Section 46 Euclidean domains Instructor: Yifan Yang Spring 2007 - - PowerPoint PPT Presentation

section 46 euclidean domains
SMART_READER_LITE
LIVE PREVIEW

Section 46 Euclidean domains Instructor: Yifan Yang Spring 2007 - - PowerPoint PPT Presentation

Section 46 Euclidean domains Instructor: Yifan Yang Spring 2007 Instructor: Yifan Yang Section 46 Euclidean domains Division algorithms for Z and F [ x ] Theorem Let a and b be two integers with b = 0 . Then there exist two


slide-1
SLIDE 1

Section 46 – Euclidean domains

Instructor: Yifan Yang Spring 2007

Instructor: Yifan Yang Section 46 – Euclidean domains

slide-2
SLIDE 2

Division algorithms for Z and F[x]

Theorem Let a and b be two integers with b = 0. Then there exist two integers q and r such that

  • a = bq + r, and
  • either r = 0 or |r| < |b|.

Theorem (23.1) Let F be a field. Then for any a(x), b(x) ∈ F[x] with b(x) = 0, there exist two polynomials q(x) and r(x) such that

  • a(x) = b(x)q(x) + r(x), and
  • either r(x) = 0 or deg r(x) < deg b(x).

Instructor: Yifan Yang Section 46 – Euclidean domains

slide-3
SLIDE 3

Division algorithms for Z and F[x]

Theorem Let a and b be two integers with b = 0. Then there exist two integers q and r such that

  • a = bq + r, and
  • either r = 0 or |r| < |b|.

Theorem (23.1) Let F be a field. Then for any a(x), b(x) ∈ F[x] with b(x) = 0, there exist two polynomials q(x) and r(x) such that

  • a(x) = b(x)q(x) + r(x), and
  • either r(x) = 0 or deg r(x) < deg b(x).

Instructor: Yifan Yang Section 46 – Euclidean domains

slide-4
SLIDE 4

Division algorithms for Z and F[x]

Theorem Let a and b be two integers with b = 0. Then there exist two integers q and r such that

  • a = bq + r, and
  • either r = 0 or |r| < |b|.

Theorem (23.1) Let F be a field. Then for any a(x), b(x) ∈ F[x] with b(x) = 0, there exist two polynomials q(x) and r(x) such that

  • a(x) = b(x)q(x) + r(x), and
  • either r(x) = 0 or deg r(x) < deg b(x).

Instructor: Yifan Yang Section 46 – Euclidean domains

slide-5
SLIDE 5

Division algorithms for Z and F[x]

Theorem Let a and b be two integers with b = 0. Then there exist two integers q and r such that

  • a = bq + r, and
  • either r = 0 or |r| < |b|.

Theorem (23.1) Let F be a field. Then for any a(x), b(x) ∈ F[x] with b(x) = 0, there exist two polynomials q(x) and r(x) such that

  • a(x) = b(x)q(x) + r(x), and
  • either r(x) = 0 or deg r(x) < deg b(x).

Instructor: Yifan Yang Section 46 – Euclidean domains

slide-6
SLIDE 6

Euclidean algorithm

Definition A Euclidean norm on an integral domain D is a function ν : D − {0} → N ∪ {0} such that the following conditions are satisfied:

  • For all a, b ∈ D with b = 0, there exist q and r in D such

that

  • a = bq + r, and
  • either r = 0 or ν(r) < ν(b).
  • For all a, b ∈ D, where neither a nor b is 0, ν(a) ≤ ν(ab).

An integral domain D is a Euclidean domain if there exists a Euclidean norm on D.

Instructor: Yifan Yang Section 46 – Euclidean domains

slide-7
SLIDE 7

Euclidean algorithm

Definition A Euclidean norm on an integral domain D is a function ν : D − {0} → N ∪ {0} such that the following conditions are satisfied:

  • For all a, b ∈ D with b = 0, there exist q and r in D such

that

  • a = bq + r, and
  • either r = 0 or ν(r) < ν(b).
  • For all a, b ∈ D, where neither a nor b is 0, ν(a) ≤ ν(ab).

An integral domain D is a Euclidean domain if there exists a Euclidean norm on D.

