# 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 ${a}_{f}\ne 0$, ${b}_{g}\ne 0$, and because $R$ has no zero divisors^{}, ${a}_{f}{b}_{g}\ne 0$. 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 $R[{X}_{1},{X}_{2},\mathrm{\dots},{X}_{n}]$.

Title | polynomial ring over integral domain |
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Canonical name | PolynomialRingOverIntegralDomain |

Date of creation | 2013-03-22 15:10:06 |

Last modified on | 2013-03-22 15:10:06 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 10 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 13P05 |

Related topic | RingAdjunction |

Related topic | FormalPowerSeries |

Related topic | ZeroPolynomial2 |

Related topic | PolynomialRingOverFieldIsEuclideanDomain |

Defines | coefficient ring |