While the formulation of unique factorization in a general integral domain is a bit more complicated than the one we used for Z, it is straightforward to show that Z satisfies the new formulation. However, the quadratic integer ring Z[√5] does not. 5. The goal of this problem is to prove that the equalities 6 = 2.3 = (1+√−5)(1 − √−5) constitute a genuine failure of unique factorization in Z[√-5]. (a) Prove that neither 2 nor 3 is the norm of an element in Z[√-5]. (b) Use (a) to prove that 2, 3, and 1 ± √-5 are prime in Z[√-5]. (Hint: Use unique factorization in Z applied to norms; part (a) will help.) (c) Verify that neither 2 nor 3 is a unit (in Z[√5]) multiplied by 1± √-5. (Hint: Problem 4 describes the unit group of Z[√-5].)
While the formulation of unique factorization in a general integral domain is a bit more complicated than the one we used for Z, it is straightforward to show that Z satisfies the new formulation. However, the quadratic integer ring Z[√5] does not. 5. The goal of this problem is to prove that the equalities 6 = 2.3 = (1+√−5)(1 − √−5) constitute a genuine failure of unique factorization in Z[√-5]. (a) Prove that neither 2 nor 3 is the norm of an element in Z[√-5]. (b) Use (a) to prove that 2, 3, and 1 ± √-5 are prime in Z[√-5]. (Hint: Use unique factorization in Z applied to norms; part (a) will help.) (c) Verify that neither 2 nor 3 is a unit (in Z[√5]) multiplied by 1± √-5. (Hint: Problem 4 describes the unit group of Z[√-5].)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![While the formulation of unique factorization in a general integral domain is a bit more
complicated than the one we used for Z, it is straightforward to show that Z satisfies the
new formulation. However, the quadratic integer ring Z[√5] does not.
5. The goal of this problem is to prove that the equalities
6=2.3=(1+√−5)(1
constitute a genuine failure of unique factorization in Z[√-5].
(a) Prove that neither 2 nor 3 is the norm of an element in Z[√-5].
(b) Use (a) to prove that 2, 3, and 1 ± √-5 are prime in Z[√-5]. (Hint: Use unique
factorization in Z applied to norms; part (a) will help.)
(c) Verify that neither 2 nor 3 is a unit (in Z[√-5]) multiplied by 1± √-5. (Hint:
Problem 4 describes the unit group of Z[√-5].)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Febb913e1-4986-4d74-b6ce-ad576ddf43d3%2F9da127a2-93c2-4023-b1d2-e339b674b617%2Fkegn9qr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:While the formulation of unique factorization in a general integral domain is a bit more
complicated than the one we used for Z, it is straightforward to show that Z satisfies the
new formulation. However, the quadratic integer ring Z[√5] does not.
5. The goal of this problem is to prove that the equalities
6=2.3=(1+√−5)(1
constitute a genuine failure of unique factorization in Z[√-5].
(a) Prove that neither 2 nor 3 is the norm of an element in Z[√-5].
(b) Use (a) to prove that 2, 3, and 1 ± √-5 are prime in Z[√-5]. (Hint: Use unique
factorization in Z applied to norms; part (a) will help.)
(c) Verify that neither 2 nor 3 is a unit (in Z[√-5]) multiplied by 1± √-5. (Hint:
Problem 4 describes the unit group of Z[√-5].)
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