9.48. Find all prime ideals in each of the following rings. 1. Z10 2. Z50
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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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!

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 \(

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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