Explanation Formula used: 1) Definition of maximal ideal: Let M be an ideal of the commutative ring R . Then M is a maximal ideal of R if M is not a proper subset of any ideal except R itself. Thus, an ideal M is a maximal ideal of R if and only if M ⊂ I ⊆ R where I is an ideal implies I = R . 2) Definition of prime ideal: An ideal I of a commutative ring R is a prime ideal if I ≠ R and if a b ∈ I implies either a ∈ I or b ∈ I . 3) Definition of an ideal: A subset I of a ring R is an ideal of R , provided the following conditions are satisfied: 1. I is non-empty. 2. x ∈ I and y ∈ I implies x + y ∈ I . 3. x ∈ I implies − x ∈ I . 4. x ∈ I and r ∈ R implies x r and r x are in I . Proof: The elements of ℤ ⊕ { 0 } are of the form ℤ ⊕ { 0 } = { ( a , 0 ) | a ∈ ℤ } . The elements of ℤ ⊕ ℤ are of the form ℤ ⊕ ℤ = { ( a , b ) | a ∈ ℤ , b ∈ ℤ } . Now, 1. 0 ∈ ℤ implies ( 0 , 0 ) ∈ ℤ ⊕ { 0 } . Therefore, ℤ ⊕ { 0 } is non-empty. 2. Let x ∈ ℤ ⊕ { 0 } and y ∈ ℤ ⊕ { 0 } Then x = ( a , 0 ) , y = ( b , 0 ) . Let x + y = ( a , 0 ) + ( b , 0 ) = ( a + b , 0 ) As a ∈ ℤ , b ∈ ℤ ⇒ a + b ∈ ℤ , ( a + b , 0 ) ∈ ℤ ⊕ { 0 } . That is, x + y ∈ ℤ ⊕ { 0 } . 3. Let x ∈ ℤ ⊕ { 0 } , Then, x = ( a , 0 ) . − x = − ( a , 0 ) = ( − a , 0 ) As a ∈ ℤ ⇒ − a ∈ ℤ , ( − a , 0 ) ∈ ℤ ⊕ { 0 } . That is, − x ∈ ℤ ⊕ { 0 } . 4. If x ∈ ℤ ⊕ { 0 } and r ∈ ℤ ⊕ ℤ . Such that x = ( a , 0 ) and r = ( c , d ) x r = ( a , 0 ) ( c , d ) = ( a c , 0 ) As a , c ∈ ℤ ⇒ a c ∈ ℤ , ( a c , 0 ) ∈ ℤ ⊕ { 0 }

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Author: Gilbert, Linda, Jimmie

Publisher: Cengage Learning,

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Chapter 6.4, Problem 25E

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To prove: ℤ⊕{0} is a prime ideal of ℤ⊕ℤ but no maximal ideal of ℤ⊕ℤ.

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Chapter 6.4, Problem 24E

Chapter 6.4, Problem 26E

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To prove: ℤ ⊕ { 0 } is a prime ideal of ℤ ⊕ ℤ but no maximal ideal of ℤ ⊕ ℤ .

Solution Summary: The author explains how mathbbZoplus is a prime ideal, but no maximal ideal.

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To prove: ℤ⊕{0} is a prime ideal of ℤ⊕ℤ but no maximal ideal of ℤ⊕ℤ.

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Chapter 6 Solutions