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Factorization in the Self-Idealization of a Pid 3 FACTORIZATION IN THE SELF-IDEALIZATION OF A PID GYU WHAN CHANG AND DANIEL SMERTNIG a b Abstract. Let D be a principal ideal domain and R(D) = { | 0 a a, b ∈ D} be its self-idealization. It is known that R(D) is a commutative noetherian ring with identity, and hence R(D) is atomic (i.e., every nonzero nonunit can be written as a finite product of irreducible elements). In this paper, we completely characterize the irreducible elements of R(D). We then use this result to show how to factorize each nonzero nonunit of R(D) into irreducible elements. We show that every irreducible element of R(D) is a primary element, and we determine the system of sets of lengths of R(D). 1. Introduction Let R be a commutative noetherian ring. Then R is atomic, which means that every nonzero nonunit element of R can be written as a finite product of atoms (irreducible elements) of R. The study of non-unique factorizations has found a lot of attention. Indeed this area has developed into a flourishing branch of Commutative Algebra (see some surveys and books [3, 6, 8, 5]). However, the focus so far was almost entirely on commutative integral domains, and only first steps were done to study factorization properties in rings with zero-divisors (see [2, 7]). In the present note we study factorizations in a subring of a matrix ring over a principal ideal domain, which will turn out to be a commutative noetherian ring with zero-divisors. To begin with, we fix our notation and terminology. Let R be a commutative ring with identity and U(R) be the set of units of R. Two elements a,b R are said ∈ to be associates if aR = bR. Clearly, if a = ub for some u U(R), then a and b are ∈ associates. An a R is said to be irreducible if a = bc implies that either b or c is ∈ associated with a. We say that R is atomic if every nonzero nonunit of R is a finite arXiv:1311.5107v1 [math.AC] 20 Nov 2013 product of irreducible elements. It is clear that noetherian rings are atomic (cf. [2, Theorem 3.2]) and that 0 R is irreducible if and only if R is an integral domain. ∈ A ring R is a half-factorial ring (HFR) (resp., bounded factorization ring (BFR)) if R is atomic and two factorizations of a nonzero nonunit into irreducible elements have the same length (resp., for each nonzero nonunit x R, there is an integer ∈ N(x) 1 so that for any factorization x = x xn, where each xi is a nonunit, ≥ 1 ··· we have n N(x)). R is a FFR (finite factorization ring) if R is atomic and each ≤ nonzero nonunit has only finitely many factorizations into irreducibles, up to order and associates. A nonzero nonunit x R is said to be prime (resp., primary) if xR ∈ Date: 24.5.2013. 2010 Mathematics Subject Classification. 13A05, 13F15, 20M13, 20M14. Key words and phrases. non-unique factorizations, matrix ring, noetherian ring, sets of lengths. 1 2 G.W.CHANGANDD.SMERTNIG is a prime (resp., primary) ideal. Hence a prime element is primary but not vice versa (for example, if Z is the ring of integers, then 4 Z is primary but not prime). ∈ We say that R is a unique factorization ring (UFR) if every nonzero principal ideal of R can be written as a product of principal prime ideals (cf. [2, Theorem 4.9]). Clearly, a prime element is irreducible, and so a UFR is atomic. For x R a nonzero nonunit, its set of lengths is defined as ∈ L(x)= k N there exist irreducibles u ,...,uk R with x = u ... uk . { ∈ | 1 ∈ 1 · · } Clearly, x is irreducible if and only if L(x)= 1 . If x U(R), we set L(x)= 0 . { } ∈ { } The system of sets of lengths is defined as (R) = L(x) x R 0 . Sets L | ∈ \{ } of lengths and invariants derived from them are some of the classical invariants considered in non-unique factorization theory (see [8, Ch. 1.4]). The reader is referred to [8] for undefined definitions and notations. Let M be an R-module. The