Question

Let \(R\) be a Euclidean domain, \(a, b, c \in R\), and \(\operatorname{gcd}(a, b)=1\). Prove the following: (i) \(a|b c \Longrightarrow a| c\), (ii) \(a \mid c\) and \(b|c \Longrightarrow a b| c\).

Short answer

(i) If \(a\mid bc\), then \(a\mid c\). (ii) If \(a\mid c\) and \(b\mid c\), then \(ab\mid c\).

Step-by-step solution

Understand the Given

We are given a Euclidean domain \(R\) and three elements \(a, b, c \in R\). Additionally, it is provided that \(\text{gcd}(a, b) = 1\), which means that \(a\) and \(b\) are coprime in the Euclidean domain.

Prove Part (i)

We need to show that if \(a\mid bc\), then \(a\mid c\). Since \(a\mid bc\), this means there exists some \(r \in R\) such that \(bc = ar\). Since \(\text{gcd}(a, b) = 1\), there are integers \(m, n\) in \(R\) such that \(am + bn = 1\). Multiply both sides by \(c\) to get \(amc + bnc = c\). Substitute \(bc = ar\) into the equation: \(amc + arnc = c\). Simplifying gives \(a(mc + rnc) = c\). Therefore, \(a\mid c\).

Prove Part (ii)

Assume \(a\mid c\) and \(b\mid c\). This implies there exist \(p, q \in R\) such that \(c = ap\) and \(c = bq\). Therefore, \(ap = bq\). Thus \(c\) can be expressed as a product of \(a\) and a term divisible by \(b\). Since \(\gcd(a, b) = 1\), you can apply the properties of coprimeness to find that for any integer \(q'\), \(bq' = c/a\) implies \(ab\mid c\). Therefore, combining these results shows \(ab\mid c\).

Conclude the Proof

From the previous steps, we've shown: (i) If \(a\mid bc\), then \(a\mid c\) due to factorization using the gcd property. (ii) If both \(a\mid c\) and \(b\mid c\), then \(ab\mid c\) follows from the fact that \(a\) and \(b\) together fully account for the factorization of \(c\) in \(R\).

Key concepts

Coprime Elements

In the context of a Euclidean domain, coprime elements play an important role. Two elements \(a\) and \(b\) in a Euclidean domain \(R\) are considered coprime if their greatest common divisor is 1, i.e., \(\text{gcd}(a, b) = 1\). This implies that there are no nontrivial common divisors of both \(a\) and \(b\) other than the units in \(R\). Therefore, if \(a\) and \(b\) are coprime, they reduce largely independent of each other's factorization over the ring.

Coprimeness is particularly useful when solving problems involving divisibility. One powerful result used in proofs involving coprime elements is the linear combination theorem, which states that if \(\text{gcd}(a, b) = 1\), then there exist \(m, n \in R\) such that \[ am + bn = 1.\]This equation indicates that the multiplication of \(a\) and \(b\) can yield other elements by linear combinations, allowing one to infer divisibility properties easily.
For example, if \(a\mid bc\), knowing that \(a\) and \(b\) are coprime helps to show \(a\mid c\). It is a powerful tool within the realm of number theory and an essential concept in solving congruence relations.

GCD (Greatest Common Divisor)

The greatest common divisor (GCD) is a measure of the largest element that divides two or more elements within a domain without leaving a remainder.

When dealing with a Euclidean domain, the process of finding the GCD is simplified through the Euclidean algorithm. This allows for a methodical approach to determine the GCD by repeated division.

Here’s a quick outline of the Euclidean algorithm:

• Identify two elements \(a\) and \(b\), such that \(a \geq b\).
• Divide \(a\) by \(b\) to find the remainder \(r\).
• Replace \(a\) with \(b\) and \(b\) with \(r\).
• Repeat the process until \(r = 0\), at which point the last nonzero remainder is the GCD.

Understanding the GCD not only helps in determining the coprimeness of two elements but also facilitates solving integer linear combinations. For example, if \(a\mid c\) and \(b\mid c\) in a setting where \(\text{gcd}(a, b) = 1\), elements \(a\) and \(b\) interact to confirm additional divisibility relationships.

Factorization Properties

Factorization in a Euclidean domain refers to breaking down elements into a product of other elements, termed as factors. This decomposition must respect the properties and divisors defined in the domain.

In Euclidean domains, every element can be expressed uniquely as a product of irreducible elements, a property that's akin to the fundamental theorem of arithmetic for integers. However, unlike integers, units may differ depending upon the specific domain in question.

Factorization becomes critical in problems involving coprime elements and divisibility analysis. Suppose you have elements \(a\), \(b\), and \(c\), and you know that \(\text{gcd}(a, b) = 1\). If \(a\mid bc\), understanding the factors involved helps to show how \(a\) could directly divide \(c\). Conversely, if \(a\mid c\) and \(b\mid c\), leveraging the factorization ensures that these divisors combine to guarantee \(ab\mid c\).
Being adept with factorization tells us not only about divisibility but also respects the multiplicative structure of the domain, crucial for solving complex algebraic problems.