Instructor: Yifan Yang Section 46 – Euclidean domains

slide-8
SLIDE 8

Euclidean algorithm

Definition A Euclidean norm on an integral domain D is a function ν : D − {0} → N ∪ {0} such that the following conditions are satisfied:

  • For all a, b ∈ D with b = 0, there exist q and r in D such

that

  • a = bq + r, and
  • either r = 0 or ν(r) < ν(b).
  • For all a, b ∈ D, where neither a nor b is 0, ν(a) ≤ ν(ab).

An integral domain D is a Euclidean domain if there exists a Euclidean norm on D.

Instructor: Yifan Yang Section 46 – Euclidean domains

slide-9
SLIDE 9

Euclidean algorithm

Definition A Euclidean norm on an integral domain D is a function ν : D − {0} → N ∪ {0} such that the following conditions are satisfied:

  • For all a, b ∈ D with b = 0, there exist q and r in D such

that

  • a = bq + r, and
  • either r = 0 or ν(r) < ν(b).
  • For all a, b ∈ D, where neither a nor b is 0, ν(a) ≤ ν(ab).

An integral domain D is a Euclidean domain if there exists a Euclidean norm on D.

Instructor: Yifan Yang Section 46 – Euclidean domains

slide-10
SLIDE 10

Euclidean algorithm

Definition A Euclidean norm on an integral domain D is a function ν : D − {0} → N ∪ {0} such that the following conditions are satisfied:

  • For all a, b ∈ D with b = 0, there exist q and r in D such

that

  • a = bq + r, and
  • either r = 0 or ν(r) < ν(b).
  • For all a, b ∈ D, where neither a nor b is 0, ν(a) ≤ ν(ab).

An integral domain D is a Euclidean domain if there exists a Euclidean norm on D.

Instructor: Yifan Yang Section 46 – Euclidean domains

slide-11
SLIDE 11

Examples

  • Define ν : Z → N ∪ {0} by ν(n) = |n|. Then ν is a

Euclidean norm.

  • Define ν : F[x] − {0} → N ∪ {0} by ν(f(x)) = deg f(x).

Then ν is a Euclidean norm.

Instructor: Yifan Yang Section 46 – Euclidean domains

slide-12
SLIDE 12

Examples

  • Define ν : Z → N ∪ {0} by ν(n) = |n|. Then ν is a

Euclidean norm.

  • Define ν : F[x] − {0} → N ∪ {0} by ν(f(x)) = deg f(x).

Then ν is a Euclidean norm.

Instructor: Yifan Yang Section 46 – Euclidean domains

slide-13
SLIDE 13

ED ⇒ PID

Theorem (46.4) Every Euclidean domain D is a principal ideal domain. Corollary (46.5) Every Euclidean domain is a unique factorization domain.

Proof.

  • Given an ideal I, if I = {0}, then I = 0 and we are done.
  • If I = {0}, let b be an element of I of minimal norm. (That

is, ν(b) = mina=0∈I ν(a).) We claim that I = b.

  • Given a ∈ I, there exist q and r in D such that a = bq + r

with r = 0 or ν(r) < ν(b).

  • The possibility ν(r) < ν(b) can not occur. Thus

a = bq ∈ b.

  • Instructor: Yifan Yang

Section 46 – Euclidean domains

slide-14
SLIDE 14

ED ⇒ PID

Theorem (46.4) Every Euclidean domain D is a principal ideal domain. Corollary (46.5) Every Euclidean domain is a unique factorization domain.

Proof.

  • Given an ideal I, if I = {0}, then I = 0 and we are done.
  • If I = {0}, let b be an element of I of minimal norm. (That

is, ν(b) = mina=0∈I ν(a).) We claim that I = b.

  • Given a ∈ I, there exist q and r in D such that a = bq + r

with r = 0 or ν(r) < ν(b).

  • The possibility ν(r) < ν(b) can not occur. Thus

a = bq ∈ b.

  • Instructor: Yifan Yang

Section 46 – Euclidean domains

slide-15
SLIDE 15

ED ⇒ PID

Theorem (46.4) Every Euclidean domain D is a principal ideal domain. Corollary (46.5) Every Euclidean domain is a unique factorization domain.

Proof.

  • Given an ideal I, if I = {0}, then I = 0 and we are done.
  • If I = {0}, let b be an element of I of minimal norm. (That

is, ν(b) = mina=0∈I ν(a).) We claim that I = b.

  • Given a ∈ I, there exist q and r in D such that a = bq + r

with r = 0 or ν(r) < ν(b).

  • The possibility ν(r) < ν(b) can not occur. Thus

a = bq ∈ b.

