Ring Homomorphism: Show Phi(1) Is Unity for R

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Homework Statement


Let R and R' are rings and phi: R to R' is a ring homomorphism such that phi[R] is not identically 0'. Show that if R has unity 1 and R' has no 0 divisors, then phi(1) is a unity for R'.

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The Attempt at a Solution



Its relatively obviously why phi(1) has to be unity for the subring phi[R]. I don't see why phi(1)r' has to be r' for every r' in R'.
 
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CAN subring have an identity other than the ring identity?
 
I don't know.

If S is a subring of R and 1 is a unity of S, then is it true that 1r=r for an r in R? I don't see why?

How do I use the fact that R' has no 0 divisors.
 
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Recall that ehrenfest is not using the usual notion of a 'ring'; he uses a variant that omits '1' from the definition. So if the ring and subring have '1', his definition of subring doesn't require them to be the same.

There is a simple example of a subring with a different unit: consider the ring of 2x2 matrices, and the subring of those whose upper-left entry is the only nonzero.
 
And in that example, there are zero divisors. Use the fact that lack of zero divisors admits a cancellation rule, i.e., r's' = r't' => s' = t'.

At least, I think that's right.
 
A ring <R,+,*> is a set R together with two binary operations + and * such that the following axioms are satisfied:
1) <R,+> is an abelian group
2) Multiplication is associative
3) For all a,b,c in R, a*(b+c)=(b+c)*a=a*b+a*c

A subring is a subset of R that is also a ring under + and *.

Now, how to solve the problem...

EDIT: sorry I didn't read the post above this when I wrote that
 
If nothing springs to mind, then just try exploring. e.g. what properties does the unit of a subring have?
 
Here is the proof that if R is a ring with no 0 divisors and S is a nonzero subring of R with unity 1, then 1 MUST be a unity of R:

11=1 implies that r11=r1 implies that r1=r by cancellation

11=1 implies that 11r=1r implies that 1r=r by cancellation

Is that right?
 
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