Field and ring homomorphism

  • Thread starter Icebreaker
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  • #1
Icebreaker
"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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  • #2
matt grime
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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?
 
  • #3
Icebreaker
It appears I assumed f is injective to begin with; circular logic, I'm afraid.
 
  • #4
matt grime
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Suppose f(x)=0, and x is not zero. What is special about fields (or any division algebras) in respect of non-zero elements in the field?
 
  • #5
Icebreaker
Let a' denote the multiplicative inverse of a. Since F is a field, f(aa')=f(a)f(a')=f(1). By hypothesis, f(1)!=f(0)=0R. Therefore, f(a) and f(a') are non-zero. So ker(f)={0F}. By theorem (6.12), f is injective.

The image of f is by def surjective. Therefore the image of f is isomorphic to F. I think.
 
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