Unique Factorisation in Abstract Algebra

Adrian Petersen Supervised by Phill Schultz University of Western Australia

1 Abstract The Fundamental Theorem of Arithmetic states that any integer can be written as the product of prime numbers and units. This decomposition is unique up to order and unit multiples. This familiar factorisation of integers can be generalised to more abstract objects such as rings, and sets of ideals or modules. The factorisation properties of each of these is investigated. Further, definitions and properties of Factorial Domains, Dedekind Domains, Factorial Monoids and Krull–Schmidt classes of modules are outlined. An effort is then made to relate each of these to a classic example of a factorial monoid, the (Ω) free commutative monoid N . We have confirmed that an isomorphism can be constructed from either a Factorial Domain, set of ideals of a Dedekind Domain, and Krull–Schmidt class of modules to this free commutative monoid (Ω) N . 1 Introduction

The Fundamental Theorem of Arithmetic states that every non-zero integer can be expressed uniquely as a product of prime numbers and units. This theorem is com- monly taught in an introduction course to pure mathematics. The aim of this project is to investigate similar factorisation behaviour in the more abstract settings of rings, monoids, ideals and modules. First we shall be considering the factorisation of elements in rings and monoids. Then we consider the factorisation of sets such as ideals and modules. Common features between the decomposition of these objects will be investigated and put into a similar framework. The reader is assumed to be familiar with concepts from an undergraduate algebra course, i.e. they are familiar with the concepts of an integral domain, multiplicative monoid, ideal of a ring and module over a ring. R is an integral domain with unit group R×. N is the natural numbers, N+ is the set of positive natural numbers, and Z represents the integers. Section two serves as an introduction to unique factorisation in a more abstract setting, but goes no further than an undergraduate algebra course. Section three introduces unique factorisation of elements in a monoid, and relates it to section two by means of an isomorphism to the set of functions f :Ω → N. It is then found similar isomorphisms can be constructed between sets of ideals, and modules with unique factorisation properties.

2 Factorial Domains

The purpose of this section is to introduce the idea of factorisation in rings. The concept of factorisation in a ring is easiest to characterise if the ring R is a domain which we shall assume from now on. Familiar examples will be discussed in this section from number theory. The idea of a Unique Factorisation Domain is analogous to the Fundamental Theorem of Arithmetic for the integers which states that every non–zero integer can be uniquely expressed as the product of prime numbers and units.

2.1 Elementary Definitions and examples Well known examples of integral domains, which we shall call simply domains, are the integers Z, the fields Zp of integers modulo a prime p, any subring containing 1 of a field F, the polynomial rings R[x1, . . . , xn] in several variables over a domain R.

1 Deﬁnition 2.1.1. Let R be a domain. Then:

(a)a unit is an invertible element of R. Let R× denote the group of units in R. a, b ∈ R are said to be associates if a = bu for some u ∈ R×.

(b) Let a ∈ R, if a = bc, then b and c are factors of a. An atom of R is a non-zero non-unit such that if a = bc then b or c is a unit. A domain is said to be atomic if every element, not zero or a unit , is a product of a ﬁnite set of atoms.

(c) Let a, b ∈ R. Then a is said to divide b , denoted a | b if and only if b = ac for some c ∈ R. p ∈ R is said to be prime if p | ab implies p | a or p | b ∀ a, b ∈ R.

Example 2.1.1. Examples of atomic domains include

(a) Z √ (b) Z[ −5] (c) R[x, y] where R is a domain

If R is not atomic then there must exist a ∈ R such that if a is a product of ﬁnitely many elements, they cannot all be atoms. For an example of a domain that is not atomic consider the ring R = {f : C → C} where C is the complex numbers, addition and and multiplication are pointwise. Then for any f ∈ R, and n ∈ N you can factorise f = gn where g(z) is an n-th root of f(z). So R has no atoms.

2.2 Factorisation into Atoms Deﬁnition 2.2.1. An element a ∈ R has a unique decomposition into atoms up to order and associates if whenever a = b1 . . . bn = c1 . . . ck for atoms bi, ci then n = k and for each i ∈ [1, n] there exists j ∈ [1, n] such that ai and bj are associates.

Factorisation√ into atoms√ in atomic domains need not be unique. Consider the domain [ −5] = a + b −5 | a, b ∈ . It is possible to show that the element √Z Z 6 ∈ Z[ −5]. 6 can be factorised into two distinct products of atoms. √ √ 6 = 2 · 3 = (1 + −5)(1 − −5) √ √ √ It is simple to prove that 2,3,(1 + −5) and (1 − −5) are atoms in Z[ −5], none of which are associates of each other.

2 √ Proof.√ Suppose that 2 can be written as a product of non-units 2 = (a + b√−5)(c + d√−5). Taking the norm of each side of this equation gives |2| = |a + b −5||c + 2 2 2 2 2 2 d −5| which simpliﬁes to 4 = |a + 5b ||c + 5d |. So |a + 5b√| = 1, 2 or 4,√ which has integral solutions (a, b) = (1, 0) or (2, 0). Hence either (a+b −5) or (c+d −5) is a unit. √ √ Similarly, it can be shown that 3,(1+ −5) and (1− −5) can only be factorised into a product of units and themselves. √ Each of these is an atom, the only units in Z[ −5] are 1 and −1, so obviously none of these atoms are associates.

