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[parent] polynomial ring over integral domain (Theorem)
Theorem 1   If the coefficient ring $ R$ is an integral domain, then so is also the 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_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.

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



"polynomial ring over integral domain" is owned by pahio.
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See Also: ring adjunction, formal power series, zero polynomial, polynomial ring over a field

Also defines:  coefficient ring

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polynomial ring which is PID (Theorem) by pahio
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Cross-references: induction, zero polynomial, product, zero divisors, leading coefficients, polynomials, proof, polynomial ring, integral domain
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This is version 6 of polynomial ring over integral domain, born on 2005-03-30, modified 2008-03-05.
Object id is 6918, canonical name is PolynomialRingOverIntegralDomain.
Accessed 2927 times total.

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AMS MSC: 13P05 (Commutative rings and algebras :: Computational aspects of commutative algebra :: Polynomials, factorization)
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