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.

  • 1
    $\begingroup$ 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 $\Phi(1)=1’$ $\endgroup$
    – rschwieb
    Feb 7, 2020 at 12:00

1 Answer 1


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!


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