Thread: Fields, 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).

Originally Posted by Coda202

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).

Hint1: The kernel of a ring homorphism gives rise to an ideal.
Hint2: What are the ideals of a field?

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.

Originally Posted by JaneBennet

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.

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 . 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 .