9.6. Let R be a commutative ring. If I = {a e R : a" = 0 for some n e N}, show that I is an ideal of R.
9.6. Let R be a commutative ring. If I = {a e R : a" = 0 for some n e N}, show that I is an ideal of R.
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 9.6 in detail? I included list of theorems and definitions from the textbook.
![**Problem 9.6:**
Let \( R \) be a commutative ring. If
\[ I = \{ a \in R : a^n = 0 \text{ for some } n \in \mathbb{N} \}, \]
show that \( I \) is an ideal of \( R \).
**Explanation:**
This problem involves proving that a specific subset \( I \) of a commutative ring \( R \) is an ideal. The subset \( I \) consists of all elements \( a \) in \( R \) such that there exists a natural number \( n \) for which the power \( a^n \) equals zero.
To show that \( I \) is an ideal, we must demonstrate two properties:
1. **Additive Closure:** For any \( a, b \in I \), the sum \( a + b \) should also be in \( I \).
2. **Absorption Property:** For any \( a \in I \) and any \( r \in R \), the product \( ra \) should be in \( I \).
The goal of this exercise is to verify these properties using the definition of \( I \) and the properties of the commutative ring \( R \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe15ed467-90ec-4e60-afef-3d3f6119f74d%2F68e03d93-00c9-43c9-a42b-946f95c2ef52%2F0nbh7w6_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 9.6:**
Let \( R \) be a commutative ring. If
\[ I = \{ a \in R : a^n = 0 \text{ for some } n \in \mathbb{N} \}, \]
show that \( I \) is an ideal of \( R \).
**Explanation:**
This problem involves proving that a specific subset \( I \) of a commutative ring \( R \) is an ideal. The subset \( I \) consists of all elements \( a \) in \( R \) such that there exists a natural number \( n \) for which the power \( a^n \) equals zero.
To show that \( I \) is an ideal, we must demonstrate two properties:
1. **Additive Closure:** For any \( a, b \in I \), the sum \( a + b \) should also be in \( I \).
2. **Absorption Property:** For any \( a \in I \) and any \( r \in R \), the product \( ra \) should be in \( I \).
The goal of this exercise is to verify these properties using the definition of \( I \) and the properties of the commutative ring \( R \).
![**Definition 9.1.** Let \( R \) be a ring. Then a subring \( I \) of \( R \) is said to be an ideal if \( ir, ri \in I \) for all \( i \in I \) and \( r \in R \). We call this the absorption property.
**Theorem 9.1.** Let \( R \) be a ring and \( I \) a subset of \( R \). Then \( I \) is an ideal if and only if
1. \( 0 \in I \);
2. \( i - j \in I \) for all \( i, j \in I \); and
3. \( ir, ri \in I \) for all \( i \in I, r \in R \).
**Example 9.1.** Let \( n \) be any integer. Then \( n\mathbb{Z} \) is an ideal of \( \mathbb{Z} \). Indeed, we already know that it is a subring. But also, if \( nk \in n\mathbb{Z} \), then for any integer \( r \), \( r(nk) = n(rk) \in n\mathbb{Z} \).
**Example 9.2.** Let \( I \) be the set of all polynomials \( f(x) \in \mathbb{R}[x] \) such that \( f(0) = 0 \). We claim that \( I \) is an ideal in \( \mathbb{R}[x] \). Certainly \( I \) contains the zero polynomial. Also, if \( f(0) = g(0) = 0 \), then \( (f-g)(0) = f(0) - g(0) = 0 \), hence \( f(x) - g(x) \in I \). Also, if \( f(0) = 0 \) and \( h(x) \in \mathbb{R}[x] \), then \( h(0)f(0) = h(0)0 = 0 \). Hence, \( h(x)f(x) \in I \).
**Example 9.3.** Let \( I \) be the set of all polynomials in \( \mathbb{Z}[x]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe15ed467-90ec-4e60-afef-3d3f6119f74d%2F68e03d93-00c9-43c9-a42b-946f95c2ef52%2F1pfu92d_processed.png&w=3840&q=75)
Transcribed Image Text:**Definition 9.1.** Let \( R \) be a ring. Then a subring \( I \) of \( R \) is said to be an ideal if \( ir, ri \in I \) for all \( i \in I \) and \( r \in R \). We call this the absorption property.
**Theorem 9.1.** Let \( R \) be a ring and \( I \) a subset of \( R \). Then \( I \) is an ideal if and only if
1. \( 0 \in I \);
2. \( i - j \in I \) for all \( i, j \in I \); and
3. \( ir, ri \in I \) for all \( i \in I, r \in R \).
**Example 9.1.** Let \( n \) be any integer. Then \( n\mathbb{Z} \) is an ideal of \( \mathbb{Z} \). Indeed, we already know that it is a subring. But also, if \( nk \in n\mathbb{Z} \), then for any integer \( r \), \( r(nk) = n(rk) \in n\mathbb{Z} \).
**Example 9.2.** Let \( I \) be the set of all polynomials \( f(x) \in \mathbb{R}[x] \) such that \( f(0) = 0 \). We claim that \( I \) is an ideal in \( \mathbb{R}[x] \). Certainly \( I \) contains the zero polynomial. Also, if \( f(0) = g(0) = 0 \), then \( (f-g)(0) = f(0) - g(0) = 0 \), hence \( f(x) - g(x) \in I \). Also, if \( f(0) = 0 \) and \( h(x) \in \mathbb{R}[x] \), then \( h(0)f(0) = h(0)0 = 0 \). Hence, \( h(x)f(x) \in I \).
**Example 9.3.** Let \( I \) be the set of all polynomials in \( \mathbb{Z}[x]
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