Let F be a field and R a ring and suppose phi: F --> R is a homomorphism.

Prove that either phi is one-to-one or phi is the trivial homomorphism (that is, phi(a) = 0 for all a in F).

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- Apr 3rd 2009, 09:24 AMCoda202Fields, Rings and Homomorphisms
Let F be a field and R a ring and suppose phi: F --> R is a homomorphism.

Prove that either phi is one-to-one or phi is the trivial homomorphism (that is, phi(a) = 0 for all a in F). - Apr 3rd 2009, 10:54 AMThePerfectHacker
- Apr 3rd 2009, 04:00 PMJaneBennet
Mind you, some authors – e.g. D.J.H. Garling in

*A Course in Galois Theory*(1986, Cambridge University Press) – insist that a ring homomorphism should map the multiplicative identity of one ring to that of the other. If you adopt such a definition, the second possibility would not be possible: all homomorphisms from a field to a ring would then have to be injective. - Apr 3rd 2009, 09:08 PMThePerfectHacker
I like to distinguish between "homomorphism between rings" and "homomorphism between commutative unitary rings". So when I see "ring homomorphism" all I think of is $\displaystyle \phi(ab) = \phi(a)\phi(b),\phi(a+b)=\phi(a)+\phi(b)$. However, if I see "commutative ring homomorphism" (it is rather common to called commutative unitary rings simply by commutative rings) then I think of the additional condition $\displaystyle \phi(1) =1$.