Explanation Given information: F is a field and θ is an epimorphism from F to ring S such that ker θ ≠ F . Formula used: 1) Kernel: If θ is a homomorphism from ring R to ring R ' , then kernel of θ is the set ker ( θ ) = { x ∈ R | θ ( x ) = 0 } . 2) Fundamental Theorem of Ring Homomorphism: If θ is an epimorphism from ring R to ring R ' , then R ' is isomorphic to R ker ( θ ) . 3) Theorem: If θ is any homomorphism from ring R to ring R ' , then ker ( θ ) is an ideal of R , and ker ( θ ) = { 0 } if and only if θ is one-to-one. 4) Definition of Field: Let F be a ring. Then, F is a field provided these conditions are satisfied: i. F is a commutative ring. ii. F has a unity e , e ≠ 0 . iii. Every non-zero element of F has a multiplicative inverse. 5) Definition of a multiplicative inverse: Let R be a ring with unity e , and let a ∈ R ; if there is an element x in R such that a x = x a = e , x is a multiplicative inverse of a . Proof: Let F be a field and θ be an epimorphism from F to ring S such that ker θ ≠ F . Consider a mapping θ : F → S is defined by θ ( x ) = s Since θ is a homomorphism from F to ring S , then ker ( θ ) is an ideal of F . F is a field, hence, it has only ideals F and { 0 } . Now ker ( θ ) = { x ∈ F | θ ( x ) = 0 } For x ∈ ker ( θ ) and x ≠ 0 since F is a field then there exists an inverse x − 1 ∈ F such that x x − 1 = 1 ∈ ker θ since ker θ is ideal of F . If I is an ideal in R with e ∈ I , then I = R . This implies ker θ = F which is contradictory since ker θ ≠ F implies ker ( θ ) = { 0 } . By using theorem, θ is one-to-one Since θ is an epimorphism and also one-to-one implies that θ is an isomorphism. Now, S is a field if it satisfies the following conditions: i. S is a commutative ring: Let s 1 , s 2 ∈ S then s 1 = θ ( a ) , s 2 = θ ( b ) for a , b ∈ F

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Author: Gilbert

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Chapter 6.2, Problem 28E

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To prove: θ is an isomorphism from F to ring S, and S is a field.

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Chapter 6.2, Problem 27E

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To prove: θ is an isomorphism from F to ring S , and S is a field.

Solution Summary: The author explains that theta is an isomorphism from F to ring S, and E is a field.

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To prove: θ is an isomorphism from F to ring S, and S is a field.

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