Expert: Paul Klarreich - 11/5/2008

Question
Hi Paul I wrote you earlier today and accidentally put in an old email address that I cannot access and followed that email up with another asking you to respond to me at my current email address. I received a reply from you that said see other answer, where is this other answer?

Aside from that I have another question I am working on that I'd humbly like some help with if you'd be willing to help me again

Let phi: F->R a homomorphism of rings, where R is a ring and F is a field. I need to show that either phi is injective or that phi(a) = 0  for all a in F. Below is the start I have for a proof but feel like I am aimlessly walking about:
how about this for a start?
suppose phi(a) = phi(b) for some a and b in F, then if phi is injective we would have
a = b and there would be only one element 0 in F that mapped to 0 in R. If phi(a) = 0 for all a in F then there would be any number of elements in F that mapped to the identity element which implies phi is not injective, (unless F ={0}). If phi(a) = 0 then phi (a b) = 0 and phi(ab) = 0 for a,b in F <> 0. In the multiplicative part, since F is a field this means that either a or b is 0 (or both)since there are no zero divisors in F, this contradicts that both a and b are nonzero thus phi(a* b) = phi(0* 0) = phi(0) = 0 = 0 0 = phi(a)* phi(b)
I don't know where to go from here, help me please?


Answer
Questioner:   Sombra
Category:  Advanced Math
Private:  No
 
Subject:  homomorphisms of rings
Question:  Hi Paul I wrote you earlier today and accidentally put in an old email address that I cannot access and followed that email up with another asking you to respond to me at my current email address. I received a reply from you that said see other answer, where is this other answer?

Aside from that I have another question I am working on that I'd humbly like some help with if you'd be willing to help me again

Let phi: F->R a homomorphism of rings, where R is a ring and F is a field. I need to show that either phi is injective (1-1?) or that phi(a) = 0  for all a in F. Below is the start I have for a proof but feel like I am aimlessly walking about:
how about this for a start?
suppose phi(a) = phi(b) for some a and b in F, then if phi is injective we would have
a = b and there would be only one element 0 in F that mapped to 0 in R. If phi(a) = 0 for all a in F then there would be any number of elements in F that mapped to the identity element which implies phi is not injective, (unless F ={0}).

If phi(a) = 0 then phi (a b) = 0 and phi(ab) = 0 for a,b in F <> 0. In the multiplicative part, since F is a field this means that either a or b is 0 (or both)since there are no zero divisors in F, this contradicts that both a and b are nonzero thus phi(a* b) = phi(0* 0) = phi(0) = 0 = 0 0 = phi(a)* phi(b)
I don't know where to go from here, help me please?
...........................
Hi, Sombra,

Sorry about the mixup with the other question.  I have it and I'll see if I can come up with something.

It has been a long time and I am not so confident about this, but:

My understanding is that F->R is a ring homomorphism if:

phi(a + b) = phi(a) + phi(b)

phi(ab) = phi(a)phi(b)

phi(1) = 1. (assuming F has a unity, which it does if it is a field.)

.......................

I think it is easy to prove that  phi(0) = 0

Is it?  For any a,

phi(0 + a) = phi(0) + phi(a) = phi(a),
so phi(0) = 0.

Yes, it is.
..............................
Now for any two elements a,b, both /= 0:  

Suppose that phi(a) = phi(b) and phi(b) /= 0

phi(a/b) = phi(a)/phi(b) = 1

So a/b = 1 and  a = b.

So either a = b or  phi(b) = 0.
I think you can do something with that.