# Fields, Rings and Homomorphisms

• Apr 3rd 2009, 10:24 AM
Coda202
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).
• Apr 3rd 2009, 11:54 AM
ThePerfectHacker
Quote:

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?
• Apr 3rd 2009, 05:00 PM
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.
• Apr 3rd 2009, 10:08 PM
ThePerfectHacker
Quote:

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 $\phi(ab) = \phi(a)\phi(b),\phi(a+b)=\phi(a)+\phi(b)$. 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 $\phi(1) =1$.