2.3 Prime Factorisation in Domains

Let a ∈ R where R is a domain, a = p1....pn is called an prime factorisation of a if p1, ..., pn are all primes. Lemma 2.3.1. Let R be a domain and let p ∈ R. If p is prime then p is an atom.

Proof. Suppose that p ∈ R is prime, by deﬁnition it is not a unit. Now if p = ab for some a, b ∈ R, then p | a or p | b. Without loss of generality suppose that p | a, then a = pc for some c ∈ R. Then p = ab = pcb, since R is a domain the cancellation law holds so p = pcb implies that 1 = cb. So b is a unit. Similarly it can be shown that if p | b then a is a unit. Beacause p is not a unit and cannot be expressed as a product of non-units, p is an atom. However, the converse of this statement is not always true. An atom in a domain need not be prime.√ An example√ is the best√ way to illustrate this. If√ we again√ consider the atoms a = 1+ −5, b = 1− −5 in Z[ −5], note that ab = (1+ −5)(1− −5) = −4. So ab | −4, but a - −4 and b - −4 so neither a or b are prime. In section 2.2. we showed by counter example that factorisation into atoms need not be unique in domains. When considering primes, it turns out that factorisation is unique up to order and unit multiple.

Lemma 2.3.2. In any domain if an element a = p1 . . . pn = q1 . . . qm where each pi and each qj are primes, then n = m and for each i there exists a unique j such that pi and qj are associates.

Proof. Suppose a = p1 . . . pn = q1 . . . qm where 1 ≤ k ≤ n for primes pi and qj.I prove by induction on n that k = n and each qj is an associate of some pi. If n = 1, then p1 = q1, so the statement is true for n = 1. Assume the statement is true for all m ≤ n. Suppose a = p1 . . . pn = q1 . . . qk where k ≤ n for primes pi and qj. Then

3 consider p1, note that p1 | q1 . . . qk so p1 | qj for some j ∈ [1, k] because p1 is prime. Since p1 and qj are prime, they are atoms, so p1 is associate to qj. Hence up1 = qj × for some u ∈ R . Relabelling qj to q1 gives us p1 . . . pn = q1 . . . qk = up1q2 . . . qk. Cancelling the p1 from each side of the equation leaves us with p2 . . . pn = uq2 . . . qk. By the induction assumption, n − 1 = m − 1 and each qj is an associate of some pi. Hence the stement is true for n. The Fundamental Theorem of Arithmetic provides a nice example of this in the integers Z. In the domain of integers every atom is a prime. It is natural to now ask the question, when are primes and atoms the same thing, and when is a factorisation into atoms unique. These two questions prove to be linked closely. A domain has unique factorisation into atoms if it is a factorial domain. Deﬁnition 2.3.1. A Factorial Domain is a domain such that every non–zero non–unit has a prime factorisation. Factorial domains are also known as Unique Factorisation Domains. Example 2.3.1. Examples of domains which are factorial are:

(a) The integers Z, this is familiar to us from the statement of the fundamental theorem of arithmetic in number theory

(b) Any ﬁeld F, for example the rational numbers Q and the complex numbers C because every element is either zero or a unit. √ (c) The Gaussian Integers, a + b −1 | a, b ∈ Z [4]. (d) R[x] is factorial if R is factorial. [2]. Not every atomic domain is factorial. Every element can be factorised into a product√ of atoms in an atomic domain, but this need not be unique. Examples such as Z[ −5] have been provided to show this. In any domain we have shown that every prime is an atom, but when is every atom a prime? Proposition 2.3.1. A domain is a factorial domain if and only if it is atomic and every atom is a prime. Proof. (⇒) Let R be a factorial domain, then R is atomic by deﬁnition. Suppose that p ∈ R is an atom, so it is not a unit. Suppose that p | ab for some a, b ∈ R, then pc = ab for some c ∈ R. Since R is a factorial domain a, b, c each have a unique factorisation into primes, so pc1....ck = a1....anb1....bm where each ai, bi, ci is a prime. Now consider a1, a1 | pc1....ck since a1 is prime a1 | cj for some j ∈ [1, k] because ai - p since

4 p is an atom. So a1 is an associate of cj. If we relabel cj to c1 we see a1 = uc1 for some × u ∈ R . Substituting into equation pc1....ck = a1....anb1....bm = uc1a2....anb1....bm. Cancelling c1 from each side leaves us with pc2....ck = ua2....anb1....bm. Repeating this process until every cj is cancelled leaves us p being associate to the last remain- ing ai or bj for some i ∈ [1, n] or j ∈ [1, m], since the number of terms is the same on both sides of the equation. Whichever term p is associate to is a factor of either a or b hence p | a or p | b.

(⇐) Suppose that R is an atomic domain and every atom is prime. Then let a = p1....pn = q1....qm is factorisation into atoms for a ∈ R. Then this is a prime factorisation since every atom is a prime. By lemma 2.3.2 prime factorisation is unique, up to order and associates. Meaning a = p1....pn = q1....qm implies n = m and each pi is associate to some qj for some i, j ∈ N. Hence, R is a factorial domain.

Lemma 2.3.2 shows that prime factorisation is unique in domains, and the remark above shows that in a factorial domains the concept of an element being prime or an atom is the same. Thus a domain is factorial if and only if it is atomic and every atom is a prime. The ideas outlined in this section provide the framework for less familiar examples to be subsequently explored. Factorisation in monoids will now be explored, and it will be shown that the multiplicative monoid of a factorial domain is a free commutative monoid.

