# UFD’s are integrally closed

Theorem: Every UFD is integrally closed^{}.

Proof: Let $R$ be a UFD, $K$ its field of fractions^{}, $u\in K,u$ integral over $R$. Then for some ${c}_{0},\mathrm{\dots},{c}_{n-1}\in R$,

$${u}^{n}+{c}_{n-1}{u}^{n-1}+\mathrm{\dots}+{c}_{0}=0$$ |

Write $u=\frac{a}{b},a,b\in R$, where $a,b$ have no non-unit common divisor (which we can assume since $R$ is a UFD). Multiply the above equation by ${b}^{n}$ to get

$${a}^{n}+{c}_{n-1}b{a}^{n-1}+\mathrm{\dots}+{c}_{0}{b}^{n}=0$$ |

Let $d$ be an irreducible^{} divisor of $b$. Then $d$ is prime since $R$ is a UFD. Now, $d|{a}^{n}$ since it divides all the other terms and thus (since $d$ is prime) $d|a$. But $a,b$ have no non-unit common divisors, so $d$ is a unit. Thus $b$ is a unit and hence $u\in R$.

Title | UFD’s are integrally closed |
---|---|

Canonical name | UFDsAreIntegrallyClosed |

Date of creation | 2013-03-22 15:49:25 |

Last modified on | 2013-03-22 15:49:25 |

Owner | rm50 (10146) |

Last modified by | rm50 (10146) |

Numerical id | 6 |

Author | rm50 (10146) |

Entry type | Theorem |

Classification | msc 13G05 |