  • Instructor: Yifan Yang

Section 46 – Euclidean domains

slide-16
SLIDE 16

ED ⇒ PID

Theorem (46.4) Every Euclidean domain D is a principal ideal domain. Corollary (46.5) Every Euclidean domain is a unique factorization domain.

Proof.

  • Given an ideal I, if I = {0}, then I = 0 and we are done.
  • If I = {0}, let b be an element of I of minimal norm. (That

is, ν(b) = mina=0∈I ν(a).) We claim that I = b.

  • Given a ∈ I, there exist q and r in D such that a = bq + r

with r = 0 or ν(r) < ν(b).

  • The possibility ν(r) < ν(b) can not occur. Thus

a = bq ∈ b.

  • Instructor: Yifan Yang

Section 46 – Euclidean domains

slide-17
SLIDE 17

ED ⇒ PID

Theorem (46.4) Every Euclidean domain D is a principal ideal domain. Corollary (46.5) Every Euclidean domain is a unique factorization domain.

Proof.

  • Given an ideal I, if I = {0}, then I = 0 and we are done.
  • If I = {0}, let b be an element of I of minimal norm. (That

is, ν(b) = mina=0∈I ν(a).) We claim that I = b.

  • Given a ∈ I, there exist q and r in D such that a = bq + r

with r = 0 or ν(r) < ν(b).

  • The possibility ν(r) < ν(b) can not occur. Thus

a = bq ∈ b.

  • Instructor: Yifan Yang

Section 46 – Euclidean domains

slide-18
SLIDE 18

ED ⇒ PID

Theorem (46.4) Every Euclidean domain D is a principal ideal domain. Corollary (46.5) Every Euclidean domain is a unique factorization domain.

Proof.

  • Given an ideal I, if I = {0}, then I = 0 and we are done.
  • If I = {0}, let b be an element of I of minimal norm. (That

is, ν(b) = mina=0∈I ν(a).) We claim that I = b.

  • Given a ∈ I, there exist q and r in D such that a = bq + r

with r = 0 or ν(r) < ν(b).

  • The possibility ν(r) < ν(b) can not occur. Thus

a = bq ∈ b.

  • Instructor: Yifan Yang

Section 46 – Euclidean domains

slide-19
SLIDE 19

Characterization of units

Theorem (46.6) Let D be a Euclidean domain with Euclidean norm ν. Then

  • ν(1) is minimal among all ν(a) for nonzero a ∈ D.
  • u ∈ D is a unit if and only if ν(u) = ν(1).

Proof.

  • By the second condition of a Euclidean domain,

ν(1) ≤ ν(1a) = ν(a).

  • Let u be a unit. Then ν(u) ≤ ν(uu−1) = ν(1). But ν(1) is
  • minimal. This shows that ν(u) = ν(1).
  • Conversely, assume that u is an element such that

ν(u) = ν(1). We have 1 = uq + r for some q, r ∈ D with r = 0 or ν(r) < ν(1). By the minimality of ν(1), we must have r = 0, i.e., 1 = uq. This shows that u has a multiplicative inverse.

  • Instructor: Yifan Yang

Section 46 – Euclidean domains

slide-20
SLIDE 20

Characterization of units

Theorem (46.6) Let D be a Euclidean domain with Euclidean norm ν. Then

  • ν(1) is minimal among all ν(a) for nonzero a ∈ D.
  • u ∈ D is a unit if and only if ν(u) = ν(1).

Proof.

  • By the second condition of a Euclidean domain,

ν(1) ≤ ν(1a) = ν(a).

  • Let u be a unit. Then ν(u) ≤ ν(uu−1) = ν(1). But ν(1) is
  • minimal. This shows that ν(u) = ν(1).
  • Conversely, assume that u is an element such that

ν(u) = ν(1). We have 1 = uq + r for some q, r ∈ D with r = 0 or ν(r) < ν(1). By the minimality of ν(1), we must have r = 0, i.e., 1 = uq. This shows that u has a multiplicative inverse.

  • Instructor: Yifan Yang

Section 46 – Euclidean domains

slide-21
SLIDE 21

Characterization of units

Theorem (46.6) Let D be a Euclidean domain with Euclidean norm ν. Then

  • ν(1) is minimal among all ν(a) for nonzero a ∈ D.
  • u ∈ D is a unit if and only if ν(u) = ν(1).

Proof.

  • By the second condition of a Euclidean domain,

ν(1) ≤ ν(1a) = ν(a).

