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9.48. Find all prime ideals in each of the following rings. 1. Z10 2. Z50

9.48. Find all prime ideals in each of the following rings. 1. Z10 2. Z50

Advanced Engineering Mathematics
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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Could you explain how to show 9.48 in great detail? I also included a list of theorems and definitions in my textbook as a reference. Really appreciate your help!

Definition 9.9. Let \( R \) be a ring. An ideal \( M \) of \( R \) is said to be **maximal** if 1. \( M \neq R \); and 2. If \( I \) is an ideal of \( R \) containing \( M \), then \( I = M \) or \( I = R \). Example 9.27. Let \( R = \mathbb{Z} \) and let \( n \) be a nonnegative integer. Then we claim that \( (n) \) is a maximal ideal of \( R \) if and only if \( n \) is prime. Indeed, \( (0) \) is certainly not maximal, as \( (0) \subset (2) \subset R \). Also, \( (1) \) is not maximal, since \( (1) = R \). If \( n \) is composite, say \( n = kl \), with \( 1 < k, l < n \), then we note that \( (n) \subset (k) \subset R \), so \( (n) \) is not maximal. Finally, let \( n \) be prime. Suppose that \( I \) is an ideal of \( R \) with \( (n) \subset I \subset R \). Take \( a \in I \setminus (n) \). Since \( a \) is not divisible by \( n \), and \( n \) is prime, we know that \( (a, n) = 1 \). Thus, by Corollary 2.1, we can find integers \( u \) and \( v \) such that \( au + nv = 1 \). But as \( a, n \in I \), this implies that \( 1 \in I \), hence \( I = R \), giving us a contradiction. (As we shall see shortly, there is another way to prove this.) Example 9.28. In any field, the ideal \( \{0\} \) is maximal! Remember, by Corollary 9.1, a field only has two ideals. In a commutative ring with identity, there is a nice test for maximality of ideals. Theorem 9.20. Let \( R \) be a commutative ring with identity, and \(
Transcribed Image Text:Definition 9.9. Let \( R \) be a ring. An ideal \( M \) of \( R \) is said to be **maximal** if 1. \( M \neq R \); and 2. If \( I \) is an ideal of \( R \) containing \( M \), then \( I = M \) or \( I = R \). Example 9.27. Let \( R = \mathbb{Z} \) and let \( n \) be a nonnegative integer. Then we claim that \( (n) \) is a maximal ideal of \( R \) if and only if \( n \) is prime. Indeed, \( (0) \) is certainly not maximal, as \( (0) \subset (2) \subset R \). Also, \( (1) \) is not maximal, since \( (1) = R \). If \( n \) is composite, say \( n = kl \), with \( 1 < k, l < n \), then we note that \( (n) \subset (k) \subset R \), so \( (n) \) is not maximal. Finally, let \( n \) be prime. Suppose that \( I \) is an ideal of \( R \) with \( (n) \subset I \subset R \). Take \( a \in I \setminus (n) \). Since \( a \) is not divisible by \( n \), and \( n \) is prime, we know that \( (a, n) = 1 \). Thus, by Corollary 2.1, we can find integers \( u \) and \( v \) such that \( au + nv = 1 \). But as \( a, n \in I \), this implies that \( 1 \in I \), hence \( I = R \), giving us a contradiction. (As we shall see shortly, there is another way to prove this.) Example 9.28. In any field, the ideal \( \{0\} \) is maximal! Remember, by Corollary 9.1, a field only has two ideals. In a commutative ring with identity, there is a nice test for maximality of ideals. Theorem 9.20. Let \( R \) be a commutative ring with identity, and \(
**Question 9.48:** Find all prime ideals in each of the following rings. 1. \(\mathbb{Z}_{10}\) 2. \(\mathbb{Z}_{50}\)
Transcribed Image Text:**Question 9.48:** Find all prime ideals in each of the following rings. 1. \(\mathbb{Z}_{10}\) 2. \(\mathbb{Z}_{50}\)
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