Question

(a) Show that the "is associate to" relation is an equivalence relation. (b) Consider an equivalence class \(C\) induced by the "is associate to" relation. Show that if \(C\) contains an irreducible element, then all elements of \(C\) are irreducible. (c) Suppose that for every equivalence class \(C\) that contains irreducibles, we choose one element of \(C,\) and call it a distinguished irreducible. Show that \(D\) is a UFD if and only if every non-zero element of \(D\) can be expressed as \(u p_{1}^{e_{1}} \cdots p_{r}^{e_{r}},\) where \(u\) is a unit, \(p_{1}, \ldots, p_{r}\) are distinguished irreducibles, and this expression is unique up to a reordering of the \(p_{i}\) 's.

Short answer

Question: Prove that the "is associate to" relation in a domain \(D\) is an equivalence relation and that if an equivalence class contains an irreducible element, then all the elements of that class are irreducible. Also, prove that a domain \(D\) is a unique factorization domain (UFD) if and only if every non-zero element of \(D\) can be expressed as \(u p_{1}^{e_{1}} \cdots p_{r}^{e_{r}}\), where \(u\) is a unit, and the \(p_{i}\) 's are distinguished irreducibles.

Step-by-step solution

Prove the reflexivity of the relation

By definition, \(a\sim b\) if and only if there exists a unit \(u\) in \(D\) such that \(a=bu\). Reflexivity means that \(a\sim a\) for any \(a\). To prove this, choose the unit as \(1\), since \(a=a\cdot 1\) and \(1\) is a unit. Step 2: Symmetric

Prove the symmetry of the relation

Symmetry means that if \(a\sim b\), then \(b\sim a\). If \(a\sim b\), then there exists a unit \(u\) in \(D\) such that \(a=bu\). Now we need to show that \(b=a u'\). Since \(u\) is a unit, there exists an inverse of \(u\), called \(u^{-1}\), which is also a unit, and we have \(u\cdot u^{-1} =1\). Therefore, multiplying \(a=bu\) from the right by \(u^{-1}\), we get \(b=(a u^{-1})\). Now, choose \(u' = u^{-1}\) and \(b=a u'\), meaning that \(b\sim a\). So the relation is symmetric. Step 3: Transitive

Prove the transitivity of the relation

Transitivity means that if \(a\sim b\) and \(b\sim c\), then \(a\sim c\). If \(a\sim b\), there exists a unit \(u\) in \(D\) such that \(a=bu\). If \(b\sim c\), there exists a unit \(v\) in \(D\) such that \(b=cv\). Substituting the expression of \(b\) from \(a=bu\) gives \(a= cv u\). Thus, we can choose the unit as \(u' = u v\). Therefore, \(a\sim c\). So the relation is transitive. Since the relation is reflexive, symmetric, and transitive, the "is associate to" relation is an equivalence relation. (b)

Proving all elements of \(C\) are irreducible

Let \(C\) be an equivalence class that contains an irreducible element \(p\). Let \(q\) be another element in \(C\). Since \(p\) and \(q\) belong to the same equivalence class, \(p\) is associate to \(q\), meaning there exists a unit \(u\) in \(D\) such that \(p=qu\). Now, suppose that \(q\) is not irreducible. Then \(q=rs\), where \(r\) and \(s\) are elements in \(D\) that are not units. Multiplying both sides by \(u\), we get \(p=ru(gs^{-1})\), where \(gs^{-1}\) is also a unit since the product of a unit and its inverse is always a unit. Thus, \(p=ru(gs^{-1})\) contradicts the fact that \(p\) is irreducible. Hence, \(q\) must be irreducible too. This proves that all elements of \(C\) are irreducible. (c)

Proving \(D\) is a UFD if and only if the given condition is met

(\(\Rightarrow\)) Suppose \(D\) is a UFD. Then, every non-zero element of \(D\) can be expressed as a product of irreducible elements and a unit. Since these irreducible elements are unique up to associates, we can pick one distinguished irreducible from each equivalence class that contains irreducibles. This fulfills the required condition. (\(\Leftarrow\)) Suppose every non-zero element of \(D\) can be expressed uniquely as \(u p_{1}^{e_{1}} \cdots p_{r}^{e_{r}}\), where \(u\) is a unit, and \(p_{1}, \ldots, p_{r}\) are distinguished irreducibles. We need to show that \(D\) is a UFD. Indeed, every non-zero element in \(D\) can be factored into irreducible elements (since the distinguished irreducibles are irreducible), and this factorization is unique as per the given condition. Hence, \(D\) is a UFD.

Key concepts

Associates in Integral Domains

In the realm of number theory, integral domains serve as a pivotal concept, particularly when we discuss associates. An integral domain is essentially a type of ring where the product of any two non-zero elements is non-zero, eliminating the possibility of divisors of zero. Within this setting, two elements are considered 'associates' if one can be obtained from the other by multiplication with a unit. A unit is an element that has a multiplicative inverse within the same domain.

For instance, if our domain is the set of integers, then the only units are 1 and -1. Hence, in this case, two integers are associates if they are the same up to a sign. Understanding associates is key when we delve into concepts like prime elements and unique factorization, as they allow us to regard slightly different elements as 'essentially the same' under multiplication by units.

Irreducible Elements

Diving into the terrain of irreducible elements unveils a fundamental aspect of number theory. An irreducible element in an integral domain is an element that is no longer 'breakable' into a product of two non-unit elements. This concept is similar to that of prime numbers within the integers, where a prime number can't be divided evenly by any other number except for itself and 1.

In the context of the previous exercise, understanding irreducibility is crucial when examining the structure of an equivalence class. If the equivalence class of an associate relation contains an irreducible element, that trait permeates through the entire class. This property is pivotal in establishing the uniqueness required for a Unique Factorization Domain (UFD). Irreducible elements function as the building blocks for other elements in such domains, analogous to the way prime numbers construct integers through multiplication.

Unique Factorization Domain (UFD)

Hand in hand with primes and irreducibles comes the concept of a Unique Factorization Domain (UFD). A UFD is a special type of integral domain where each element (except for zero and units) can be broken down into a unique product of irreducible elements, up to reordering and multiplication by units. In simpler terms, within such a domain, an element can be factorized into 'primes' in only one way, mirroring the prime factorization of integers.

The exercise under scrutiny showcases the requirement for a domain to be classified as a UFD. Emphasizing the need for uniqueness in factorization brings to light the core principle of UFDs: while multiple factorizations might look different, they are fundamentally the same when we account for the associates and the order of factors. This feature of UFDs allows for a stable and predictable structure that is crucial in many areas of number theory and beyond.

Equivalence Class

The last piece of the puzzle is the equivalence class. In mathematics, an equivalence class aggregates elements that are considered equivalent based on a defined equivalence relation. In this case, the relation is 'is associate to'. For example, in an integral domain, all elements that are associates of each other form a single equivalence class. It's a way of partitioning a set into distinctly categorized groups where each member within a group is interchangeable under the equivalence relation.

Referring back to the exercise, an equivalence class containing an irreducible element is like a tight-knit family where every member shares a defining characteristic: irreducibility. This revelation further cements our understanding of factorizations within UFDs, as choosing a representative or 'distinguished irreducible' from each equivalence class allows for unique expressions of other elements. It's a harmonious symphony where every note, though unique, contributes to the grand melody of the domain's structure.