# polynomial ring which is PID

Theorem. If a polynomial ring $D[X]$ over an integral domain^{} $D$ is a principal ideal domain^{}, then coefficient ring $D$ is a field. (Cf. the corollary 4 in the entry polynomial ring over a field.)

Proof. Let $a$ be any non-zero element of $D$. Then the ideal $(a,X)$ of $D[X]$ is a principal ideal^{} $(f(X))$ with $f(X)$ a non-zero polynomial (http://planetmath.org/ZeroPolynomial2). Therefore,

$$a=f(X)g(X),X=f(X)h(X)$$ |

with $g(X)$ and $h(X)$ certain polynomials in $D[X]$. From these equations one infers that $f(X)$ is a polynomial $c$ and $h(X)$ is a first degree polynomial ${b}_{0}+{b}_{1}X$ (${b}_{1}\ne 0$). Thus we obtain the equation

$$c{b}_{0}+c{b}_{1}X=X,$$ |

which shows that $c{b}_{1}$ is the unity 1 of $D$. Thus $c=f(X)$ is a unit of $D$, whence

$$(a,X)=(f(X))=(1)=D[X].$$ |

So we can write

$$1=a\cdot u(X)+X\cdot v(X),$$ |

where $u(X),v(X)\in D[X]$. This equation cannot be possible without that $a$ times the constant term of $u(X)$ is the unity. Accordingly, $a$ has a multiplicative inverse^{} in $D$. Because $a$ was arbitrary non-zero elenent of the integral domain $D$, $D$ is a field.

## References

- 1 David M. Burton: A first course in rings and ideals. Addison-Wesley Publishing Company. Reading, Menlo Park, London, Don Mills (1970).

Title | polynomial ring which is PID |
---|---|

Canonical name | PolynomialRingWhichIsPID |

Date of creation | 2013-03-22 17:53:04 |

Last modified on | 2013-03-22 17:53:04 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 8 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 13P05 |

Related topic | PolynomialRingOverFieldIsEuclideanDomain |