[proofplan] The hypothesis says exactly that every element killed in the quotient is also killed by $\varphi$. We therefore define the candidate map on a coset $r+I$ by sending it to $\varphi(r)$, and the main point is to check that this value does not depend on the representative $r$. Once well-definedness is established, preservation of addition, multiplication, and identity follows directly from the [quotient ring](/page/Quotient%20Ring) operations and the homomorphism properties of $\varphi$. Uniqueness follows because every element of $R/I$ is a coset. [/proofplan] [step:Record what the kernel containment gives] The theorem assumes \begin{align*} I \subseteq \ker \varphi. \end{align*} Thus, for every $a \in I$, the definition $\ker \varphi=\{r \in R : \varphi(r)=0_S\}$ gives \begin{align*} \varphi(a)=0_S. \end{align*} This is the only kernel fact used below. [/step] [step:Define the induced map on cosets] Define a function \begin{align*} \widetilde{\varphi}: R/I \to S \end{align*} by the rule \begin{align*} \widetilde{\varphi}(r+I)=\varphi(r) \end{align*} for a representative $r \in R$. Since every element of the quotient ring $R/I$ has the form $r+I$ for some $r \in R$, this formula defines a candidate function on all of $R/I$ once well-definedness is proved. [/step] [step:Prove that the induced formula is independent of representatives] Let $r,s \in R$ and suppose that $r+I=s+I$ in $R/I$. Since $I \trianglelefteq R$ is a two-sided ideal, it is in particular an additive subgroup of $R$, so the hypotheses of [citetheorem:8254] apply. Therefore this equality of cosets is equivalent to \begin{align*} r-s \in I. \end{align*} Since $I \subseteq \ker \varphi$, we have $r-s \in \ker \varphi$, so \begin{align*} 0_S=\varphi(r-s)=\varphi(r)-\varphi(s). \end{align*} Adding $\varphi(s)$ to both sides gives $\varphi(r)=\varphi(s)$. Hence $\widetilde{\varphi}(r+I)$ is independent of the representative $r$, so $\widetilde{\varphi}:R/I \to S$ is well-defined. [guided] We must check the only possible ambiguity in the definition. The element $r+I \in R/I$ can have many representatives, so we need to prove that if $r+I=s+I$, then the proposed values $\varphi(r)$ and $\varphi(s)$ are equal. Let $r,s \in R$ and assume \begin{align*} r+I=s+I. \end{align*} Since $I \trianglelefteq R$ is a two-sided ideal, it is in particular an additive subgroup of $R$, so the hypotheses of [citetheorem:8254] apply. That theorem says that equality of cosets in a quotient ring is equivalent to the difference of representatives lying in the ideal. Therefore \begin{align*} r-s \in I. \end{align*} The hypothesis $I \subseteq \ker \varphi$ now applies: because $r-s \in I$, we also have $r-s \in \ker \varphi$. By the definition of the kernel of the ring homomorphism $\varphi:R \to S$, \begin{align*} \varphi(r-s)=0_S. \end{align*} Since $\varphi$ preserves subtraction, this becomes \begin{align*} \varphi(r)-\varphi(s)=0_S. \end{align*} Adding $\varphi(s)$ in the additive group of $S$ gives \begin{align*} \varphi(r)=\varphi(s). \end{align*} Thus the value assigned to the coset $r+I$ depends only on the coset, not on the chosen representative. Therefore the formula \begin{align*} \widetilde{\varphi}(r+I)=\varphi(r) \end{align*} defines a well-defined function $\widetilde{\varphi}:R/I \to S$. [/guided] [/step] [step:Verify that the induced map is a ring homomorphism] Let $r,s \in R$. Using the addition operation in the quotient ring and the fact that $\varphi$ is a ring homomorphism, we obtain \begin{align*} \widetilde{\varphi}((r+I)+(s+I))=\widetilde{\varphi}((r+s)+I)=\varphi(r+s)=\varphi(r)+\varphi(s)=\widetilde{\varphi}(r+I)+\widetilde{\varphi}(s+I). \end{align*} Using the multiplication operation in the quotient ring and again using that $\varphi$ is a ring homomorphism, we obtain \begin{align*} \widetilde{\varphi}((r+I)(s+I))=\widetilde{\varphi}((rs)+I)=\varphi(rs)=\varphi(r)\varphi(s)=\widetilde{\varphi}(r+I)\widetilde{\varphi}(s+I). \end{align*} Since the ring homomorphisms in this statement are unital, the identity element of $R/I$ is $1_R+I$, and \begin{align*} \widetilde{\varphi}(1_R+I)=\varphi(1_R)=1_S. \end{align*} Therefore $\widetilde{\varphi}:R/I \to S$ is a unital ring homomorphism. [/step] [step:Prove uniqueness and recover the factorization] Let \begin{align*} \psi: R/I \to S \end{align*} be any ring homomorphism satisfying \begin{align*} \psi(r+I)=\varphi(r) \end{align*} for every $r \in R$. For an arbitrary element $x \in R/I$, choose $r \in R$ such that $x=r+I$. Then \begin{align*} \psi(x)=\psi(r+I)=\varphi(r)=\widetilde{\varphi}(r+I)=\widetilde{\varphi}(x). \end{align*} Thus $\psi$ and $\widetilde{\varphi}$ agree on every element of $R/I$, so $\psi=\widetilde{\varphi}$. Let \begin{align*} \pi: R \to R/I \end{align*} denote the quotient homomorphism defined by $\pi(r)=r+I$. For every $r \in R$, \begin{align*} (\widetilde{\varphi}\circ \pi)(r)=\widetilde{\varphi}(r+I)=\varphi(r). \end{align*} Therefore $\varphi=\widetilde{\varphi}\circ \pi$. Hence the induced ring homomorphism is unique and satisfies the equivalent factorization statement, completing the proof. [/step]