3 Monoids

The aim of this section is to introduce the idea of a monoid and illustrate the ideas presented using examples, using both additive and multiplicative notation.

3.1 Elementary Deﬁnitions Deﬁnition 3.1.1. A monoid (M, ∗, e) is a non-empty set M with an associative binary operation ∗ and an identity e. It is commutative if the operation is commutative. A subset N ⊆ M is a submonoid if N is closed under its binary operation and contains the identity.

Clearly a submonoid N of a monoid M is itself a monoid under the restriction of the operation

5 Example 3.1.1. (a) (N, +, 0) is a commutative monoid with identity 0 and for any k > 0 the set of integers ≥ k together with 0 is a submonoid.

(b) (N+, ·, 1) is a commutative monoid with identity 1 and for any k > 0 the set of multiples of k together with 1 is a submonoid.

(c) Let R be a commutative unital ring. Then (R, ·, 1R) is a commutative monoid where 1R is the multiplicative identity in the ring and any subring containing 1 is a submonoid. It is easy to show that in any monoid, the identity is unique. Deﬁnition 3.1.2. (a)A monoid homomorphism is a function f from a monoid M to a monoid N such that f(a ∗ b) = f(a) ∗ f(b) ∀a, b ∈ M Let f : M → N be a monoid homomorphism. The kernel of f, denoted ker(f) ,is the subset K ⊆ M such that f : m 7→ e. ker f) is a submonoid of M and the image of f is a submonoid of N.

(b) Suppose M is a commutative monoid. Let a, , b, c ∈ M. M is said to be a cancellation monoid if a ∗ b = a ∗ c =⇒ b = c ∀ a, b, c ∈ M ∗

(c)A unit or an invertible element is an element a such that there exists b ∈ M, called the inverse of a, with a ∗ b = e. It is easy to show that if a has an inverse, it is unique. Let M × denotes the set of units in M. a, b ∈ M are said to be associates if a = b ∗ u for some u ∈ M ×.

(d) Let a ∈ M, a monoid. If a = b ∗ c, then b and c are factors of a. An atom is a non–unit a whose only factors are units A monoid is said to be atomic if every element, not zero or a unit , is a product of a ﬁnite set of atoms.

Example 3.1.2. Let R be a domain. Then (R, ·, 1R) forms a commutative, cancellation monoid whose units are the units of R. Deﬁnition 3.1.3. Let (M, ∗, 1) be a monoid. Let a, b ∈ M, a is said to split b denoted a|b, if b = a ∗ c for some c ∈ M A non–unit p ∈ M is said to be prime if p|a ∗ b ⇒ p|a or p|b ∀ a, b ∈ M. When studying rings the idea of ‘splitting’is usually associated with the ‘divides’ relation. However in our context it is common for our monoid to have addition as its binary operation. In this case a|b, means b = a + c for some c ∈ M It is routine to check that units divide all elements of a monoid and an element divides a unit if and only if it is a unit.

6 3.2 Reduced Monoids When we investigated factorisation in factorial domains, we discovered that factorisation was unique up to order and unit multiples. In commutative cancellation monoids it turns out that unique factorisation follows the same pattern. However in a reduced commutative cancellation monoid we can ignore the statement ‘up to unit multiples’, the purpose of this section is to introduce the idea of reduced commutative cancellation monoids and discuss how to it is possible to force a monoid to be reduced. Deﬁnition 3.2.1. Let (M, ∗, e) be a commutative monoid. (M, ∗, e) is reduced if a ∗ b = e =⇒ a = e and b = e. Equivalently, M contains only one unit.

Example 3.2.1. (a) (N, +, 0) is reduced since if x, y ∈ (N, +, 0). If x + y = 0, then x = y = 0.

(b) Similarly, (N+, ·, 1) is a reduced monoid. Definition 3.2.2. Let ∼ be an equivalence relation on a commutative monoid M such that a1 ∼ a2 and b1 ∼ b2 =⇒ a1b1 ∼ a2b2 ∀ a1, a2, b1, b2 ∈ M. Then the equivalence relation is said to be a congruence relation on M. [1] The idea of a congruence relation is important because it enables us to omit the phrase ‘up to unit multiples’ in the definition of unique factorisation. In order to do this we factor out unit multiples with the congruence relation associativity. Recall that two elements a, b ∈ M are associates if a = bu for some u ∈ M ×. Lemma 3.2.1. Let (M, ∗, e) be a commutative monoid, then the relation ∼ defined by a ∼ b if a and b are associates forms a congruence relation. Proof. It is routine to check that ∼ is an equivalence relation. To check that ∼ is a congruence, suppose that a1 ∼ a2 and b1 ∼ b2. Then a1 = u ∗ a2 and b1 = v ∗ b2 for × some u, v ∈ M . So a1 ∗ b1 = (u ∗ a2) ∗ (v ∗ b2) = (u ∗ v) ∗ (a2 ∗ b2) where u ∗ v is a unit, thus a1b1 ∼ a2b2. Theorem 3.2.1. Let (S, ∗, e) be a monoid and let ∼ be a congruence on S. For a ∈ S, denote the congruence class of a by [a] and let M be the set of congruence classes. Define an operation ◦ on M by [a] ◦ [b] = [a ∗ b]. Then (a) (M, ◦, [e]) is a monoid;

(b) If (S, ∗, e) is commutative, then so is (M, ◦, [e]);

7 (c) If (S, ∗, e) is cancellative, then so is (M, ◦, [e]);

(d) (S, ∗, e) is commutative and cancellative and ∼ is the relation ‘is an associate of’then (M, ◦, [e[) is reduced.