  • Let u be a unit. Then ν(u) ≤ ν(uu−1) = ν(1). But ν(1) is
  • minimal. This shows that ν(u) = ν(1).
  • Conversely, assume that u is an element such that

ν(u) = ν(1). We have 1 = uq + r for some q, r ∈ D with r = 0 or ν(r) < ν(1). By the minimality of ν(1), we must have r = 0, i.e., 1 = uq. This shows that u has a multiplicative inverse.

  • Instructor: Yifan Yang

Section 46 – Euclidean domains

slide-22
SLIDE 22

Characterization of units

Theorem (46.6) Let D be a Euclidean domain with Euclidean norm ν. Then

  • ν(1) is minimal among all ν(a) for nonzero a ∈ D.
  • u ∈ D is a unit if and only if ν(u) = ν(1).

Proof.

  • By the second condition of a Euclidean domain,

ν(1) ≤ ν(1a) = ν(a).

  • Let u be a unit. Then ν(u) ≤ ν(uu−1) = ν(1). But ν(1) is
  • minimal. This shows that ν(u) = ν(1).
  • Conversely, assume that u is an element such that

ν(u) = ν(1). We have 1 = uq + r for some q, r ∈ D with r = 0 or ν(r) < ν(1). By the minimality of ν(1), we must have r = 0, i.e., 1 = uq. This shows that u has a multiplicative inverse.

  • Instructor: Yifan Yang

Section 46 – Euclidean domains

slide-23
SLIDE 23

Characterization of units

Theorem (46.6) Let D be a Euclidean domain with Euclidean norm ν. Then

  • ν(1) is minimal among all ν(a) for nonzero a ∈ D.
  • u ∈ D is a unit if and only if ν(u) = ν(1).

Proof.

  • By the second condition of a Euclidean domain,

ν(1) ≤ ν(1a) = ν(a).

  • Let u be a unit. Then ν(u) ≤ ν(uu−1) = ν(1). But ν(1) is
  • minimal. This shows that ν(u) = ν(1).
  • Conversely, assume that u is an element such that

ν(u) = ν(1). We have 1 = uq + r for some q, r ∈ D with r = 0 or ν(r) < ν(1). By the minimality of ν(1), we must have r = 0, i.e., 1 = uq. This shows that u has a multiplicative inverse.

  • Instructor: Yifan Yang

Section 46 – Euclidean domains

slide-24
SLIDE 24

Characterization of units

Theorem (46.6) Let D be a Euclidean domain with Euclidean norm ν. Then

  • ν(1) is minimal among all ν(a) for nonzero a ∈ D.
  • u ∈ D is a unit if and only if ν(u) = ν(1).

Proof.

  • By the second condition of a Euclidean domain,

ν(1) ≤ ν(1a) = ν(a).

  • Let u be a unit. Then ν(u) ≤ ν(uu−1) = ν(1). But ν(1) is
  • minimal. This shows that ν(u) = ν(1).
  • Conversely, assume that u is an element such that

ν(u) = ν(1). We have 1 = uq + r for some q, r ∈ D with r = 0 or ν(r) < ν(1). By the minimality of ν(1), we must have r = 0, i.e., 1 = uq. This shows that u has a multiplicative inverse.

  • Instructor: Yifan Yang

Section 46 – Euclidean domains

slide-25
SLIDE 25

Characterization of units

Theorem (46.6) Let D be a Euclidean domain with Euclidean norm ν. Then

  • ν(1) is minimal among all ν(a) for nonzero a ∈ D.
  • u ∈ D is a unit if and only if ν(u) = ν(1).

Proof.

  • By the second condition of a Euclidean domain,

ν(1) ≤ ν(1a) = ν(a).

  • Let u be a unit. Then ν(u) ≤ ν(uu−1) = ν(1). But ν(1) is
  • minimal. This shows that ν(u) = ν(1).
  • Conversely, assume that u is an element such that

ν(u) = ν(1). We have 1 = uq + r for some q, r ∈ D with r = 0 or ν(r) < ν(1). By the minimality of ν(1), we must have r = 0, i.e., 1 = uq. This shows that u has a multiplicative inverse.

  • Instructor: Yifan Yang

Section 46 – Euclidean domains

slide-26
SLIDE 26

Characterization of units

Theorem (46.6) Let D be a Euclidean domain with Euclidean norm ν. Then

  • ν(1) is minimal among all ν(a) for nonzero a ∈ D.
  • u ∈ D is a unit if and only if ν(u) = ν(1).

