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let $R$ be a ring with identity element 1 and $\psi$ a non-trivial homomorphism of $R$ in an integer domain $D$. Show that $\psi(1)$ is the identity element of $D$

My try: I assumed that there exists in $D$ another element $e$ which is also identity, so $e=e\psi(1)=\psi(1)$ and as $D$ but I don't know if that is enough for ensure that $\psi(1)$ is the identity element in $D$.

JAISON ALEXANDER MUNOZ HORMIGA
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  • I don't know how you are showing $e=e\psi(1)$. – Brian Moehring May 05 '23 at 04:47
  • If your definition of "integral domain" requires the existence of multiplicative identity, then there is a somewhat more direct proof that comes from $\psi(1)(e-\psi(1)) = 0$, but as you see in the linked question, this assumption isn't necessary to show $\psi(1)$ is a multiplicative identity in $D$. – Brian Moehring May 05 '23 at 05:05

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