[proofplan] The proof is a direct comparison between the definition of injectivity and the definition of the kernel. If $\varphi$ is injective, then the only element that can map to $0_S$ is $0_R$, because ring homomorphisms send $0_R$ to $0_S$. Conversely, if the kernel is trivial, then equality $\varphi(a)=\varphi(b)$ forces $\varphi(a-b)=0_S$, so $a-b$ lies in the kernel and hence $a=b$. [/proofplan] [step:Show that injectivity forces the kernel to be trivial] Assume that $\varphi$ is injective. Since $\varphi: R \to S$ is a ring homomorphism, it preserves addition, so \begin{align*} \varphi(0_R) = \varphi(0_R + 0_R) = \varphi(0_R) + \varphi(0_R). \end{align*} Adding the additive inverse of $\varphi(0_R)$ in $S$ to both sides gives $\varphi(0_R)=0_S$. Let $x \in \ker \varphi$. By definition of the kernel, $\varphi(x)=0_S$. Since $\varphi(0_R)=0_S$, we have \begin{align*} \varphi(x)=\varphi(0_R). \end{align*} Injectivity of $\varphi$ gives $x=0_R$. Hence every element of $\ker \varphi$ equals $0_R$, and since $\varphi(0_R)=0_S$, we also have $0_R \in \ker \varphi$. Therefore \begin{align*} \ker \varphi = \{0_R\}. \end{align*} [/step] [step:Use the trivial kernel to prove injectivity] Assume that \begin{align*} \ker \varphi = \{0_R\}. \end{align*} Let $a,b \in R$ satisfy $\varphi(a)=\varphi(b)$. Since $\varphi$ is a ring homomorphism, it preserves addition and additive inverses. Therefore \begin{align*} \varphi(a-b) = \varphi(a) - \varphi(b). \end{align*} Using $\varphi(a)=\varphi(b)$, the right-hand side is $0_S$, so \begin{align*} \varphi(a-b)=0_S. \end{align*} Thus $a-b \in \ker \varphi$. By the assumption on the kernel, $a-b=0_R$. Adding $b$ to both sides in the additive group of $R$ gives $a=b$. Hence $\varphi$ is injective. [guided] We want to prove that $\varphi$ is injective, so we start with the defining test for injectivity: take arbitrary elements $a,b \in R$ and assume that their images agree, \begin{align*} \varphi(a)=\varphi(b). \end{align*} Our goal is to prove $a=b$. The hypothesis about the kernel can only be used on elements that map to $0_S$. Therefore we convert the equality of two images into a statement about an element mapping to zero. Since $\varphi: R \to S$ is a ring homomorphism, it preserves addition and additive inverses. Applying this to the element $a-b \in R$ gives \begin{align*} \varphi(a-b) = \varphi(a) - \varphi(b). \end{align*} Because $\varphi(a)=\varphi(b)$, subtraction in the additive group of $S$ gives \begin{align*} \varphi(a)-\varphi(b)=0_S. \end{align*} Hence \begin{align*} \varphi(a-b)=0_S. \end{align*} By definition of the kernel, \begin{align*} \ker \varphi = \{x \in R : \varphi(x)=0_S\}. \end{align*} The previous display therefore says exactly that $a-b \in \ker \varphi$. The kernel is assumed to be $\{0_R\}$, so \begin{align*} a-b=0_R. \end{align*} Finally, adding $b$ to both sides in the additive group of the ring $R$ yields \begin{align*} a=b. \end{align*} Since arbitrary $a,b \in R$ with $\varphi(a)=\varphi(b)$ must be equal, $\varphi$ is injective. [/guided] [/step]