The subset 2 of C is closed under the addition and multiplication operations on C; it is also closed under subtraction, and contains 0 and 1.¹ These properties make Z an example of a subring of C. A large class of subrings of C can be constructed as follows. For every integer d € Z - {0}, the polynomial X²-d has exactly two roots in C (by the fundamental theorem of algebra!): if we denote one of the roots by √d, then the other is -√d. Define the subset Z[√d] of C by Z[√d] = {z € C: z = a +b√d for some a, b = Z}. The notation Z[√d] is read "Z adjoin √d." We use this terminology because (it turns out) Z[√d] is the smallest subring of C that contains Z and √d. By definition, the elements of Z[√d] are the complex numbers having a particular form: a +b√d with a, b € Z. This means that, to prove that a complex number is an element of Z[√d], we must show that the number can be expressed in the indicated form. Keep this in mind when solving the first problem below. 1. Let de Z-{ (a) Using the definition of Z[√d], explain why Z is a subset of Z[√d]. (b) Prove that Z[√d] is closed under the addition operation on C. (c) Prove that Z[√d] is closed under the multiplication operation on C.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Problem 1 

The subset Z of C is closed under the addition and multiplication operations on C; it is
also closed under subtraction, and contains 0 and 1. These properties make Z an example
of a subring of C. A large class of subrings of C can be constructed as follows. For every
integer d € Z - {0}, the polynomial X²-d has exactly two roots in C (by the fundamental
theorem of algebra!): if we denote one of the roots by √d, then the other is -√d. Define
the subset Z[√d] of C by
Z[√d] = {z € C: z = a +b√d for some a, b = Z}.
The notation Z[√d] is read "Z adjoin √d." We use this terminology because (it turns out)
Z[√d is the smallest subring of C that contains Z and √d. By definition, the elements
of Z[√d] are the complex numbers having a particular form: a+b√d with a, b € Z. This
means that, to prove that a complex number is an element of Z[√d], we must show that the
number can be expressed in the indicated form. Keep this in mind when solving the first
problem below.
1. Let de Z-{0}.
(a) Using the definition of Z[√d], explain why Z is a subset of Z[√d].
(b) Prove that Z[√d] is closed under the addition operation on C.
(c) Prove that Z[√d] is closed under the multiplication operation on C.
(You will not actually need the assumption that d 0 in your arguments; but if d = 0,
then Z[√d] = Z, and we are most interested in the case where √d & Q, as this ensures that
Z[√d properly contains Z.)
2. Let d € Z - {0} and assume that √d & Q (we will eventually prove that this is true if
and only if d is not the square of an integer, i.e., if and only if √d & Z). In this case, the
elements of Z[√d are called quadratic integers.
(a) Prove that, if z € Z[√d], then there are unique integers a and b such that z = a+b√d.
(Hint: Such integers exist by the definition of Z[√d]. Assuming that there are integers
a₁, a2, b₁, and by such that z = a₁ + b₁ √d and z= a₂ + b₂√d, we need to show that
0₁ 0₂ and b₁ b₂. Ignoring the first equality, suppose for the sake of a contradiction
that b₁ b₂. Use the irrationality of √d to reach a contradiction. How does the
desired result follow?)
(b) We would like to define a function
=
: Z[√d] → Z[√d] by
a(z) = a - b√d,
where a, b € Z satisfy 2 = a +b√d. Why is there a potential issue with trying to
define a function using this formula? Explain how (a) resolves this issue, so that the
displayed formula does indeed give a well-defined function, which we call conjugation.
(Hint: compare with the first problem of Homework 1.)
(c) Prove that the function o defined in (b) is bijective.
Transcribed Image Text:The subset Z of C is closed under the addition and multiplication operations on C; it is also closed under subtraction, and contains 0 and 1. These properties make Z an example of a subring of C. A large class of subrings of C can be constructed as follows. For every integer d € Z - {0}, the polynomial X²-d has exactly two roots in C (by the fundamental theorem of algebra!): if we denote one of the roots by √d, then the other is -√d. Define the subset Z[√d] of C by Z[√d] = {z € C: z = a +b√d for some a, b = Z}. The notation Z[√d] is read "Z adjoin √d." We use this terminology because (it turns out) Z[√d is the smallest subring of C that contains Z and √d. By definition, the elements of Z[√d] are the complex numbers having a particular form: a+b√d with a, b € Z. This means that, to prove that a complex number is an element of Z[√d], we must show that the number can be expressed in the indicated form. Keep this in mind when solving the first problem below. 1. Let de Z-{0}. (a) Using the definition of Z[√d], explain why Z is a subset of Z[√d]. (b) Prove that Z[√d] is closed under the addition operation on C. (c) Prove that Z[√d] is closed under the multiplication operation on C. (You will not actually need the assumption that d 0 in your arguments; but if d = 0, then Z[√d] = Z, and we are most interested in the case where √d & Q, as this ensures that Z[√d properly contains Z.) 2. Let d € Z - {0} and assume that √d & Q (we will eventually prove that this is true if and only if d is not the square of an integer, i.e., if and only if √d & Z). In this case, the elements of Z[√d are called quadratic integers. (a) Prove that, if z € Z[√d], then there are unique integers a and b such that z = a+b√d. (Hint: Such integers exist by the definition of Z[√d]. Assuming that there are integers a₁, a2, b₁, and by such that z = a₁ + b₁ √d and z= a₂ + b₂√d, we need to show that 0₁ 0₂ and b₁ b₂. Ignoring the first equality, suppose for the sake of a contradiction that b₁ b₂. Use the irrationality of √d to reach a contradiction. How does the desired result follow?) (b) We would like to define a function = : Z[√d] → Z[√d] by a(z) = a - b√d, where a, b € Z satisfy 2 = a +b√d. Why is there a potential issue with trying to define a function using this formula? Explain how (a) resolves this issue, so that the displayed formula does indeed give a well-defined function, which we call conjugation. (Hint: compare with the first problem of Homework 1.) (c) Prove that the function o defined in (b) is bijective.
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