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Solution to Exercise 26.18 Show that each homomorphism from a field to a ring is either injective or maps everything onto 0. Proof. Suppose we have a homomorphism φ : F → R where F is a field and R is a ring (for example R itself could be a field). The exercise asks us to show that either the kernel of φ is equal to {0} (in which case φ will be injective) or to F (meaning precisely that φ(x) = 0 for all x ∈ F ). The field F has a unity 1. We will investigate what sort of value φ can take on this special element. Suppose that 1 ∈ ker φ. Then φ(1) = 0. For any x ∈ F we use the fact that 1 is identity for multiplication in F and the fact that φ is assumed to be a homomorphism to compute that φ(x) = φ(x1) = φ(x)φ(1) = φ(x)0 = 0. Thus in this case φ maps everything into 0. Suppose now that 1 ∈ / ker φ. Let x ∈ ker φ. Thus we know that φ(x) = 0. If x 6= 0, then φ(1) = φ(xx−1 ) = φ(x)φ(x−1 ) = 0φ(x−1 ) = 0, contradicting our assumption. Thus x must be 0 and so ker φ = {0}. 1