# Homomorphism from a ring with unity to a integral domain maps unity to unity?

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.

• Also a little more direct is noticing that $\Phi(1)=\Phi(1)^2$. Then either $\Phi(1)=0$ and you have the zero ring, or $\Phi(1)$ cancels from $1’\Phi(1)=\Phi(1)\Phi(1)$ to get $\Phi(1)=1’$ Feb 7, 2020 at 12:00
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!