(2) If a, b = Z, we set N(a+b√-5) = (a + b√−5)(a − b√-5) = a² + 5b². Let R = Z[√5] and R* = R-0. Consider the nomm map N : R* → Z*. (d) Show that 2,3,1+√-5, 1-√-5 are irreducible elemetns in R. (2) Observe 6 = 2.3 = (1+√√-5). (1-√-5). Show R is not a unique factorization domain.
(2) If a, b = Z, we set N(a+b√-5) = (a + b√−5)(a − b√-5) = a² + 5b². Let R = Z[√5] and R* = R-0. Consider the nomm map N : R* → Z*. (d) Show that 2,3,1+√-5, 1-√-5 are irreducible elemetns in R. (2) Observe 6 = 2.3 = (1+√√-5). (1-√-5). Show R is not a unique factorization domain.
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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![Let R=1
2 = Z[√5] and R* = R-0. Consider the nomm map N : R* → Z*.
If a, b = Z, we set N(a+b√-5) = (a + b√−5)(a − b√-5) = a² + 5b².
(d) Show that 2, 3, 1+√√-5, 1-√√–5 are irreducible elemetns in R.
(2) Observe 6 = 2 · 3 = (1 + √−5) . (1-√-5). Show R is not a unique factorization
domain.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffa923b6f-81dd-482c-8885-6de6bc295751%2Fee4b62a9-f0af-4891-89ed-52c1cdb35e92%2Fbf2a5uc_processed.png&w=3840&q=75)
Transcribed Image Text:Let R=1
2 = Z[√5] and R* = R-0. Consider the nomm map N : R* → Z*.
If a, b = Z, we set N(a+b√-5) = (a + b√−5)(a − b√-5) = a² + 5b².
(d) Show that 2, 3, 1+√√-5, 1-√√–5 are irreducible elemetns in R.
(2) Observe 6 = 2 · 3 = (1 + √−5) . (1-√-5). Show R is not a unique factorization
domain.
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