(e) In this case, if a is an atom in S then [a] is an atom in M and if a is a prime in S then [a] is a prime in M;

Proof. The proof is a routine application of the deﬁnitions.

Lemma 3.2.2. The mapping (S, ∗, e) −→ (S/ ∼, ◦, [e]) where a 7−→ [a] is a surjective homomorphism (epimorphism). The kernel of this mapping is the set {a ∈ S | a ∼ i}.

Proof. The proof again is a routine application of the deﬁnitions. Now we see that we can impose the reduced property onto any monoid by fac- toring out unit multiples. For example, Let ∼ be the congruence relation of associativity on the monoid (Z, ·, 1). The congruence class of n ∈ Z is {n, −n}, and the image of the epimorphism is the reduced monoid (N, ·, 1).

3.3 Free commutative monoids

For any set Ω, NΩ denotes the set of functions f :Ω −→ N. Under pointwise addition the set of functions (NΩ, +, 0) forms a commutative monoid with identity 0, the function 0 : ω 7−→ 0 , ∀ ω ∈ Ω

Proof. Let f, g, h ∈ NΩ, then ((f + g) + h)(ω) = (f + g)(ω) + h(ω) = f(ω) + g(ω) + h(ω) = f(ω) + (g + h)(ω) = (f + (g + h))(ω) by associativity of addition in the integers. Let 0 : ω 7−→ 0 , ∀ ω ∈ Ω, then (f + (0))(ω) = f(ω) + (0)(ω) = f(ω) for all ω ∈ Ω.

Deﬁnition 3.3.1. (a) The support of a function f : A −→ B is the set

supp(f) = {a ∈ A : f(a) 6= 0}

(b) The function f is said the have ﬁnite support if supp(f) is ﬁnite.

8 Every element in a factorial monoid can be factorised into prime elements uniquely up to unit multiples. If this factorial monoid is reduced the factorisation will be unique, as only one unit exists. The following sequence of lemmas show that a free commutative monoid is a reduced factorial monoid, hence has unique factorisation.

Lemma 3.3.1. In the free commutative monoid N(Ω), f ∈ N(Ω) is an atom if and only if α ∈ Ω such that f(α) = 1 and f(β) = 0, ∀ β 6= α.

Proof. (⇒) Suppose f ∈ N(Ω) is an atom, and f is some other function other than f(α) = 1 for some α ∈ Ω and f(β) = 0, ∀ β 6= α. Since f is an atom it is non–zero.

Let {ωi : i = 1 . . . m} be the set of elements of Ω where f(ωi) 6= 0. Then f : ωi 7→ nωi . Pm Then f = i=1 nωi fωi and fωi : ωi 7→ 1 and fωi : a 7→ 0 for all other a ∈ Ω. Which is a contradiction since f is meant to be an atom.

(⇐) Suppose f(α) = 1 for some α ∈ Ω and f(β) = 0, ∀ β 6= α. Then if f = g + h, g or h is the function 0 : ω 7→ 0 for every ω ∈ Ω because the image of f is an atom in (N, +, 0). Lemma 3.3.2. If Ω is an infinite set, NΩ is non atomic and N(Ω) is atomic. Ω Proof. Atoms in N have the form fa : a 7→ 1 and fa : b 7→ 0 for all other b ∈ Ω. (Ω) Suppose f ∈ N and let {ωi : i = 1 . . . m} be the set of elements of Ω where Pm f(ωi) 6= 0. Then f : ωi 7→ nωi . Then f = i=1 nωi fωi and fωi : ωi 7→ 1 and fωi : a 7→ 0 for all other a ∈ Ω, is an atomic decomposition for f. This is a finite product of atoms since only a finite number of elements in Ω have non-zero image under f. Consequently if f ∈ NΩ and supp(f) is infinite, f cannot be expressed as a finite product of atoms fωi because there are infinite elements in ω that have non-zero image under f.

Lemma 3.3.3. Atoms in N(Ω) are prime Proof. Suppose f ∈ N(Ω) such that f is an atom, and f | g + h for some g, h ∈ N(Ω). Then f +j = g+h for some j ∈ N(Ω). Now suppose f 6= g,(f +j)(ω) = f(ω)+j(ω) = (g + h)(ω) = g(ω) + h(ω). Since f is an atom f(a) = 1 for some a ∈ Ω and f(b) = 0 for all other b ∈ Ω. So at least g(a) or h(a) 6= 0, hence f | g or f | h.

3.4 Factorisation in Monoids When investigating factorisation in domains, we found that elements can be factorised into atoms or primes. Similarly we can factorise elements in our monoids into atoms or primes.

9 Let a ∈ M. a = p1....pn is called an atomic factorisation of a if the p1, ..., pn are all atoms, and a prime factorisation of a if p1, ..., pn are all primes. M is atomic if we can factorise every element√ into a ﬁnite product of atoms. For example, the monoids (N, +, 0) and (Z[ −5], ·, 1) are atomic. It has been previously shown that factorisation√ into atoms need not be unique, the same√ example provided for the domain Z[ −5] demonstrates this for the monoid (Z[ −5], ·, 1). Deﬁnition 3.4.1. An element a ∈ M has a unique decomposition into atoms up to order and associates if whenever a = b1 . . . bn = c1 . . . ck for atoms bi, ci then n = k and for each i ∈ [1, n] there exists j ∈ [1, n] such that ai and bj are associates. Contrary to factorisation into atoms, it can be shown that factorisation into prime elements is unique in reduced commutative cancellation monoids. The proof is similar to that for a domain (see lemma 2.3.2).

