# irreducible of a UFD is prime

Any irreducible element^{} of a factorial ring $D$ is a prime element^{} of $D$.

Proof. Let $p$ be an arbitrary irreducible element of $D$. Thus $p$ is a non-unit. If $ab\in (p)\setminus \{0\}$, then $ab=cp$ with $c\in D$. We write $a,b,c$ as products of irreducibles:

$$a={p}_{1}\mathrm{\cdots}{p}_{l},b={q}_{1}\mathrm{\cdots}{q}_{m},c={r}_{1}\mathrm{\cdots}{r}_{n}$$ |

Here, one of those first two products may me empty, i.e. it may be a unit. We have

${p}_{1}\mathrm{\cdots}{p}_{l}{q}_{1}\mathrm{\cdots}{q}_{m}={r}_{1}\mathrm{\cdots}{r}_{n}p.$ | (1) |

Due to the uniqueness of prime factorization^{}, every factor ${r}_{k}$ is an associate^{} of certain of the $l+m$ irreducibles on the left hand side of (1). Accordingly, $p$ has to be an associate of one of the ${p}_{i}$’s or ${q}_{j}$’s. It means that either $a\in (p)$ or $b\in (p)$. Thus, $(p)$ is a prime ideal^{} of $D$, and its generator^{} must be a prime element.

Title | irreducible of a UFD is prime |
---|---|

Canonical name | IrreducibleOfAUFDIsPrime |

Date of creation | 2013-03-22 18:04:35 |

Last modified on | 2013-03-22 18:04:35 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 7 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 13G05 |

Classification | msc 13F15 |

Related topic | PrimeElementIsIrreducibleInIntegralDomain |