1. We proved thta the ideal < x + 1 > is prime in Z[x] by definition. In this problem, show this ideal is prime, but not maximal, in Z[x] using the FHT as follows: (a) Define the substitution homomorphism μ-1: Z[x] → Z by μ-1 (p(x)) = p(-1) - you can assume this is a homomorphism. Show it is onto (b) Prove keru_1 =< x + 1 > (c) Then apply the FHT to answer the question (d) Now use the FHT again, to explain why is a maximl ideal of F[x], where F is any field

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Certainly! Here's a transcription of the text that could appear on an educational website:

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### Exploring the Ideal \(\langle x + 1 \rangle\) in \(\mathbb{Z}[x]\)

We previously proved that the ideal \(\langle x + 1 \rangle\) is prime in \(\mathbb{Z}[x]\) by definition. In this problem, we will demonstrate that this ideal is prime, but not maximal, in \(\mathbb{Z}[x]\) using the First Isomorphism Theorem (FHT) as follows:

1. **Define the Substitution Homomorphism:**
   - Define the homomorphism \(\mu_{-1}: \mathbb{Z}[x] \to \mathbb{Z}\) by \(\mu_{-1}(p(x)) = p(-1)\). You may assume this is a homomorphism. Demonstrate that it is onto.

2. **Prove \( \ker \mu_{-1} = \langle x + 1 \rangle \):**
   - Show that the kernel of this homomorphism is precisely \(\langle x + 1 \rangle\).

3. **Apply the FHT:**
   - Use the First Isomorphism Theorem to resolve the question of whether the ideal is prime but not maximal.

4. **Use the FHT Again:**
   - Apply the FHT once more to explain why \(\langle x + 1 \rangle\) is a maximal ideal of \(F[x]\), where \(F\) is any field.

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This content helps you understand the concepts of substitution homomorphisms, kernels of homomorphisms, and how the First Isomorphism Theorem applies to determine the nature of ideals in polynomial rings.
Transcribed Image Text:Certainly! Here's a transcription of the text that could appear on an educational website: --- ### Exploring the Ideal \(\langle x + 1 \rangle\) in \(\mathbb{Z}[x]\) We previously proved that the ideal \(\langle x + 1 \rangle\) is prime in \(\mathbb{Z}[x]\) by definition. In this problem, we will demonstrate that this ideal is prime, but not maximal, in \(\mathbb{Z}[x]\) using the First Isomorphism Theorem (FHT) as follows: 1. **Define the Substitution Homomorphism:** - Define the homomorphism \(\mu_{-1}: \mathbb{Z}[x] \to \mathbb{Z}\) by \(\mu_{-1}(p(x)) = p(-1)\). You may assume this is a homomorphism. Demonstrate that it is onto. 2. **Prove \( \ker \mu_{-1} = \langle x + 1 \rangle \):** - Show that the kernel of this homomorphism is precisely \(\langle x + 1 \rangle\). 3. **Apply the FHT:** - Use the First Isomorphism Theorem to resolve the question of whether the ideal is prime but not maximal. 4. **Use the FHT Again:** - Apply the FHT once more to explain why \(\langle x + 1 \rangle\) is a maximal ideal of \(F[x]\), where \(F\) is any field. --- This content helps you understand the concepts of substitution homomorphisms, kernels of homomorphisms, and how the First Isomorphism Theorem applies to determine the nature of ideals in polynomial rings.
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