Lemma 3.4.1. In any commutative cancellation monoid if an element a = p1 . . . pn = q1 . . . qm where each pi and each qj are primes, then n = m and for each i there exists a unique j such that pi and qj are associates. Lemma 2.3.1 states that every prime is an atom in a domain. The same can be shown to be true in a cancellation monoid.

Lemma 3.4.2. Let (M, ∗, e) be a cancellation monoid and let a ∈ M. If a is a prime then a is an atom.

Proof. Suppose M is a monoid, and p ∈ M is a prime element, so not a unit . Now suppose that p = a ∗ b for some a, b ∈ R, since p is prime, p|a or p|b Without loss of generality let us suppose that p|a, then a = p ∗ c for some c ∈ M. Now p = a ∗ b = p ∗ c ∗ b, hence p = p ∗ c ∗ b implies 1R = cb because M is a cancellation monoid. If you cancel the p you get e = c ∗ b so b is a unit. Similarly, if p|b then a is a unit. Since p is not a unit and cannot be expressed as a product of non units, p is an atom.

Deﬁnition 3.4.2. A Factorial Monoid is a commutative cancellation monoid M in which every non-unit has a prime factorisation.

Hence a factorial monoid is characterised by the property that every non unit has a decomposition as a product of primes unique up to order and associates. Similar to the deﬁnition of a factorial domain, the second statement can be altered to state that every atom is prime. Since it has been proven that prime factorisation is unique in commutative cancellation monoids.

10 Deﬁnition 3.4.3 (simpler deﬁnition). A Factorial Monoid is a commutative cancellation monoid M such that: (a) M is atomic;

(b) Every atom is prime So far nothing new has been explored in terms of monoids. It has just been shown that all of the results from section 1 hold in terms of the more general concept of a monoid. It has been established that in a factorial monoid every element can be decomposed into a product of prime elements, however this factorisation is only unique up to unit multiples and order. To illustrate this consider the integers. (Z, ·, 1) forms a commutative, cancellation monoid. Consider the element 30, now 30 = 1 · 5 · 2 · 3 = −1 · −5 · 2 · 3. We say that this factorisation is unique up to unit multiple. Since 5 is an associate of −5 and 1 and −1 are units in Z. Note that in a reduced factorial monoid, prime factorisations are actually unique up to order, since the only unit is the identity. The following proposition is an application of this result to factorial domains. In the previous section we showed that the congruence classes (under congruence relation of ‘is an associate of ’) of a commutative cancellation monoid form a reduced commutative cancellation monoid. Hence we can force a monoid to be reduced, moreover have factorisation up to order, rather than up to order and associates. Lemma 3.4.3. Let (S, ∗, e) be a monoid and let ∼ be a congruence on S deﬁned by a ∼ b if a and b are associates. For a ∈ S, denote the congruence class of a by [a] and let M be the set of congruence classes. Deﬁne an operation ◦ on M by [a] ◦ [b] = [a ∗ b]. If (S, ∗, e) is factorial, then (M, ◦, [e]) is a reduced factorial monoid. For an example of this consider a factorial domain (R, ·, 1). Let (R, ·, 1) be the multiplicative monoid of a factorial domain R, and let (M, ·, 1) be the factor monoid with respect to the congruence ‘associate’. Then (M, ·, 1) is a reduced factorial monoid and hence every element has a prime factorisation unique up to order. By Lemma 3.3.3 we conclude:

Theorem 3.4.1. N(Ω) is a reduced factorial monoid.

3.5 Connection between free monoids and factorial domains Finally, it is now clear that commutative cancellation monoids have the same factorisation properties as domains. A factorial monoid has unique factorisation and every

11 atom is prime, just like a domain. Moreover, if a factorial monoid is reduced, the factorisation is actually unique. The free commutative monoid has been established as a reduced factorial monoid. So we may express a domain as a free commutative monoid. The connection between factorial domains and monoids is that they both determine a monoid (M, +, 0) which is

1. A factorial monoid

2. Isomorphic to the monoid (N(Ω), +, 0) so the multiplicative structure of a factorial domain determines a free commutative monoid in the sense that if R is a factorial domain and [R] is the set of congruence classes under the relation of unit multiples, then ([R], ·, [1]) is a free commutative monoid.

Theorem 3.5.1. Let R be a Factorial domain. Let Ω be the set of congruence classes of associates of primes. Deﬁne φ : R → N(Ω) by φ(r) = f where for each r ∈ R deﬁne fr :Ω → N by fr(p) = the number of occurrences of p in the unique prime decomposition of r if r 6= 1 and f1(p) = 0. Since each r is a product of primes, (Ω) fr ∈ N . Then φ is a monoid isomorphism.

4 Ideals in a Domain

4.1 Elementary Deﬁnitions and Examples Let R be a domain, and let I denote the set of ideals of R.

Deﬁnition 4.1.1. (a) Let P ∈ I. P is a prime ideal if for all a, b ∈ R, ab ∈ P implies a ∈ P or b ∈ P . Let P denote the set of prime ideals of R.

(b) Let I/R. R is a maximal ideal if I 6= R and for every ideal J such that I ⊂ J ⊂ R, either I = J or J = R.

