[proofplan] We prove directly from the definitions. First we name the kernel and show it is an additive subgroup of $R$ by checking nonemptiness and closure under subtraction. Then we verify the two absorption properties: multiplying an element of the kernel on either side by an arbitrary element of $R$ keeps it in the kernel. These two properties are exactly the definition of a two-sided ideal. [/proofplan]

[step:Define the kernel and show it is nonempty] Define the subset $K\subset R$ by \begin{align*} K=\ker\varphi=\{a\in R:\varphi(a)=0_S\}. \end{align*} Since $\varphi$ is a ring homomorphism, it preserves the additive identity, so $\varphi(0_R)=0_S$. Hence $0_R\in K$, and therefore $K$ is nonempty. [/step]

[step:Show the kernel is closed under subtraction]Let $a,b\in K$. By definition of $K$, we have $\varphi(a)=0_S$ and $\varphi(b)=0_S$. Since $\varphi$ is additive and preserves additive inverses, \begin{align*} \varphi(a-b)=\varphi(a)+\varphi(-b)=\varphi(a)-\varphi(b). \end{align*} Substituting $\varphi(a)=0_S$ and $\varphi(b)=0_S$ gives \begin{align*} \varphi(a-b)=0_S-0_S=0_S. \end{align*} Thus $a-b\in K$. Since $K$ is nonempty and closed under subtraction, $K$ is an additive subgroup of $R$.[/step]

[guided]We need to prove that $K$ is an additive subgroup of the additive group of $R$. A convenient [subgroup criterion](/theorems/932) says that a nonempty subset of an additive group is a subgroup if it is closed under subtraction. The previous step already showed that $K$ is nonempty, so it remains to check subtraction. Take arbitrary elements $a,b\in K$. The meaning of $a\in K$ is exactly that $\varphi(a)=0_S$, and the meaning of $b\in K$ is exactly that $\varphi(b)=0_S$. Because $\varphi:R\to S$ is a ring homomorphism, it is in particular a homomorphism of additive groups. Therefore it preserves addition and additive inverses, so \begin{align*} \varphi(a-b)=\varphi(a+(-b))=\varphi(a)+\varphi(-b)=\varphi(a)-\varphi(b). \end{align*} Using the two kernel equalities, \begin{align*} \varphi(a-b)=0_S-0_S=0_S. \end{align*} Therefore $a-b\in K$ by the definition of $K$. Since $a$ and $b$ were arbitrary elements of $K$, the subset $K$ is closed under subtraction. Together with nonemptiness, this proves that $K$ is an additive subgroup of $R$.[/guided]

[step:Verify absorption under left and right multiplication] Let $r\in R$ and $a\in K$. Since $a\in K$, we have $\varphi(a)=0_S$. Using multiplicativity of the ring homomorphism $\varphi$, \begin{align*} \varphi(ra)=\varphi(r)\varphi(a)=\varphi(r)0_S=0_S. \end{align*} Thus $ra\in K$. Similarly, \begin{align*} \varphi(ar)=\varphi(a)\varphi(r)=0_S\varphi(r)=0_S. \end{align*} Thus $ar\in K$. Hence $K$ is closed under multiplication by arbitrary elements of $R$ on both the left and the right. [/step]

[step:Conclude that the kernel is a two-sided ideal] We have shown that $K=\ker\varphi$ is an additive subgroup of $R$ and that, for every $r\in R$ and every $a\in K$, both $ra$ and $ar$ lie in $K$. By the definition of a two-sided ideal, this proves \begin{align*} \ker\varphi\trianglelefteq R. \end{align*} [/step]