# 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  $a_{f}\neq 0$,  $b_{g}\neq 0$,  and because $R$ has no zero divisors,  $a_{f}b_{g}\neq 0$.  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  $R[X_{1},\,X_{2},\,\ldots,\,X_{n}]$.

Title polynomial ring over integral domain PolynomialRingOverIntegralDomain 2013-03-22 15:10:06 2013-03-22 15:10:06 pahio (2872) pahio (2872) 10 pahio (2872) Theorem msc 13P05 RingAdjunction FormalPowerSeries ZeroPolynomial2 PolynomialRingOverFieldIsEuclideanDomain coefficient ring