Example 4.1.1. (a) In the ring of integers Z, any subring nZ = {nx : x ∈ Z} is an ideal of Z. Moreover nZ is a prime ideal if and only if n is a prime number.

(b) In the ring of polynomials Z[x]. The ideal generated by 2 and x, is a prime ideal.

(c) In the ring Z of integers, 3Z is a prime and maximal ideal, but 4Z is neither since 4Z ⊂ 2Z and 2 × 2 ∈ 4Z but 2 6∈ 4Z.

12 Lemma 4.1.1. Let P/R. Then P is a prime ideal if and only if R/P is a domain. Moreover P is a maximal proper ideal if and only if R/P is a ﬁeld.

It follows that that maximal ideals are prime. The ring Z has the property that prime ideals are maximal, but this is false in general, for example in the ring R = F[x, y] of polynomials in two variables over a ﬁeld F, the principle ideal (x) is prime but not maximal, since it is contained in the ideal (x, y) generated by x and y. Later I shall consider a class of domains, called Dedekind domains, having the property that all prime ideals are maximal.

4.2 Factorisation of Ideals To investigate factorisation of ideals, we first must define a product of ideals. Let I,J be two ideals of a domain R. Then the product of I and J denoted IJ is Pn IJ = { i=1 aibi | ai ∈ I, bi ∈ J} for some positive integer n The following Lemma shows that the subset IJ of R described above, together with the intersection I ∩ J, are ideals. The proof is omitted as it is a routine consequence of the definitions. Lemma 4.2.1. Let I,J/R. Then IJ and I ∩ J/R. Furthermore, IJ ⊆ I ∩ J, I ∩ J is contained in every ideal containing I and J and I + J contains every ideal containing I and J.

Example 4.2.1. Consider the domain Z. Let I = 2Z and J = 3Z be ideals of Z. Pn Then IJ = { i=1 aibi | ai ∈ I, bi ∈ J} = 6Z Definition 4.2.1. Let I,J/R, where R s a domain. We say I splits J, denoted I|J, if there exists K/R such that J = IK. Lemma 4.2.2. Let P ∈ I, then P is prime if and only if P | IK =⇒ P | I or P | K for any I,K ∈ I [2, page m]. Just as in factorial domains and free monoids, we are interested with domains in which every ideal has a prime decomposition. Let I ∈ I. Then I = P1....Pn where Pi ∈ P are proper ideals is called a prime ideal factorisation of I. An ideal I ∈ I is said to be proper if R 6= {0} and I 6= R. Definition 4.2.2. A Dedekind Domain is an integral domain R in which every proper ideal is the product of a finite number of proper prime ideals [1]. Lemma 4.2.3. If R is a Dedekind domain, then under the operation of ideal multiplication, I is a reduced commutative cancellation monoid with identity R.

13 Proof. Associativity Pn Pn Let I, J, K ∈ I. Then (IJ)K = { i=1(aibi)ci | ai ∈ I, bi ∈ J, ci ∈ K} = { i=1 ai(bici) | ai ∈ I, bi ∈ J, ci ∈ K} = I(JK) since I, J, K are all commutative rings because they are ideals of a domain.

R is the identity Pn Suppose I ∈ I, then claim RI = { i=1 aibi | ai ∈ R, bi ∈ I} = I. Suppose x ∈ RI, Pn then x = i=1 aibi for some ai ∈ R, bi ∈ I. Each product is in the ideal I because it is an ideal, hence the sum is in the ideal. Thus RI ⊆ I. Suppose x ∈ I, then x = ex which is a product of elements from R and I, hence x ∈ RI. Therefore we have RI ⊆ I, and I ⊆ RI, meaning RI = I.

Commutative and Cancellation I is obviously commutative since multiplication of elements is commutative. It can be shown ([1, Chapter 8, Section 6]) that multiplication of ideals in a Dedekind ring is cancellative.

Lemma 4.2.4. In any Dedekind domain if an element I = P1 ...Pn = Q1 ...Qm where each Pi and Qj are prime ideals, then m = n and for every i ∈ [1, n] there exists j ∈ [1, n] such that Pi = Qj. The proof of this Lemma is once again very similar to that of lemma 2.3.2. Which has been a common theme throughout this paper. A domain is factorial if its elements have prime decompositions and it is Dedekind if its ideals have prime decompositions. Dedekind domains need not be factorial and factorial domains need not be Dedekind.

Example 4.2.2. (a) The integers Z are a Dedekind domain √ (b) Z[ −5] is a Dedekind domain which is not factorial. (c) R[x, y] is a factorial domain which is not Dedekind [1].

4.3 Expressing a Set of Ideals as a Free Commutative Monoid Recall lemma 4.2 which states that the set of ideals I of a Dedekind domain R form a commutative cancellation monoid under ideal multiplication with identity R. It turns out that this forms the free monoid N(Ω). Theorem 4.3.1. Let R be a Dedekind domain. Let I be the monoid of ideals of R, and Ω be the set of prime ideals of R. Deﬁne φ : I → N(Ω) by φ(I) = f where for

14 each I ∈ I deﬁne fI : P → N by fI (P ) = the number of occurrences of P in the unique prime decomposition of I if I 6= R and fR(P ) = 0. Since each I is a product (Ω) of proper prime ideals, fI ∈ N . Then φ is a monoid isomorphism. Proof. φ is a homomorphism Let I,J ∈ I, then φ(I) + φ(J) = fI + fJ , and φ(IJ) = fIJ . We need to show that fI +fJ = fIJ . Let P ∈ P, then (fI +fJ )(P ) = (fI )(P )+(fJ )(P ) this is simply the sum of the number of times P occurs in the unique factorisation of I and J, and fIJ (P ) is the number of times P occurs in the factorisation of IJ, which is the sum of the number of occurrences of P in I and J. Hence fI + fJ = fIJ =⇒ φ(I) + φ(J) = φ(IJ).

