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# 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 $R[X_{1},\,X_{2},\,\ldots,\,X_{n}]$.

Defines:

coefficient ring

Related:

RingAdjunction, FormalPowerSeries, ZeroPolynomial2, PolynomialRingOverFieldIsEuclideanDomain

Type of Math Object:

Theorem

Major Section:

Reference

Parent:

Groups audience:

## Mathematics Subject Classification

13P05*no label found*

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