Let $G$ and $K$ be two groups and let $\varphi:G\rightarrow K$ be a group homomorphism.
Suppose that the kernel of $\varphi$ consists of just the identity element $e_G$ of $G$. We will prove that this implies that $\varphi$ is injective. Let $g$ and $h$ be elements of $G$ such that $\varphi(g)=\varphi(h)$. Multiplying the inverse of $\varphi(h)$ by both sides of the equation, we get $\varphi(g)\varphi(h)^{-1}=e_H$, but since $\varphi$ is a homomorphism, then $\varphi(g)\varphi(h)^{-1}=\varphi(gh^{-1})=e_H$. This means that the element $gh^{-1}$ is an element of the kernel of $\varphi$. But remember that the only element in the kernel is $e_G$. This means that $gh^{-1}=e_G$. Therefore $g=h$ and so $\varphi$ is injective.
On the other hand, suppose that $\varphi$ is injective. Let $g$ and $h$ be elements of the kernel of $\varphi$. This means that $\varphi(g)=\varphi(h)=e_H$. Since $\varphi$ is injective, then $g=h$. This means that there is only one element in the kernel of $\varphi$ and since $\varphi(e_G)=e_H$, then $e_G$ is the only element in the kernel.
Thus we have proven that a group homomorphism being injective is the same as it having a trivial kernel.
Since the kernel of a ring homomorphism is the same the as the kernel of the corresponding group homomorphism, then this also applies to ring homomorphisms.