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Abstract Algebra II Homework 5B: Problems on Field Extensions and Linear Transformations, Assignments of Abstract Algebra

Solutions to problem 4, 5, and 6 from homework 5b of the abstract algebra ii course (math 252). The problems involve proving equalities of field extensions, finding irreducible polynomials, and showing that the 'multiplication by √d' map is a linear transformation. Problem 7 is a hint for constructing regular 5-gons using trigonometric identities or primitive roots of unity.

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2009/2010

Uploaded on 02/25/2010

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MATH 252: ABSTRACT ALGEBRA II
HOMEWORK #5B
Problem 4 (DF 13.2.7). Prove that Q(2 + 3) = Q(2,3). [Hint: Consider (2 + 3)2.] Conclude
that [Q(2 + 3) : Q] = 4. Find an irreducible polynomial over Qsatisfied by 2 + 3.
Problem 5 (DF 13.2.14). Prove that if [F(α) : F] is odd then F(α) = F(α2).
Problem 6 (DF 13.2.21). Let DZbe squarefree and let K=Q(D). Let α=a+bDK.
(a) Show that the “multiplication by α map
φ:KK
β7→ φ(β) = αβ
is a linear transformation (of vector spaces over Q).
(b) Compute the matrix of φon the basis 1,Dof K.
Problem 7 (sorta DF 13.3.5).
(a) Show that α= 2 cos(2π /5) satisfies the equation x2+x1 = 0. [Hint: Use a trigonometric identity
or the fact that α=ζ5+ 15where ζ5= exp(2πi/5) = cos(2π/5) + isin(2π/5) is a primitive fifth
root of unity.]
(b) Conclude that the regular 5-gon is constructible by straightedge and compass.
Date: 24 March 2008; due 4 April 2008.
1

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MATH 252: ABSTRACT ALGEBRA II

HOMEWORK #5B

Problem 4 (DF 13.2.7). Prove that Q(

3) = Q(

3). [Hint: Consider (

2 .] Conclude

that [Q(

  1. : Q] = 4. Find an irreducible polynomial over Q satisfied by

Problem 5 (DF 13.2.14). Prove that if [F (α) : F ] is odd then F (α) = F (α^2 ).

Problem 6 (DF 13.2.21). Let D ∈ Z be squarefree and let K = Q(

D). Let α = a + b

D ∈ K.

(a) Show that the “multiplication by α” map

φ : K → K

β 7 → φ(β) = αβ

is a linear transformation (of vector spaces over Q).

(b) Compute the matrix of φ on the basis 1,

D of K.

Problem 7 (sorta DF 13.3.5).

(a) Show that α = 2 cos(2π/5) satisfies the equation x^2 + x − 1 = 0. [Hint: Use a trigonometric identity or the fact that α = ζ 5 + 1/ζ 5 where ζ 5 = exp(2πi/5) = cos(2π/5) + i sin(2π/5) is a primitive fifth root of unity.] (b) Conclude that the regular 5-gon is constructible by straightedge and compass.

Date: 24 March 2008; due 4 April 2008. 1