φ is injective Suppose that φ(I) = φ(J). Then fI = fJ =⇒ fI (P ) = fJ (P ), ∀ P ∈ P. Then the factorisation into prime ideals for I and J are the same, since each prime ideal occurs the same number of times. This factorisation is unique since R is a Dedekind domain, hence I = J

φ is surjective (Ω) Let f ∈ N , where f : Pin 7→ n where n, for Pin ∈ P and n ∈ N. Then choose each

Pin to occur n times in the factorisation of some ideal K. Then φ(K) = fK = f

5 Modules

We have seen that the Fundamental Theoreom of Arithmetic can be generalised to factorial domains, factorial monoids, and Dedekind domains. In this section we see that unique factorisation can be generalised to decomposition of modules into direct sums of indecomposable modules. The reader is assumed to be familiar with the concepts of left module over a ring, homomorphisms and isomorphisms of modules. Familiar examples of R-Modules which will be used include any vector space over a field F, and a finite abelian group over Z. So far we have seen in domains, monoids and ideals, elements can be built up by combining atoms. The idea of indecomposable modules is analogous to this. ∼ Just as for vector spaces, we can form the direct sum M = M1 ⊕ .... ⊕ Mn of a family of modules. This just the set of n-tuples (x1, ...., xn) with xi ∈ Mi, with obvious coordinate-wise operations [3]. Definition 5.0.1. A non-zero R-module M is decomposable provided there exists ∼ nonzero R-modules M1 and M2 such that M = M1 ⊕ M2; otherwise M is indecomposable [3].

15 Note that the 0 module is not considered to be indecomposable, for the same reason that 1 is not regarded to be a prime number [3]. Sometimes, it’s better to work with internal decompositions; a module M is decomposable if an only if there exist nonzero submodules M1 and M2 of M such that M = M1 ⊕ M2[3]. Deﬁnition 5.0.2. A set M of modules is called Noetherian if every M ∈ M is atomic and M is closed under direct sums and summands, i.e., for all modules M and N,M ⊕ N ∈ M if and only if N and M ∈ M.

Definition 5.0.3. A Noetherian set M of modules is called a Krull-Schmidt class if indecomposable decompositions are unique up to isomorphism. i.e. if N1 ⊕ .... ⊕ ∼ Nn = M1 ⊕ .... ⊕ Mm then m = n and for each i ∈ [1, n] there exists j ∈ [1, n] such ∼ that Ni = Mj. Example 5.0.1. (a) Let M be the set of finite dimensional vector spaces over the field F. Then M is a Krull-Schmidt class

(b) Let M be a class of ﬁnite abelian groups. Every indecomposable in M is Zpn for integers p and n where p is prime.

The entire aim of this section is to show the link between Krull-Schmidt monoids and the free commutative monoid N(Ω). It follows that the set of all R-Modules forms a commutative monoid.

Lemma 5.0.1. (M, ⊕, 0) is a commutative monoid.

Proof. Associativity Let L, M, N ∈ M, then (L ⊕ M) ⊕ N = {(l, m, n): l ∈ L, m ∈ M, n ∈ N} = L ⊕ (M ⊕ N) since = ((l, m), n) = (l, (m, n)) = (l, m, n)

0 is the identity Let M ∈ M, and 0 is the zero module. Then M + 0 = {(m, 0) : m ∈ M} ∼= {(0, m): m ∈ M} = 0 + M ∼= M. Trivial isomorphism via φ :(m, 0) 7→ (0, m) 7→ m.

Commutativity Let M,N ∈ M, then M ⊕ N = {(m, n): m ∈ M, n ∈ N} ∼= N ⊕ M = {(n, m): n ∈ N, m ∈ M}. This isomorphism is trivial via the function φ :(m, n) 7→ (n, m). If we impose a congruence relation on M similar to section 3.3. we see that we can force the reduced property on the set of R-Modules, moreover we can construct an isomorphism to the free commutative monoid N(Ω).

16 Lemma 5.0.2. Let M be a set of R-Modules. The relation M ∼= N forms a congruence relation. Proof. The proof that the relation is an equivalence is a routine application of the definition of isomorphism. To see that it is a congruence let M1,M2,N1,N2 ∈ M and ∼ ∼ suppose that M1 = M2, and N1 = N2. Then there exists isomorphisms f : M1 → M2 and g : N1 → N2. Need to find isomorphism h : M1 ⊕ N1 → M2 ⊕ N2. This function h :(m1, n1) 7→ (f(m1), g(n1)) is the isomorphism because it is injective, surjective and preserves addition and scalar multiplication. Similarly to direct sums of modules, we can define a direct sum on a family of congruence classes of modules. Let [Mi] represent the congruence class of modules isomorphic to Mi. Then the direct sum is defined as [M1] ⊕ [M2] = [M1 ⊕ M2]. M will be taken to mean the set of isomorphism classes of the set of all R-Modules M. Theorem 5.0.2. The factor monoid (M, ⊕, [0]) forms a reduced commutative cancellation monoid. Proof. (M, ⊕, [0]) is a commutative monoid follows directly from the proof of lemma 5.0.1.

