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Expert: - 11/4/2008


Question
Let phi: F->R, F a field, R a ring and phi a homomorphism of rings. How do I show either phi(a) = 0 for all a in F or phi is injective?
I have a possible proof but I can't make the conclusion with
what have, perhaps you have one I can follow, please?

Answer
Let σ: F->R
F a field
R a ring
σ a homomorphism of rings

How do I show either σ(a) = 0 for all a in F or σ is injective?

In ring theory or abstract algebra, a ring homomorphism is a function between two rings which respects the operations of addition and multiplication.

More precisely, if R and S are rings, then a ring homomorphism is a function f : R → S such that

f(a + b) = f(a) + f(b) for all a and b in R
f(ab) = f(a) f(b) for all a and b in R
f(1) = 1

Naturally, if one does not require rings to have a multiplicative identity then the last condition is dropped.

Directly from these definitions, one can deduce:

f(0) = 0

f(−a) = −f(a)

If a has a multiplicative inverse in R, then f(a) has a multiplicative inverse in S and we have f(a−1) = (f(a))−1. Therefore, f induces a group homomorphism from the group of units of R to the group of units of S.

The kernel of f, defined as ker(f) = {a in R : f(a) = 0} is an ideal in R. Every ideal in a commutative ring R arises from some ring homomorphism in this way. For rings with identity the kernel of a ring homomorphism is a subring without identity.

The homomorphism f is injective if and only if the ker(f) = {0}.

I took this out of the artice
http://en.wikipedia.org/wiki/Ring_homomorphism

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