Proof.

  • By the second condition of a Euclidean domain,

ν(1) ≤ ν(1a) = ν(a).

  • Let u be a unit. Then ν(u) ≤ ν(uu−1) = ν(1). But ν(1) is
  • minimal. This shows that ν(u) = ν(1).
  • Conversely, assume that u is an element such that

ν(u) = ν(1). We have 1 = uq + r for some q, r ∈ D with r = 0 or ν(r) < ν(1). By the minimality of ν(1), we must have r = 0, i.e., 1 = uq. This shows that u has a multiplicative inverse.

  • Instructor: Yifan Yang

Section 46 – Euclidean domains

slide-27
SLIDE 27

Characterization of units

Theorem (46.6) Let D be a Euclidean domain with Euclidean norm ν. Then

  • ν(1) is minimal among all ν(a) for nonzero a ∈ D.
  • u ∈ D is a unit if and only if ν(u) = ν(1).

Proof.

  • By the second condition of a Euclidean domain,

ν(1) ≤ ν(1a) = ν(a).

  • Let u be a unit. Then ν(u) ≤ ν(uu−1) = ν(1). But ν(1) is
  • minimal. This shows that ν(u) = ν(1).
  • Conversely, assume that u is an element such that

ν(u) = ν(1). We have 1 = uq + r for some q, r ∈ D with r = 0 or ν(r) < ν(1). By the minimality of ν(1), we must have r = 0, i.e., 1 = uq. This shows that u has a multiplicative inverse.

  • Instructor: Yifan Yang

Section 46 – Euclidean domains

slide-28
SLIDE 28

Characterization of units

Theorem (46.6) Let D be a Euclidean domain with Euclidean norm ν. Then

  • ν(1) is minimal among all ν(a) for nonzero a ∈ D.
  • u ∈ D is a unit if and only if ν(u) = ν(1).

Proof.

  • By the second condition of a Euclidean domain,

ν(1) ≤ ν(1a) = ν(a).

  • Let u be a unit. Then ν(u) ≤ ν(uu−1) = ν(1). But ν(1) is
  • minimal. This shows that ν(u) = ν(1).
  • Conversely, assume that u is an element such that

ν(u) = ν(1). We have 1 = uq + r for some q, r ∈ D with r = 0 or ν(r) < ν(1). By the minimality of ν(1), we must have r = 0, i.e., 1 = uq. This shows that u has a multiplicative inverse.

  • Instructor: Yifan Yang

Section 46 – Euclidean domains

slide-29
SLIDE 29

Greatest common divisor

Definition Let D be an integral domain, and a, b ∈ D be two elements. A common divisor d of a and b is a greatest common divisor of a and b if every common divisor of a and b divides d. Remark GCD’s may not exist in general. For example, consider a = 4 and b = 2(1 + √ −3) in Z[ √ −3] we have 4 = 2 · 2 = (1 + √ −3)(1 − √ −3). Thus, 1 + √ −3 and 2 both divide a and b. However, there is no divisor of a and b that is divisible by both 2 and 1 + √ −3.

Instructor: Yifan Yang Section 46 – Euclidean domains

slide-30
SLIDE 30

Greatest common divisor

Definition Let D be an integral domain, and a, b ∈ D be two elements. A common divisor d of a and b is a greatest common divisor of a and b if every common divisor of a and b divides d. Remark GCD’s may not exist in general. For example, consider a = 4 and b = 2(1 + √ −3) in Z[ √ −3] we have 4 = 2 · 2 = (1 + √ −3)(1 − √ −3). Thus, 1 + √ −3 and 2 both divide a and b. However, there is no divisor of a and b that is divisible by both 2 and 1 + √ −3.

Instructor: Yifan Yang Section 46 – Euclidean domains

slide-31
SLIDE 31

Euclidean algorithm

Theorem (46.9) Let D be a Euclidean domain, and a and b be two elements of D.

  • Then a GCD of a and b can be obtained by the Euclidean

algorithm.

  • a + b = gcd(a, b).

Remark In a Euclidean domain, d is a GCD of a and b if and only if d has the largest Euclidean norm among all common divisors of a and b.

Instructor: Yifan Yang Section 46 – Euclidean domains

slide-32
SLIDE 32

Euclidean algorithm

Theorem (46.9) Let D be a Euclidean domain, and a and b be two elements of D.

