I'm studying an ideal $I \trianglelefteq R$ and noticed that for a certain non-injective, non-zero homomorphism $\varphi: R \rightarrow S$ I can show that $\varphi(I)=\varphi(R)$. I'm wondering if this implies that $I=R$.
It holds for the one little example I could think of. Let $R=\mathbb{Z}$, $S=\mathbb{Z}$, $I=k\mathbb{Z}$, and $\varphi_{m}(n)=mn$, $m > 1$. Suppose $\varphi_{m}(I)=\varphi_{m}(R)$. Then $\varphi_{m}(I)=\varphi_{m}(k\mathbb{Z})=mk\mathbb{Z}=\varphi_{m}(R)=\varphi_{m}(\mathbb{Z})=m\mathbb{Z}$. Thus, $mk\mathbb{Z}=m\mathbb{Z} \Rightarrow k=1 \Rightarrow I=\mathbb{Z}=R$.
(Rings not necessarily commutative and/or may not contain 1.)