5. Show that Z[V-6] is not a UFD by factoring 10 in two ways. Why does this prove that Z[V-6] is not a PID?
5. Show that Z[V-6] is not a UFD by factoring 10 in two ways. Why does this prove that Z[V-6] is not a PID?
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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![**Problem 5:**
**Objective:**
Show that \(\mathbb{Z}[\sqrt{-6}]\) is not a Unique Factorization Domain (UFD) by factoring 10 in two different ways. Analyze why this result implies that \(\mathbb{Z}[\sqrt{-6}]\) is not a Principal Ideal Domain (PID).
**Explanation:**
- **Unique Factorization Domain (UFD):** A ring is a UFD if every element can be factored into irreducible elements uniquely, up to units and order.
- **Principal Ideal Domain (PID):** A ring is a PID if every ideal can be generated by a single element.
**Instructions:**
1. **Factor 10 in Two Ways in \(\mathbb{Z}[\sqrt{-6}]\):**
- Try to find two distinct factorizations of the number 10 using elements of the ring \(\mathbb{Z}[\sqrt{-6}]\).
- Verify that these factorizations cannot be obtained from each other by simply reordering or multiplying by units.
2. **Explain the Implication for PID:**
- Discuss how the failure of unique factorization indicates that \(\mathbb{Z}[\sqrt{-6}]\) is not a PID, often linked through results in ring theory indicating that UFDs are PIDs and vice versa under certain conditions.
**Goals:**
Understand the relationship between the structure of a ring and its factorization properties, and how these properties interrelate in algebraic number theory.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa9774f45-b264-467e-8b06-716b402d428d%2F1f3a8211-f8f7-44da-a997-7fb5bf124a4c%2Fmna2yf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 5:**
**Objective:**
Show that \(\mathbb{Z}[\sqrt{-6}]\) is not a Unique Factorization Domain (UFD) by factoring 10 in two different ways. Analyze why this result implies that \(\mathbb{Z}[\sqrt{-6}]\) is not a Principal Ideal Domain (PID).
**Explanation:**
- **Unique Factorization Domain (UFD):** A ring is a UFD if every element can be factored into irreducible elements uniquely, up to units and order.
- **Principal Ideal Domain (PID):** A ring is a PID if every ideal can be generated by a single element.
**Instructions:**
1. **Factor 10 in Two Ways in \(\mathbb{Z}[\sqrt{-6}]\):**
- Try to find two distinct factorizations of the number 10 using elements of the ring \(\mathbb{Z}[\sqrt{-6}]\).
- Verify that these factorizations cannot be obtained from each other by simply reordering or multiplying by units.
2. **Explain the Implication for PID:**
- Discuss how the failure of unique factorization indicates that \(\mathbb{Z}[\sqrt{-6}]\) is not a PID, often linked through results in ring theory indicating that UFDs are PIDs and vice versa under certain conditions.
**Goals:**
Understand the relationship between the structure of a ring and its factorization properties, and how these properties interrelate in algebraic number theory.
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