given nonzero nonunit element have the same length, though the factors themselves may differ.^[34] Every UFD is an HFD, but HFDs provide a bridge to broader classes like bounded factorization domains, with applications in number theory for analyzing tame factorization behaviors in rings of algebraic integers.^[36]

References

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Section 10.120 (034O): Factorization—The Stacks project
A unique factorization domain, abbreviated UFD, is a domain R such that if x \in R is a nonzero, nonunit, then x has a factorization into irreducibles.
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[PDF] NOTES ON UNIQUE FACTORIZATION DOMAINS Alfonso Gracia ...
Jan 21, 2016 · In this section we want to define what it means that “every” element can be written as product of “primes” in a “unique” way (as we normally ...
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[PDF] Unique Factorization Domains
A Unique Factorization Domain (UFD) is an integral domain R in which every nonzero element r ∈ R which is not a unit has the following properties. 1 r can be ...
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[PDF] Polynomial rings and unique factorization domains
A UFD is an integral domain where every non-zero non-unit is a product of irreducibles, and the decomposition is unique up to order and units.
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Unique Factorization | Department of Mathematical Sciences
Let D be a principal ideal domain. If a and b are nonzero elements of D, then D contains a greatest common divisor of a and b, of the form as+bt for s,t in D.
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[PDF] Abstract Algebra
... ABSTRACT ALGEBRA. Third Edition. David S. Dummit. University of Vermont. Richard M. Foote. University of Vermont john Wiley & Sons, Inc. Page 6. ASSOCIATE ...
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(a) Definition: Suppose D is an integral domain and a, b ∈ D. Then a and b are associates if a = ub for some unit u. Example: In Z, a = 5 and b = -5 are ...
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[PDF] 6.6. Unique Factorization Domains
In a unique factorization domain, any finite set of nonzero elements has a greatest common divisor, which is unique up to multiplication by units. Proof. Let a1 ...
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[PDF] 6 Gaussian Integers and Rings of Algebraic Integers
As in the integers, unique factorization will follow from the equivalence of primes and irreducibles. Definition 6.12. Let π be a Gaussian integer such that N(π) ...
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[PDF] Polynomial rings and unique factorization domains
A ring is a unique factorization domain, abbreviated UFD, if it is an integral domain such that (1) Every non-zero non-unit is a product of irreducibles. (2) ...<|separator|>
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[PDF] Andrew Adair Math 676 Fall 2022 Homework 3 Solutions 2022 ...
Sep 20, 2022 · To see this is not a UFD, use the example from the hint, or something similar: x2 and x3 are now both irreducibles, and x2x2x2 = x6 = x3x3 ...
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Jan 13, 2021 · ZC is not atomic, and so it is not a UFD. Z[. √. −5] is an atomic domain that is not a UFD: for instance,. 6=2 · 3 = (1 −. √. −5)(1 +. √. −5 ...
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unique factorization domain. One can show, however, that every non-unit in R ... tored into irreducibles is called an atomic domain. Prove that every.<|control11|><|separator|>
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Dec 7, 2013 · A commutative domain R is a unique factorization domain, or. UFD, if ... Show that R is a UFD if and only if every irreducible is prime.
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[PDF] GCD domains
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A domain R is a unique factorization domain (UFD) if any two factorizations are equivalent. [1.0.1] Theorem: (Gauss) Let R be a unique factorization domain.
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Adp-if v(a) <= v(b), and to Ad if v,(a) > v,(b). COROLLARY 1.6. Let A be a domain, a an element of A, and y a product of prime elements ...
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May 12, 2008 · A unique factorization domain is an integral domain where every element can be factorized uniquely as a product of irreducible elements. gcd ...
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Unique factorization domain that is not a Principal ideal domain
Apr 24, 2016 · It is also known that for a UFD R, R[X] is also a UFD. Therefore the polynomial ring F[X1,X2] in two variables is a UFD as F[X1,X2]=F[X1][X2].
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Mar 22, 2013 · We define a ring R=Z[√−5]={n+m√−5:n,m∈Z} R = ℤ ⁢ [ - 5 ] = { n + m ⁢ - 5 : n , m ∈ ℤ } with addition and multiplication inherited from C ℂ ...
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Overrings of Half-Factorial Domains | Canadian Mathematical Bulletin
Nov 20, 2018 · An atomic integral domain D is a half-factorial domain (HFD) if for any irreducible elements α1,..., αn, β1,..., βm of D with α1... αn = β1 ...