polynomial ring over integral domain

Theorem.

If the coefficient ring $R$ is an integral domain, then so is also its polynomial ring $R[X]$.

Proof.  Let $f(X)$ and $g(X)$ be two non-zero polynomials in $R[X]$ and let $a_f$ and $b_g$ be their leading coefficients, respectively.  Thus  $a_f\ne 0$,  $b_g\ne 0$,  and because $R$ has no zero divisors,  $a_f{b}_g\ne 0$.  But the product $a_f{b}_g$ is the leading coefficient of $f(X)g(X)$ and so $f(X)g(X)$ cannot be the zero polynomial.  Consequently, $R[X]$ has no zero divisors, Q.E.D.

Remark.  The theorem may by induction be generalized for the polynomial ring  $R[X_1,X_2,\mathrm{\dots },X_n]$.