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Chapter 3: Problem 2

Let \(R\) be an integral domain. Show that $$ a \sim b \Longleftrightarrow(a \mid b \text { and } b \mid a) \Longleftrightarrow\langle a\rangle=\langle b\rangle, $$ where \(\langle a\rangle=R a=\\{r a: r \in R\\}\) is the ideal generated by \(a\).

Short Answer

Expert verified
The condition \(a \mid b\) and \(b \mid a\) is equivalent to \(\langle a \rangle = \langle b \rangle\) in an integral domain.

Step by step solution

01

Understand the Notation

In this problem, we are asked to show an equivalence of three conditions about elements \(a\) and \(b\) in an integral domain \(R\). The notation \(a \sim b\) means \(a\) and \(b\) are associates, or each divides the other. Additionally, \(\langle a \rangle = \langle b \rangle\) signifies that the ideals generated by \(a\) and \(b\) are identical.
02

Show that Divisibility Implies Equal Ideals

Assume \(a \mid b\) and \(b \mid a\). This means there exist elements \(r, s \in R\) such that \(b = ra\) and \(a = sb\). Consider the ideal \(\langle a \rangle\), which contains elements of the form \(ra\). Since \(b = ra\), \(b \in \langle a \rangle\). Similarly, because \(a = sb\), we have \(a \in \langle b \rangle\). Hence, every element in \(\langle a \rangle\) is in \(\langle b \rangle\) and vice versa, so \(\langle a \rangle = \langle b \rangle\).
03

Show that Equal Ideals Imply Divisibility

Now assume \(\langle a \rangle = \langle b \rangle\). This implies \(a = rb\) for some \(r \in R\) and \(b = sa\) for some \(s \in R\), i.e., \(a \mid b\) and \(b \mid a\). Thus, if the ideals are the same, \(a\) and \(b\) are associates, satisfying \(a \mid b\) and \(b \mid a\).
04

Conclude the Proof

Since we have shown both directions of implication, \(a \sim b \Longleftrightarrow \langle a \rangle = \langle b \rangle\). Therefore, the statement is proved.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Associates
In the context of an integral domain, two elements are called **associates** if they can be expressed as multiples of each other by a unit of the ring. A unit is an element that has a multiplicative inverse.It's like saying two elements are the same up to multiplication by a non-zero divisor, which in the case of numbers, could be factors like 1 or -1. So, for elements in an integral domain:
  • If you can find a unit, say \(u\), such that \(a = ub\), then \(a\) and \(b\) are associates.
  • This is similar to saying they're essentially the 'same' in how they contribute to the structure of the ring.
Thus, in the mathematical notation \(a \sim b\), it informs you that "a" and "b" are associated in this way within the integral domain.
Divisibility
Divisibility in integral domains can be akin to our common understanding of dividing numbers. An element \(a\) divides another element \(b\) if there is some element \(r\) in the domain such that \(b = ra\).It means that \(b\) is a scaled or "stretched" version of \(a\). Importantly:
  • If \(a \mid b\), then there is no remainder when \(b\) is 'divided' by \(a\).
  • The reverse, \(b \mid a\), also holds in this associative virtue, which results in them being multiples of each other, just scaled differently.
This concept becomes critical in determining when two elements are associates, as it rephrases to each element being a divisor of the other.
Ideal
The term "ideal" in ring theory has a distinct interpretation. In a ring like an integral domain, an ideal is a set that absorbs multiplication by elements of the ring. When you hear of the ideal \(\langle a \rangle\), think of it as a collection of all possible products of \(a\) with every element in the ring.Let's break it down:
  • The ideal \(\langle a \rangle\) contains all elements of the form \(ra\) where \(r\) is any element from the ring \(R\).
  • Now, when two ideals \(\langle a \rangle\) and \(\langle b \rangle\) are equal, every element produced by multiplying \(a\) with elements of \(R\) can also be produced by multiplying \(b\) with elements of \(R\), and vice versa.
This equivalence forms a basis to conclude similarities or associations between \(a\) and \(b\), as the entire "reach" of these elements within the ring is identical.
Equivalence
In mathematical terms, equivalence often means there is a bidirectional relationship or mutual association between concepts. For this exercise, when associating elements in an integral domain, we use conditions of equivalence.Equivalence Structures:
  • Starting with the equivalence relation \(a \sim b\), if \(a\) divides \(b\) and \(b\) divides \(a\), then we're looking at them as interchangeable through associates.
  • Also, equivalence is seen in ideal terms: when \(\langle a \rangle = \langle b \rangle\), demonstrating how their generated 'power' in the ring is indistinguishable.
Understanding these equivalences proves that the relationships involved preserve the structure and properties of the integral domain, ensuring that either view, whether through divisibility or ideal equality, leads to the same conclusion.

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Most popular questions from this chapter

(i) Show that the norm \(N: \mathbb{Z}[i] \longrightarrow \mathbb{N}\) with \(N(\alpha)=\alpha \bar{\alpha}\) on the ring of Gaussian integers \(\mathbb{Z}[i]\) is a Euclidean function. Hint: Consider the exact quotient of two Gaussian integers \(\alpha, \beta \in \mathbb{Z}[i]\) in \(\mathbb{C}\). (ii) Show that the units in \(\mathbb{Z}[i]\) are precisely the elements of norm 1 and enumerate them. (iii) Prove that there is no multiplicative normal form on \(\mathbb{Z}[i]\) which extends the usual normal form \(\operatorname{normal}(a)=|a|\) on \(\mathbb{Z}\). Hint: Consider normal \(\left((1+i)^{2}\right)\). Why is normal \((a+i b)=|a|+i|b|\) for \(a, b \in \mathbb{Z}\) not a normal form? (iv) Compute all greatest common divisors of 6 and \(3+i\) in \(\mathbb{Z}[i]\) and their representations as a linear combination of 6 and \(3+i\). (v) Compute a gcd of 12277 and \(399+20 i\).

Are there \(s, t \in \mathbb{Z}\) such that \(24 s+14 t=1\) ?

For each of the following pairs of integers, find their greatest common divisor using the Euclidean Algorithm: (i) 34,21 ; (ii) 136,51 ; (iii) 481,325 ; (iv) \(8771,3206 .\)

Prove that two odd integers whose difference is 32 are coprime.

Prove that \(\mathbb{Z}[x]\) is not a Euclidean domain. Hint: If it were, then we could compute \(s, t \in \mathbb{Z}[x]\) such that \(s \cdot 2+t \cdot x=\operatorname{gcd}(2, x)\), using the Extended Euclidean Algorithm.
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