[proofplan] We verify directly that $R[x]$ has no zero divisors. Given nonzero polynomials $f,g \in R[x]$, choose their highest nonzero coefficients and compute the coefficient of the highest possible power in $fg$. Since $R$ is an [integral domain](/page/Integral%20Domain), the product of those leading coefficients is nonzero, so $fg \ne 0$. The remaining ring axioms required for an integral domain are inherited by the [polynomial ring](/page/Polynomial%20Ring) construction from the commutative ring $R$. [/proofplan]

[step:Record the ring structure inherited by $R[x$]] Since $R$ is an integral domain, $R$ is a nonzero commutative ring with multiplicative identity $1_R$ and no zero divisors. By the definition of the polynomial ring over a commutative ring, $R[x]$ is a commutative ring with multiplicative identity the constant polynomial $1_R$. Its zero element is the constant polynomial $0_R$. Because $1_R \ne 0_R$ in $R$, the constant polynomial $1_R$ is not equal to the constant polynomial $0_R$ in $R[x]$. Thus $R[x]$ is a nonzero commutative ring with identity. It remains only to prove that $R[x]$ has no zero divisors. [/step]

[step:Compute the leading coefficient of a product of nonzero polynomials]Let $f,g \in R[x]$ be nonzero polynomials. Since $f$ is nonzero, there exist an integer $m \ge 0$ and coefficients $a_0,\dots,a_m \in R$ with $a_m \ne 0_R$ such that \begin{align*} f = \sum_{i=0}^{m} a_i x^i. \end{align*} Similarly, since $g$ is nonzero, there exist an integer $n \ge 0$ and coefficients $b_0,\dots,b_n \in R$ with $b_n \ne 0_R$ such that \begin{align*} g = \sum_{j=0}^{n} b_j x^j. \end{align*} By the definition of multiplication in $R[x]$, the coefficient of $x^{m+n}$ in $fg$ is \begin{align*} \sum_{i+j=m+n} a_i b_j. \end{align*} For every pair $(i,j)$ with $0 \le i \le m$, $0 \le j \le n$, and $i+j=m+n$, we must have $i=m$ and $j=n$. Indeed, if $i<m$, then $j=m+n-i>n$, which is impossible; similarly, if $j<n$, then $i>m$, which is impossible. Hence the coefficient of $x^{m+n}$ in $fg$ is exactly $a_m b_n$.[/step]

[guided]The point of choosing $m$ and $n$ is that they mark the highest powers of $x$ that actually occur in the two nonzero polynomials. We are not assigning a degree to the zero polynomial: the assumption $f \ne 0$ gives a highest index $m \ge 0$ with nonzero coefficient, and the assumption $g \ne 0$ gives a highest index $n \ge 0$ with nonzero coefficient. Thus we may write \begin{align*} f = \sum_{i=0}^{m} a_i x^i \end{align*} with $a_m \ne 0_R$, and \begin{align*} g = \sum_{j=0}^{n} b_j x^j \end{align*} with $b_n \ne 0_R$. Now examine the coefficient of the highest possible power $x^{m+n}$ in the product. By the multiplication rule for polynomials, this coefficient is the sum of all products $a_i b_j$ whose exponents add to $m+n$: \begin{align*} \sum_{i+j=m+n} a_i b_j. \end{align*} But among the available indices $0 \le i \le m$ and $0 \le j \le n$, the only way to have $i+j=m+n$ is to take $i=m$ and $j=n$. If $i<m$, then the equation $i+j=m+n$ forces $j>n$, outside the range of coefficients of $g$. If $j<n$, it forces $i>m$, outside the range of coefficients of $f$. Therefore the coefficient of $x^{m+n}$ in $fg$ is precisely \begin{align*} a_m b_n. \end{align*}[/guided]

[step:Use the domain property of $R$ to rule out zero divisors in $R[x]$] The coefficients $a_m$ and $b_n$ are nonzero elements of $R$. Since $R$ is an integral domain, it has no zero divisors, so \begin{align*} a_m b_n \ne 0_R. \end{align*} Therefore the coefficient of $x^{m+n}$ in $fg$ is nonzero, and hence $fg$ is not the zero polynomial. We have shown that whenever $f,g \in R[x]$ are nonzero, their product $fg$ is nonzero. Thus $R[x]$ has no zero divisors. Since $R[x]$ is also a nonzero commutative ring with identity, $R[x]$ is an integral domain. [/step]