Cancellation Suppose [M] ⊕ [N] = [M] ⊕ [L], then [M ⊕ N] = [M ⊕ L] then M ⊕ N ∼= M ⊕ L. So there exists an isomorphism φ :(m, n) 7→ (m, l), if we limit this function to the second component of the 2-tuple, φrestricted : n 7→ l is still an isomorphism. Hence N ∼= L, so that [N] = [L].

Reduced Suppose [M] ⊕ [N] = [0], then [M ⊕ N] = [0]. So M ⊕ N ∼= 0, there exists an isomorphism φ :(m, n) 7→ 0. Since φ is an isomorphism it must be surjective, the 0 module has only one element, so M ⊕ N must have only one element. To satisfy module properties M ⊕ N must be closed under addition so its only element can be (0, 0). Hence M = N = 0 so that [M] = [N] = [0]. Lemma 5.0.3. Let [M] ∈ M. M is indecomposable if and only if [M] is an atom in the monoid (M, ⊕, [0]) [3].

Proof. (⇒) Suppose M is indecomposable. Now suppose [M] = [M1]⊕[M2] for some ∼ [M1], [M2] ∈ M. Then M = M1 ⊕ M2, since M is indecomposable M1 or M2 is the ∼ ∼ ∼ 0 module. So we have M = M1 ⊕ 0 or M = 0 ⊕ M2 which leaves us with M = M1 ∼ or M = M2. Thus [M] = [M1] or [M] = [M2], [M] is an atom.

∼ (⇐) Suppose [M] is an atom, and M = M1 ⊕ M2.

17 5.1 Expressing a Krull–Schmidt Class of Modules as the Free Commutative Monoid If we consider the set M of isomorphism classes of all R-Modules we see that the statement of deﬁnition 5.0.3 becomes stronger.

Deﬁnition 5.1.1. M is Krull-Schmidt if [N1] ⊕ .... ⊕ [Nn] = [M1] ⊕ .... ⊕ [Mm] then m = n and for each i ∈ [1, n] there exists j ∈ [1, n] such that [Ni] = [Mj]. Theorem 5.1.1. Let M be a Krull-Schmidt class of modules and let M be the set of congruence classes of M. Then under the operation of direct sum, M is a free commutative monoid.

Proof. Let Ω be the set of indecomposable elements of M. Define φ : M → N(Ω) by φ([M]) = f where for each [M] ∈ M define f[M] :Ω → N by f[M](P ) = the number of occurrences of P in the unique indecomposable decomposition of [M]. The proof that φ defines an isomorphism is based upon the proof of theorem 4.3.1.

φ is a homomorphism Let [M], [N] ∈ M, then φ([M]) + φ([N]) = f[M] + f[M], and φ([M] ⊕ [N]) = f[M]⊕[N]. We need to show that f[M] + f[N] = f[M]⊕[N]. Let [P ] ∈ Ω, then (f[M] + f[N])([P ]) = (f[M])(P ) + (f[N])(P ) this is simply the sum of the number of times [P ] occurs in the unique factorisation of [M] and [N], and f[M]⊕[N]([P ]) is the number of times [P ] occurs in the factorisation of [M]⊕[N], which is the sum of the number of occurrences of [P ] in [M] and [N]. Hence f[M]+f[N] = f[M]⊕[N] =⇒ φ([M])+φ([N]) = φ([M]⊕[N]).

φ is injective Suppose that φ([M]) = φ([N]). Then f[M] = f[N] =⇒ f[M]([P ]) = f[N]([P ]), ∀ [P ] ∈ Ω. Then the factorisation into indecomposable modules for [M] and [N] are the same, since each indecomposable module occurs the same number of times. This factorisation is unique since M is a Krull–Schmidt class, hence [M] = [N]

φ is surjective (Ω) Let f ∈ N , where f :[P ]in 7→ n where n, for [P ]in ∈ Ω and n ∈ N. Then choose each [P ]in to occur n times in the factorisation of some ideal K. Then φ([M]) = f[M] = f

18 6 Conclusion

In conclusion, we have seen that structures with known unique factorisation properties can be expressed as a free commutative monoid. I have successfully shown that an isomorphism can be constructed from a reduced factorial domain to the free commutative monoid. Similarly an isomorphism can be constructed from either a set of ideals of a Dedekind domain or set of modules from a Krull–Schmidt class to the free commutative monoid. Elucidation of the connections between these three sorts of decompositions may lead to progress in the mysterious area of the structure of modules over a commutative ring based on the well known properties of decompositions of elements and ideals of the ring.

7 Acknowledgements

I would like the thank Professor Phill Schultz for his guidance and supervision over the course of writing this project. I would also like to extend my thanks to the Australian Mathematical Sciences Institute for my scholarship and the opportunity to travel to the Big Day In 2015.

References

[1] T.W. Hungerford, Algebra, University of Washington, 1974

[2] N. Jacobson, Basic Algebra, Yale University, 1974

[3] N.R. Baeth, R. Wiegand, Factorisation Theory and Decomposition in Modules, The American Mathematical Monthly, Vol. 120, no. 1 (January 2013), pp.3-34

[4] P.M. Cohn, Unique Factorisation Domains, The American Mathematical Monthly, Vol. 80, No. 6, pp 117