  • Then a GCD of a and b can be obtained by the Euclidean

algorithm.

  • a + b = gcd(a, b).

Remark In a Euclidean domain, d is a GCD of a and b if and only if d has the largest Euclidean norm among all common divisors of a and b.

Instructor: Yifan Yang Section 46 – Euclidean domains

slide-33
SLIDE 33

Euclidean algorithm

Theorem (46.9) Let D be a Euclidean domain, and a and b be two elements of D.

  • Then a GCD of a and b can be obtained by the Euclidean

algorithm.

  • a + b = gcd(a, b).

Remark In a Euclidean domain, d is a GCD of a and b if and only if d has the largest Euclidean norm among all common divisors of a and b.

Instructor: Yifan Yang Section 46 – Euclidean domains

slide-34
SLIDE 34

Euclidean algorithm

Theorem (46.9) Let D be a Euclidean domain, and a and b be two elements of D.

  • Then a GCD of a and b can be obtained by the Euclidean

algorithm.

  • a + b = gcd(a, b).

Remark In a Euclidean domain, d is a GCD of a and b if and only if d has the largest Euclidean norm among all common divisors of a and b.

Instructor: Yifan Yang Section 46 – Euclidean domains

slide-35
SLIDE 35

Proof of Theorem 46.9

  • Assume b = 0. Consider the Euclidean algorithm

a = bq1 + r1 b = r1q2 + r2 . . . . . . rn−1 = rnqn+1 + 0. (The process must stop because ν(r1), ν(r2), · · · is a series

  • f strictly decreasing non-negative integers.) We claim that

rn is a GCD of a and b.

Instructor: Yifan Yang Section 46 – Euclidean domains

slide-36
SLIDE 36

Proof of Theorem 46.9

  • Assume b = 0. Consider the Euclidean algorithm

a = bq1 + r1 b = r1q2 + r2 . . . . . . rn−1 = rnqn+1 + 0. (The process must stop because ν(r1), ν(r2), · · · is a series

  • f strictly decreasing non-negative integers.) We claim that

rn is a GCD of a and b.

Instructor: Yifan Yang Section 46 – Euclidean domains

slide-37
SLIDE 37

Proof of Theorem 46.9

  • Assume b = 0. Consider the Euclidean algorithm

a = bq1 + r1 b = r1q2 + r2 . . . . . . rn−1 = rnqn+1 + 0. (The process must stop because ν(r1), ν(r2), · · · is a series

  • f strictly decreasing non-negative integers.) We claim that

rn is a GCD of a and b.

Instructor: Yifan Yang Section 46 – Euclidean domains

slide-38
SLIDE 38

Proof of Theorem 46.9, continued

  • It is straightforward to show that c is a common divisor of

rk−1 and rk if and only it is a common divisor of rk and rk+1. Thus, rn is a common divisor of a and b. Also, every common divisor of a and b must divide rn. We now show that a + b = gcd(a, b).

  • Since gcd(a, b)|a, b, we have a, b ⊂ gcd(a, b). It

follows that a + b ⊂ gcd(a, b).

  • Conversely, the Euclidean algorithm shows that there

exists elements r and s such that gcd(a, b) = ra + sb. It follows that gcd(a, b) ∈ a + b.

  • Therefore, a + b = gcd(a, b).
  • Instructor: Yifan Yang

Section 46 – Euclidean domains

slide-39
SLIDE 39

Proof of Theorem 46.9, continued

  • It is straightforward to show that c is a common divisor of

rk−1 and rk if and only it is a common divisor of rk and rk+1. Thus, rn is a common divisor of a and b. Also, every common divisor of a and b must divide rn. We now show that a + b = gcd(a, b).

  • Since gcd(a, b)|a, b, we have a, b ⊂ gcd(a, b). It

follows that a + b ⊂ gcd(a, b).

  • Conversely, the Euclidean algorithm shows that there

exists elements r and s such that gcd(a, b) = ra + sb. It follows that gcd(a, b) ∈ a + b.

  • Therefore, a + b = gcd(a, b).
  • Instructor: Yifan Yang

Section 46 – Euclidean domains

slide-40
SLIDE 40

Proof of Theorem 46.9, continued

  • It is straightforward to show that c is a common divisor of

rk−1 and rk if and only it is a common divisor of rk and rk+1. Thus, rn is a common divisor of a and b. Also, every common divisor of a and b must divide rn. We now show that a + b = gcd(a, b).

