I know a similar question was asked here. But my exercise is asking me to do without a hypothesis.
If R and R' are rings with unity(denote $1$ and $1'$ for the $R$ and $R'$ identities, respectively), R' integral domain and $\Phi:R\rightarrow R'$ a ring homomorphism, then $\Phi(1)=1'$.
My attempt:
$\Phi(a)=\Phi(1\cdot a)=\Phi(1)\cdot\Phi(a)\Rightarrow\Phi(a)=0, \forall a\!\in\!R$ or $\Phi(1)=1'$
With the similar question hypothesis I can conclude, since for a $\Phi(r)\neq0$ for a nonzero $r\!\in\!R$, then $\Phi(a)\neq0$ if $a=r$. But, without this hypothesis I can't think of a solution.
Thanks in advance.
Best Answer
If $\Phi(r)=0$ for all $r\in R$, then you have the zero homomorphism. As is stated, the question is false, because $\Phi(1)\neq 1'$ in that case. So you should add the hypothesis $\Phi$ is not the zero ring homomorphism (or $\Phi(r)\neq 0$ for some $r\in R$). I wouldn't worry because the zero homomorphism is a "boring" case.
If $\Phi(r)\neq 0$ for some $r\in R$, your proof is fine!