[proofplan] We verify the defining properties of an [integral domain](/page/Integral%20Domain) for $S$. The subring hypothesis gives that $S$ is a ring under the operations inherited from $R$, while commutativity and the identity element are inherited from $R$. The nonzero hypothesis supplies the nonzero ring condition. Finally, any zero-divisor equation in $S$ is also a zero-divisor equation in $R$, so the absence of zero divisors in $R$ forces the absence of zero divisors in $S$. [/proofplan] [step:Inherit the commutative ring structure and identity from $R$] Since $S \subset R$ is a subring, the addition map $+: S \times S \to S$ and multiplication map $\cdot: S \times S \to S$ are the restrictions of the corresponding operations on $R$, and $S$ is a ring under these operations. Because $R$ is an integral domain, multiplication in $R$ is commutative. Hence for all $a,b \in S$, regarding $a$ and $b$ as elements of $R$, we have \begin{align*} a b = b a. \end{align*} Thus multiplication in $S$ is commutative. The hypothesis $1_S = 1_R$ gives a multiplicative identity for $S$ compatible with the identity of $R$. [/step] [step:Verify that $S$ is nonzero] The theorem assumes $S \ne \{0_R\}$. Since the zero element of the subring $S$ is the same element as the zero element of $R$, this says precisely that $S$ is not the zero ring. [/step] [step:Exclude zero divisors in $S$ by viewing the equation inside $R$] Let $a,b \in S$ satisfy $a \ne 0_R$, $b \ne 0_R$, and \begin{align*} a b = 0_R \end{align*} as an equality in $S$. Since the multiplication and zero element of $S$ are inherited from $R$, the same equality holds in $R$. But $R$ is an integral domain, so it has no zero divisors. Therefore an equality $a b = 0_R$ in $R$ with $a,b \in R$ forces $a = 0_R$ or $b = 0_R$, contradicting the assumptions $a \ne 0_R$ and $b \ne 0_R$. Hence no two nonzero elements of $S$ multiply to $0_R$. [guided] We need to prove that $S$ has no zero divisors. The only available source of this property is that $R$ has no zero divisors, so the key point is that equations in $S$ are also equations in $R$. Take arbitrary elements $a,b \in S$ and assume that both are nonzero: \begin{align*} a \ne 0_R \end{align*} and \begin{align*} b \ne 0_R. \end{align*} Suppose also that their product is zero in $S$: \begin{align*} a b = 0_R. \end{align*} This equality can be read inside $R$ because $S$ is a subring of $R$: the multiplication on $S$ is the restriction of the multiplication on $R$, and the zero element of $S$ is the same element as $0_R$. Now use the defining zero-divisor property of an integral domain. Since $R$ is an integral domain, whenever $x,y \in R$ satisfy $xy = 0_R$, one has $x = 0_R$ or $y = 0_R$. Applying this to $x=a$ and $y=b$ gives \begin{align*} a = 0_R \quad \text{or} \quad b = 0_R. \end{align*} This contradicts the assumptions that $a$ and $b$ were both nonzero. Therefore there are no nonzero elements $a,b \in S$ with $ab=0_R$, so $S$ has no zero divisors. [/guided] [/step] [step:Conclude that $S$ is an integral domain] We have shown that $S$ is a nonzero commutative ring with multiplicative identity $1_S$ and with no zero divisors. These are exactly the defining properties of an integral domain. Therefore $S$ is an integral domain. [/step]