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

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