1. In this exercise we will prove that Z[√-5] is an integral domain but not a unique factorisation domain (UFD). Note that in S we have the following factorisations of the element 6: 6=2.3=(1+√√−5) · (1−√−5). It turns out that these are factorisations into irreducible elements. To see this, prove that the element 1+ √-5 is irreducible, by assuming that 1+√√-5= (a+b√√-5). (c+d√√−5) and taking modulus squared. Explain briefly why a similar argument shows that the other three ele- ments 1-5, 2, and 3 are also irreducible.
1. In this exercise we will prove that Z[√-5] is an integral domain but not a unique factorisation domain (UFD). Note that in S we have the following factorisations of the element 6: 6=2.3=(1+√√−5) · (1−√−5). It turns out that these are factorisations into irreducible elements. To see this, prove that the element 1+ √-5 is irreducible, by assuming that 1+√√-5= (a+b√√-5). (c+d√√−5) and taking modulus squared. Explain briefly why a similar argument shows that the other three ele- ments 1-5, 2, and 3 are also irreducible.
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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![1. In this exercise we will prove that Z[√-5] is an integral domain but not a
unique factorisation domain (UFD).
Note that in S we have the following factorisations of the element 6:
6=2.3=(1+√√−5) · (1−√−5).
It turns out that these are factorisations into irreducible elements. To
see this, prove that the element 1+ √-5 is irreducible, by assuming
that
1+√√-5= (a+b√√-5). (c+d√√−5)
and taking modulus squared.
Explain briefly why a similar argument shows that the other three ele-
ments 1-5, 2, and 3 are also irreducible.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0de9773e-39c1-4df6-a7d6-864501c7f552%2Fa0d4012f-1040-467a-ad26-15ac5d404cdc%2F1dcqfxe_processed.png&w=3840&q=75)
Transcribed Image Text:1. In this exercise we will prove that Z[√-5] is an integral domain but not a
unique factorisation domain (UFD).
Note that in S we have the following factorisations of the element 6:
6=2.3=(1+√√−5) · (1−√−5).
It turns out that these are factorisations into irreducible elements. To
see this, prove that the element 1+ √-5 is irreducible, by assuming
that
1+√√-5= (a+b√√-5). (c+d√√−5)
and taking modulus squared.
Explain briefly why a similar argument shows that the other three ele-
ments 1-5, 2, and 3 are also irreducible.
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