Abstract Algebra

An Interactive Approach, Second Edition

William Paulsen

• Publisher: Chapman and Hall/CRC
• Year: 2018
• ISBN:
  9781498719797
• Edition: 2
• Topic: Matemáticas
• Subtopic: Matemáticas general

The new edition of Abstract Algebra: An Interactive Approach presents a hands-on and traditional approach to learning groups, rings, and fields. It then goes further to offer optional technology use to create opportunities for interactive learning and computer use.

This new edition offers a more traditional approach offering additional topics to the primary syllabus placed after primary topics are covered. This creates a more natural flow to the order of the subjects presented. This edition is transformed by historical notes and better explanations of why topics are covered.

This innovative textbook shows how students can better grasp difficult algebraic concepts through the use of computer programs. It encourages students to experiment with various applications of abstract algebra, thereby obtaining a real-world perspective of this area.

Each chapter includes, corresponding Sage notebooks, traditional exercises, and several interactive computer problems that utilize Sage and Mathematica ® to explore groups, rings, fields and additional topics.

This text does not sacrifice mathematical rigor. It covers classical proofs, such as Abel's theorem, as well as many topics not found in most standard introductory texts. The author explores semi-direct products, polycyclic groups, Rubik's Cube ® -like puzzles, and Wedderburn's theorem. The author also incorporates problem sequences that allow students to delve into interesting topics, including Fermat's two square theorem.

Answers to Odd-Numbered Problems

Section 0.1

1) q = 25, r = 15

3) q = –19, r = 22

5) q = 166, r = 13

7) q = 0, r = 35

9) q = 0, r = 0

11) 2ⁿ = 2 · 2^{n – 1} < 2(n – 1)! < n(n – 1)! = n!

13) If (n – 1)³ + 2(n – 1) = 3k, then n² + 2n = 3(k + n² + n + 1).

15) If 6^{n – 1} + 4 = 20k, then 6ⁿ + 4 = 20(6k – 1).

17) (n – 1)((n – 1) + 1)/2 + n = n(n + 1)/2.

19) (n – 1)((n – 1) + 1)(2(n – 1) + 1)/6 + n² = n(n + 1)(2n + 1)/6.

21) (n – 1)((n – 1) + 1)((n – 1) + 2)/3 + n(n + 1) = n(n + 1)(n + 2)/3.

23) 2 · 24 + (–1) · 42 = 6.

25) 2 · 102 + (–3) · 66 = 6.

27) 14 · 1999 +(–965) · 29 = 1.

29) 5 · (–602)+ 12 · 252 = 14.

31) 0 · 0 + 1 · 7 = 7.

33) Since xy is a common multiple, by the well-ordering axiom there is a least common multiple, say z = ax = by. Note that gcd(a, b) = 1, else we can divide by gcd(a, b) to produce an even smaller common multiple. Then there is a u and v such that ua + vb = 1, so uaxy + vbxy = xy, hence z(uy + vx) = xy.

35) 2⁸ · 5³.

37) 2² · 3 · 5² · 19.

39) 7⁴ · 11.

41) u = –13717445541839, v = 97393865569283.

43) 3² · 17² · 379721.

45) 449 · 494927 · 444444443.

Section 0.2

1) {e, n, o, r, t, x, y}.

3) a) Not one-to-one, f(–1) = f(1) = 1. b) Not onto, f(x) ≠ –1.

5) a) One-to-one, x³ = y³ ⇒ x = y. b) Onto, $f\left(\sqrt[3]{y}\right)=y$ .

7) a) Not one-to-one, f(0) = f(4) = 0. b) Not onto, f(x) ≠ –5, since x² – 4x + 5 has complex roots.

9) a) One-to-one, if x even, y odd, then y = x + 1/2. b) Not onto, f(x) ≠ 3.

11) a) One-to-one, if x even, y odd, then x = 2y – 1 is odd. b) Not onto, f(x) ≠ 4.

13) a) Not one-to-one f(0) = f(3) = 1. b) Onto, either f(2y – 2) = y or f(2y + 1) = y.

15) If 2x² + x = 2y² + y = c, then x...