polynomial ring over integral domainPolynomialRingOverIntegralDomain2005-03-30 06:20:172008-08-02 07:52:15Theorempolynomial ringcoefficient ring% this is the default PlanetMath preamble. as your knowledge
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\newtheorem*{thmplain}{Theorem}\begin{thmplain}
\, If the {\em coefficient ring} $R$ is an integral domain, then so is also its polynomial ring $R[X]$.
\end{thmplain}
{\em 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\, $a_f \neq 0$,\, $b_g \neq 0$,\, and because $R$ has no zero divisors,\, $a_fb_g \neq 0$.\, But the product $a_fb_g$ 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.\\
\textbf{Remark.}\, The theorem may by induction be generalized for the polynomial ring\, $R[X_1,\,X_2,\,\ldots,\,X_n]$.