idealization R(+)M of M is a ring, which is defined as an abelian group R M, whose ring multiplication is given by (a,b) (x, y) = ⊕ · (ax,ay + bx) for all a, x R and b,y M. It is known that R(+)M is a noetherian ∈ ∈ ring if and only if R is noetherian and M is finitely generated [4, Theorem 4.8]. Let D be an integral domain, Mat (D) be the ring of 2 2 matrices over D, and 2×2 × a b R(D) = a,b D . It is easy to show that R(D) is a commutative { 0 a | ∈ } ring with identity under the usual matrix addition and multiplication; in particular, a 0 R(D) is a subring of Mat (D). Clearly, the map a embeds D into 2×2 7→ 0 a a b R(D), and the map ϕ : D(+)D R(D), given by ϕ(a,b) = , is a ring → 0 a isomorphism. In view of this isomorphism, R(D) is called the self-idealization of D (cf. [13]). There is also an isomorphism D[X]/ X2 R(D) mapping a+bX+ X2 h i → h i a b to . Some factorization properties of R(+)M have been studied in [2, 0 a Theorem 5.2]. For more on basic properties of R(+)M (and hence of R(D)), see [4] or [11, Section 25]. Let D be a principal ideal domain (PID). Then R(D) is noetherian, and thus R(D) is atomic. In Section 2, we first characterize the irreducible elements of R(D), and we then use this result to show how to factorize each nonzero nonunit of R(D) 0 1 into irreducible elements via the factorization of D. We show that is 0 0 the unique prime element (up to associates) of R(D). We prove that every nonzero nonunit of R(D) can be written as a product of primary elements. Finally, in Section 3, we completely describe the system of sets of lengths (R(D)). L We write N = 1, 2, 3,... for the set of positive integers, and N = N 0 for { } 0 ∪{ } the set of non-negative integers. 2. Factorization in R(D) when D is a PID Let D be an integral domain, and a b R(D)= a,b D { 0 a | ∈ } FACTORIZATION IN THE SELF-IDEALIZATION OF A PID 3 1 0 be the self-idealization of D. Clearly, is the identity of R(D). 0 1 a b If α = R(D), then nr(α) = a is the norm, and this is a ring homo- 0 a ∈ morphism R(D) D. Observe that α is a zero-divisor if and only if a = 0. We → write R(D)• for the monoid of non-zero-divisors of R(D). We begin this paper by characterizing the units of R(D), which is very useful in the proof of Theorem 5. Lemma 1. (cf. [11, Theorem 25.1(6)]) If D is an integral domain, then α = a b R(D) is a unit of R(D) if and only if a is a unit of D. 0 a ∈ x y 1 0 x y Proof. If α is a unit, then α = for some R(D). · 0 x 0 1 0 x ∈ Thus ax = 1, and so a U(D). Conversely, assume that a is a unit, and let ∈ a b u bu2 1 0 u bu2 u = a−1. Then − = and − R(D). 0 a 0 u 0 1 0 u ∈ Thus α is a unit. For an arbitrary commutative ring R, there can be two elements a,b R such ∈ that a and b are associates but a = ub for all u U(R) (see, for example, [2, 6 ∈ Example 2.3]). This cannot happen in the self-idealization of an integral domain. Lemma 2. Let D be an integral domain and α, β R(D) and let a,b,x,y D ∈ ∈ a b x y such that α = and β = . The following statements are equivalent. 0 a 0 x (1) α and β are associates. (2) There exists θ U(R(D)) such that β = θα. ∈ (3) There exists u U(D) such that x = au and y bu mod a. ∈ ≡ Proof. (1) (2): If α and β are associates, then there are some γ,δ R(D) so ⇒ ∈ a b x y that α = βγ and β = αδ. Hence if γ = 1 1 and δ = 1 1 , then 0 a 0 x 1 1 a = xa and x = ax , and so a , x U(D). Thus γ,δ U(R(D)) by Lemma 1. 1 1 1 1 ∈ ∈ u v (2) (3): Let θ = . By Lemma 1, u U(D). From β = θα it follows ⇒ 0 u ∈ that x = au and y = av + bu bu mod a. ≡ u v (3) (2) and (1): Let v D be such that y = bu + av. Define θ = . ⇒ ∈ 0 u Then θ U(R(D)) by Lemma 1 and β = θα. Thus, α and β are associates. ∈ We write α β if α, β R(D) are associates. ≃ ∈ a b Lemma 3. Let D be a PID and let α = R(D)•. If a = cd with coprime 0 a ∈ c, d D, then there exist γ,δ R(D) with α = γδ and nr(γ) = c, nr(δ) = d.
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