Question

Let F be a field, and let R be a ring with unity 1 R ne0. Let1s be the unity element of F, and let : F -> R be a homomorphism ofrings such that phi (1 F )=1 R

(a ) Let A∈F and. B∈Kerϕ Prove that
ab∈Kerϕi
(b) Prove that Kerϕ = {0}
(c) Prove that delta is injective.

Answer 1

(a) If A ∈ F and B ∈ Kerϕ, then AB ∈ Kerϕ.

(b) Prove that Kerϕ = {0}.

(c) Prove that δ is injective.

Step-by-step explanation:

(a) To prove AB ∈ Kerϕ, let's consider AB for A ∈ F and B ∈ Kerϕ. By the definition of Kerϕ, ϕ(B) = 0 in R. As ϕ is a ring homomorphism, ϕ(AB) = ϕ(A) ⋅ ϕ(B). Since ϕ(B) = 0, then ϕ(AB) = ϕ(A) ⋅ 0 = 0, implying that AB ∈ Kerϕ.

(b) To prove Kerϕ = {0}, assume x is an element in Kerϕ such that x ≠ 0. As x ∈ Kerϕ, ϕ(x) = 0 in R. However, ϕ(1_F) = 1_R by the given condition. As ϕ is a ring homomorphism, ϕ(1_F) ⋅ ϕ(x) = ϕ(x) = 0. This contradicts the fact that ϕ(x) = 0 and ϕ(1_F) = 1_R unless x = 0. Hence, Kerϕ = {0}.

(c) To prove that δ is injective, consider δ(x) = δ(y) for x, y ∈ F. Assume δ is not injective, so there exist x ≠ y in F such that δ(x) = δ(y). As δ(x) = ϕ(x) and δ(y) = ϕ(y), it follows that ϕ(x) = ϕ(y). As ϕ is a ring homomorphism, ϕ(x - y) = ϕ(x) - ϕ(y) = 0. Since ϕ is injective, x - y = 0, implying that x = y, which contradicts the assumption. Hence, δ is injective