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'polynomial ring over integral domain'
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| Title of object: |
polynomial ring over integral domain |
| Canonical Name: |
PolynomialRingOverIntegralDomain |
| Type: |
Theorem |
| Created on: |
2005-03-30 06:20:17 |
| Modified on: |
2005-03-30 06:23:21 |
| Classification: |
msc:13P05 |
| Defines: |
coefficient ring |
Revision comment (for changes between this and next version):
| Changes for correction #9547 ('spelling'). |
Preamble:
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\usepackage{amssymb}
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%\usepackage{psfrag}
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\newtheorem*{thmplain}{Theorem} |
Content:
\begin{thmplain}
\, If the {\em coefficient ring} $R$ is an integral domain, then so is also the 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 $f(X)g(X)$ can not 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 $R[X_1,\,X_2,\,...,\,X_n]$. |
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