Let $\Phi : R \rightarrow R'$ be a ring homomorphism, where $R,R'$ are rings with unity. Then which of these is true? Let $\Phi : R \rightarrow R'$ be a ring homomorphism, where $R,R'$ are rings with unity. Then which of these is true : 
$(i)~\Phi(1)=1 ~\forall~$ rings $R,R~'$ with unity
$(ii)~\Phi(1) \ne 1 $ for any rings $R,R~'$
$(iii)~ \Phi(1) \ne 1 $ if $R~'$ is an integral domain or if $\Phi$ is onto.
$(iv)~ \Phi(1) = 1 $ if $R~'$ is an integral domain or if $\Phi$ is onto.
Attempt:
$\Phi(1\cdot1)= [ \Phi(1)]^2 $
$\implies \Phi(1) [\Phi(1) - 1]=0$
Only when $R~'$ is an integral domain can we infer that the above statement means either $\Phi(1)=0$ or $\Phi(1)=1$
If $\Phi(1)=0,$ then $\Phi$ represents the trivial homomorphism as $\Phi(x) = \Phi(x) \Phi(1) = 0$.
Else, if $\Phi$ is onto and $R~'$ is an integral domain, then the only option for $\Phi(1)$ is being equal to $1$.
So, $(iv)$ should be the right option.
Could someone please confirm if I am right?
Thank you for your help.

Here's the problem with (i):
Suppose $R' = R = \Bbb Z \times \Bbb Z$, with component-wise addition and multiplication and:
$\Phi((a,b)) = (a,0)$. We have:
$\Phi((a,b) + (a',b')) = \Phi((a+a',b+b'))$
$ = (a+a',0) = (a,0) + (a',0) = \Phi((a,b)) + \Phi((a',b'))$
and: $\Phi((a,b)(a',b')) = \Phi((aa',bb')) = (aa',0) = (a,0)(a',0) = \Phi((a,b))\Phi((a',b'))$,
but $\Phi((1,1)) = (1,0)$ is not the multiplicative identity of $R$ (although it is the multiplicative identity of $\Phi(R)$).
