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(a) We indicated the codomain of the norm function as Z, but it is not quite obvious that Na actually takes values in Z (as opposed to Z[√d]). Writing z = a +b√d for (unique!) a, b = Z, give an explicit formula for Na(z) in terms of a and b, and explain why this formula confirms that Na(z) does in fact lie in Z. (b) Prove that, for all 2₁, 22 € Z[√d], Na(2122) = Na(21)Na(22). (Hint: if you use Home- work 2.5(b), this should be painless.) (c) Let z € Z[√d]. Prove that ze Z[√d] if and only if Na(z) € {±1}. (Hint: If z € Z[√d]x, use the multiplicativity of Na established in (b) to show that Na(z) € ZX. For the converse, you should be able to exhibit an inverse for z in Z[√d using od.)

(a) We indicated the codomain of the norm function as Z, but it is not quite obvious that Na actually takes values in Z (as opposed to Z[√d]). Writing z = a +b√d for (unique!) a, b = Z, give an explicit formula for Na(z) in terms of a and b, and explain why this formula confirms that Na(z) does in fact lie in Z. (b) Prove that, for all 2₁, 22 € Z[√d], Na(2122) = Na(21)Na(22). (Hint: if you use Home- work 2.5(b), this should be painless.) (c) Let z € Z[√d]. Prove that ze Z[√d] if and only if Na(z) € {±1}. (Hint: If z € Z[√d]x, use the multiplicativity of Na established in (b) to show that Na(z) € ZX. For the converse, you should be able to exhibit an inverse for z in Z[√d using od.)

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter6: Vector Spaces
Section6.2: Linear Independence, Basis, And Dimension
Problem 61EQ
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2. Let d € Z - {0}, assume that √d & Q, and let od : Z[√d] → Z[√d] be the conjugation function. We define the norm function Na : Z[√d] → Z by Na(z) = zoa(z). (For example, when d = -1, σa(z) = z is the usual complex conjugate, and Na(z) = |z|² is the square of the complex absolute value.) (a) We indicated the codomain of the norm function as Z, but it is not quite obvious that Na actually takes values in Z (as opposed to Z[√d]). Writing z = a +b√d for (unique!) a, b = Z, give an explicit formula for Na(z) in terms of a and b, and explain why this formula confirms that Na(z) does in fact lie in Z. (b) Prove that, for all 2₁, 22 € Z[√d], Na(2122) = Na(2₁)Na(22). (Hint: if you use Home- work 2.5(b), this should be painless.) (c) Let z € Z[√d]. Prove that ze Z[√d] if and only if Na(z) € {±1}. (Hint: If z € Z[√d]x, use the multiplicativity of Na established in (b) to show that Na(z) € ZX. For the converse, you should be able to exhibit an inverse for z in Z[√d] using od.)
Transcribed Image Text:2. Let d € Z - {0}, assume that √d & Q, and let od : Z[√d] → Z[√d] be the conjugation function. We define the norm function Na : Z[√d] → Z by Na(z) = zoa(z). (For example, when d = -1, σa(z) = z is the usual complex conjugate, and Na(z) = |z|² is the square of the complex absolute value.) (a) We indicated the codomain of the norm function as Z, but it is not quite obvious that Na actually takes values in Z (as opposed to Z[√d]). Writing z = a +b√d for (unique!) a, b = Z, give an explicit formula for Na(z) in terms of a and b, and explain why this formula confirms that Na(z) does in fact lie in Z. (b) Prove that, for all 2₁, 22 € Z[√d], Na(2122) = Na(2₁)Na(22). (Hint: if you use Home- work 2.5(b), this should be painless.) (c) Let z € Z[√d]. Prove that ze Z[√d] if and only if Na(z) € {±1}. (Hint: If z € Z[√d]x, use the multiplicativity of Na established in (b) to show that Na(z) € ZX. For the converse, you should be able to exhibit an inverse for z in Z[√d] using od.)
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