**Proof plan.** The additive group structure is already handled by the [First Isomorphism Theorem for Groups](/theorems/842). We only need to additionally verify that the induced map respects multiplication and sends $1_R$ to $1_S$. **Step 1: $\ker(\varphi) \trianglelefteq R$ as an ideal.** [claim: Kernel Is Ideal] $\ker(\varphi)$ is an ideal of $R$. [/claim] [proof] Since $\varphi : (R, +) \to (S, +)$ is a group homomorphism, $\ker(\varphi)$ is a subgroup of $(R, +, 0_R)$. For strong closure: if $a \in \ker(\varphi)$ and $b \in R$, then $\varphi(ab) = \varphi(a)\varphi(b) = 0 \cdot \varphi(b) = 0$, so $ab \in \ker(\varphi)$. [/proof] **Step 2: Define the quotient map.** By the [First Isomorphism Theorem for Groups](/theorems/842) applied to the additive [groups](/page/Group), the map \begin{align*} \Phi : R/\ker(\varphi) &\to \operatorname{im}(\varphi) \\ r + \ker(\varphi) &\mapsto \varphi(r) \end{align*} is a well-defined bijective additive group homomorphism. **Step 3: $\Phi$ respects multiplication.** [claim: Multiplicativity] $\Phi$ is a ring homomorphism. [/claim] [proof] \begin{align*} \Phi\bigl((r + \ker\varphi)(t + \ker\varphi)\bigr) = \Phi(rt + \ker\varphi) = \varphi(rt) = \varphi(r)\varphi(t) = \Phi(r + \ker\varphi)\Phi(t + \ker\varphi). \end{align*} Also $\Phi(1_R + \ker\varphi) = \varphi(1_R) = 1_S$. [/proof] **Step 4: Conclusion.** Since $\Phi$ is a bijective ring homomorphism, it is a ring isomorphism: \begin{align*} R/\ker(\varphi) \cong \operatorname{im}(\varphi) \leq S. \qquad \square \end{align*}