# Field and ring homomorphism

"Let F be a field and let f:F->R be a ring homomorphism satisfying f(0) != f(1). Show that f is necessarily injective."

Assume f(a)=f(b), then f(a)-f(b)=0R => f(a-b)=0R. f(0F)=0R and therefore a=b.

But this implies that every homomorphism is injective. How can that be?

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Homework Helper
It at best implies every *field* homomorphism is either the zero map or an injection. This is not surprising: the kernel of a ring homomorphism (so in particular a field homomorphism) is an ideal, but fields have no nontrivial ideals thus the kernel is either trivial or all of the field.

However your proof doesn't show even that.

How did you deduce that f(a-b)=0 implies a-b=0?

It appears I assumed f is injective to begin with; circular logic, I'm afraid.