  • Since gcd(a, b)|a, b, we have a, b ⊂ gcd(a, b). It

follows that a + b ⊂ gcd(a, b).

  • Conversely, the Euclidean algorithm shows that there

exists elements r and s such that gcd(a, b) = ra + sb. It follows that gcd(a, b) ∈ a + b.

  • Therefore, a + b = gcd(a, b).
  • Instructor: Yifan Yang

Section 46 – Euclidean domains

slide-41
SLIDE 41

Proof of Theorem 46.9, continued

  • It is straightforward to show that c is a common divisor of

rk−1 and rk if and only it is a common divisor of rk and rk+1. Thus, rn is a common divisor of a and b. Also, every common divisor of a and b must divide rn. We now show that a + b = gcd(a, b).

  • Since gcd(a, b)|a, b, we have a, b ⊂ gcd(a, b). It

follows that a + b ⊂ gcd(a, b).

  • Conversely, the Euclidean algorithm shows that there

exists elements r and s such that gcd(a, b) = ra + sb. It follows that gcd(a, b) ∈ a + b.

  • Therefore, a + b = gcd(a, b).
  • Instructor: Yifan Yang

Section 46 – Euclidean domains

slide-42
SLIDE 42

Proof of Theorem 46.9, continued

  • It is straightforward to show that c is a common divisor of

rk−1 and rk if and only it is a common divisor of rk and rk+1. Thus, rn is a common divisor of a and b. Also, every common divisor of a and b must divide rn. We now show that a + b = gcd(a, b).

  • Since gcd(a, b)|a, b, we have a, b ⊂ gcd(a, b). It

follows that a + b ⊂ gcd(a, b).

  • Conversely, the Euclidean algorithm shows that there

exists elements r and s such that gcd(a, b) = ra + sb. It follows that gcd(a, b) ∈ a + b.

  • Therefore, a + b = gcd(a, b).
  • Instructor: Yifan Yang

Section 46 – Euclidean domains

slide-43
SLIDE 43

Proof of Theorem 46.9, continued

  • It is straightforward to show that c is a common divisor of

rk−1 and rk if and only it is a common divisor of rk and rk+1. Thus, rn is a common divisor of a and b. Also, every common divisor of a and b must divide rn. We now show that a + b = gcd(a, b).

  • Since gcd(a, b)|a, b, we have a, b ⊂ gcd(a, b). It

follows that a + b ⊂ gcd(a, b).

  • Conversely, the Euclidean algorithm shows that there

exists elements r and s such that gcd(a, b) = ra + sb. It follows that gcd(a, b) ∈ a + b.

  • Therefore, a + b = gcd(a, b).
  • Instructor: Yifan Yang

Section 46 – Euclidean domains

slide-44
SLIDE 44

Proof of Theorem 46.9, continued

  • It is straightforward to show that c is a common divisor of

rk−1 and rk if and only it is a common divisor of rk and rk+1. Thus, rn is a common divisor of a and b. Also, every common divisor of a and b must divide rn. We now show that a + b = gcd(a, b).

  • Since gcd(a, b)|a, b, we have a, b ⊂ gcd(a, b). It

follows that a + b ⊂ gcd(a, b).

  • Conversely, the Euclidean algorithm shows that there

exists elements r and s such that gcd(a, b) = ra + sb. It follows that gcd(a, b) ∈ a + b.

  • Therefore, a + b = gcd(a, b).
  • Instructor: Yifan Yang

Section 46 – Euclidean domains

slide-45
SLIDE 45

Proof of Theorem 46.9, continued

  • It is straightforward to show that c is a common divisor of

rk−1 and rk if and only it is a common divisor of rk and rk+1. Thus, rn is a common divisor of a and b. Also, every common divisor of a and b must divide rn. We now show that a + b = gcd(a, b).

  • Since gcd(a, b)|a, b, we have a, b ⊂ gcd(a, b). It

follows that a + b ⊂ gcd(a, b).

  • Conversely, the Euclidean algorithm shows that there

exists elements r and s such that gcd(a, b) = ra + sb. It follows that gcd(a, b) ∈ a + b.

  • Therefore, a + b = gcd(a, b).
  • Instructor: Yifan Yang

Section 46 – Euclidean domains

slide-46
SLIDE 46

Homework

Problems 2, 5, 8, 9, 16, 17, 18 of Section 46.

Instructor: Yifan Yang Section 46 – Euclidean domains