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Chapter 3: Problem 16

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 \(b|c \Longrightarrow a b| c\). a+t b=1$.

Short Answer

Expert verified
For (i), given \(a|bc\) and \(\gcd(a, b)=1\), \(a|c\). For (ii), given \(a|c\) and \(b|c\), \(ab|c\).

Step by step solution

01

Understanding the Problem

We need to prove two statements involving divisibility in a Euclidean domain, where the greatest common divisor (gcd) of two elements is one.
02

Proof of Part (i)

We are given that \( a | bc \) and \( \operatorname{gcd}(a, b) = 1 \). According to B茅zout's identity, since \( a \) and \( b \) are coprime, there exist integers \( s \) and \( t \) such that \( as + bt = 1 \). Multiplying the entire equation by \( c \), we get \( acs + btc = c \). Since \( a | bc \), it follows that \( a \) divides \( btc \). Thus, \( a \) divides the entire left side \( acs + btc \), which means \( a | c \).
03

Assertion Verification for Part (i)

The result holds because \( acs \) is divisible by \( a \) trivially, and since \( a | bc \), \( a | btc \) as well. Therefore, \( a \) divides the whole expression and hence, **\( a | c \).**
04

Proof of Part (ii)

We need to show that if \( a \mid c \) and \( b \mid c \), then \( ab \mid c \). Since \( a \mid c \), we can say \( c = aq \) for some \( q \in R \). Also, since \( b \mid c \), \( c = br \) for some \( r \in R \). Therefore, \( aq = br \). The gcd condition is irrelevant here. To show \( ab \mid c \), since \( a \mid c \) and \( b \mid q \), using the identity \( aq = br \), we have \( q = bs \) for some \( s \), hence \( c = a(bs) = (ab)s \), proving \( ab \mid c \).
05

Summarizing Part (ii)

With \( aq = br \), and knowing \( a \mid c \) permits expressing \( q = bs \) which confirms \( c = (ab)s \). Thus, \( ab \) divides \( c \) since \( ab \mid c = (ab)s \), proving the second part. **\( ab | c \).**

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Key Concepts

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

Divisibility
In mathematics, divisibility is a fundamental concept that determines whether one integer is a factor of another integer. Specifically, in a Euclidean domain, when we say that an element \( a \) divides another element \( bc \), it means there exists some element \( k \) in the domain such that \( bc = ak \). This simple principle forms the backbone of many proofs and problem-solving approaches in number theory.
  • The notation we use for divisibility is \( a | bc \), meaning "\( a \) divides \( bc \)" without leaving a remainder.
  • If \( a | bc \) and \( \operatorname{gcd}(a, b) = 1 \), then \( a \) must divide \( c \). This is because \( a \) and \( b \) share no common factors other than 1.
  • The property \( a | bc \) allows us to conclude that any multiple of \( a \) is also divisible by \( a \).
This property becomes particularly powerful in a Euclidean domain where specific and structured division enables algorithmic solutions for finding factors and common divisors.
B茅zout's Identity
B茅zout's identity is a key concept in understanding relationships between coprime elements in algebra. For any two integers \( a \) and \( b \), B茅zout's identity asserts that if their greatest common divisor is 1, there exist integers \( s \) and \( t \) such that the linear combination \( as + bt = 1 \). This identity is instrumental in various proofs concerning divisibility and prime factors.
  • B茅zout's identity guarantees an expression for 1 as a linear combination of \( a \) and \( b \) when \( \operatorname{gcd}(a, b) = 1 \).
  • This linear combination property can be used to simplify many complex expressions, as seen in proving \( a | c \) when \( a | bc \).
  • Using B茅zout's identity, multiplying the equation by another element \( c \) provides the flexibility necessary to conclude that \( a \) divides \( c \) under the given conditions.
The identity helps illustrate why coprime elements can achieve certain properties when combined in particular ways, fundamentally linking divisibility and algebraic operations.
Greatest Common Divisor
The greatest common divisor (GCD) is the largest number that divides two numbers without leaving a remainder. In a Euclidean domain, the gcd is particularly important when assessing divisibility and when using properties like B茅zout's identity.
  • The gcd of two elements \( a \) and \( b \), \( \operatorname{gcd}(a, b) \), can be computed using the Euclidean algorithm, which exploits the division algorithm of the domain.
  • When \( \operatorname{gcd}(a, b) = 1 \), \( a \) and \( b \) are considered coprime, meaning they have no common divisors except for the number 1.
  • Knowing the gcd is essential in proving divisibility results such as \( ab | c \) if \( a \mid c \) and \( b \mid c \), as it emphasizes that the combination of \( a \) and \( b \) encapsulates all necessary divisors of \( c \).
Understanding the GCD helps to create more intuitive number relationships, offering deeper insight into how numbers can be broken down into their prime factors and how they relate to one another in terms of divisibility.

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Most popular questions from this chapter

Show that \(\operatorname{gcd}(a, b)=1\) and \(\operatorname{gcd}(a, c)=1\) imply that \(\operatorname{gcd}(a, b c)=1\).

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 .

Use the Extended Euclidean Algorithm to find \(\operatorname{ged}(f, g)\), for \(f, g \in \mathbb{Z}_{p}[x]\) in each of the following examples (arithmetic in \(\mathrm{Z}_{p}=\\{0, \ldots, p-1\\}\) is done modulo \(p\) ). In each case compute the corresponding polynomials \(s\) and \(t\) such that \(\operatorname{gcd}(f, g)=s f+t \mathrm{~g}\). (i) \(f=x^{3}+x+1, g=x^{2}+x+1\) for \(p=2\) and \(p=3\). (ii) \(f=x^{4}+x^{3}+x+1, g=x^{3}+x^{2}+x+1\) for \(p=2\) and \(p=3\). (iii) \(f=x^{5}+x^{4}+x^{3}+x+1, g=x^{4}+x^{3}+x^{2}+x+1\) for \(p=5\). (iv) \(f=x^{5}+x^{4}+x^{3}-x^{2}-x+1, g=x^{3}+x^{2}+x+1\) for \(p=3\) and \(p=5\).

(i) Show that for polynomials \(f, g \in F[x]\) of degrees \(n \geq m\), where \(F\) is a field, computing all entries \(s_{i}\) in the traditional Extended Euclidean Algorithm from the quotients \(q_{1}\) takes at most \(2 m^{2}+2 m\) additions and multiplications in \(F\). Hint: Exhibit the bound for the normal case and prove that this is the worst case. (ii) Prove that the corresponding estimate for the Extended Euclidean Algorithm is \(2 m^{2}+3 m+1\).

Let \(R\) be a Euclidean domain, with a Euclidean function \(d: R \rightarrow N \cup\\{-\infty\\}\) that has the additional properties o \(d(a b)=d(a)+d(b) .\) o \(d(a+b) \leq \max \\{d(a), d(b)\\}\), with equality if \(d(a) \neq d(b)\). o \(d\) is surjective. for all \(a, b \in R\). Prove that \(R\) is a polynomial ring with \(d\) as degree function. Proceed as follows: (i) Prove that \(d(a)=-\infty\) if and only if \(a=0\). (ii) Show that \(F=\\{a \in R: d(a) \leq 0\\}\) is a subfield of \(R\). (iii) Let \(x \in R\) be such that \(d(x)=1\), and prove that every nonzero \(a \in R\) has a unique representation $$ a=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0} $$ where \(n=d(a), a_{0} \ldots, a_{n} \in F_{\text {, and }} a_{n} \neq 0\).
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