# Eisenstein criterion

###### Theorem (Eisenstein criterion).

Let $f$ be a primitive polynomial over a commutative^{} unique factorization domain^{} $R$, say

$$f(x)={a}_{0}+{a}_{1}x+{a}_{2}{x}^{2}+\mathrm{\dots}+{a}_{n}{x}^{n}.$$ |

If $R$ has an irreducible element^{} $p$ such that

$$p\mid {a}_{m}\mathit{\hspace{1em}\hspace{1em}}0\le m\le n-1$$ |

$${p}^{2}\nmid {a}_{0}$$ |

$$p\nmid {a}_{n}$$ |

then $f$ is irreducible^{}.

###### Proof.

Suppose

$$f=({b}_{0}+\mathrm{\dots}+{b}_{s}{x}^{s})({c}_{0}+\mathrm{\dots}+{c}_{t}{x}^{t})$$ |

where $s>0$ and $t>0$. Since ${a}_{0}={b}_{0}{c}_{0}$, we know that $p$ divides one but not both of ${b}_{0}$ and ${c}_{0}$; suppose $p\mid {c}_{0}$. By hypothesis^{}, not all the ${c}_{m}$ are divisible by $p$; let $k$ be the smallest index such that $p\nmid {c}_{k}$. We have ${a}_{k}={b}_{0}{c}_{k}+{b}_{1}{c}_{k-1}+\mathrm{\dots}+{b}_{k}{c}_{0}$.
We also have $p\mid {a}_{k}$, and $p$ divides every summand except one on the right side, which yields a contradiction^{}. QED
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Title | Eisenstein criterion |
---|---|

Canonical name | EisensteinCriterion |

Date of creation | 2013-03-22 12:16:32 |

Last modified on | 2013-03-22 12:16:32 |

Owner | Daume (40) |

Last modified by | Daume (40) |

Numerical id | 13 |

Author | Daume (40) |

Entry type | Theorem |

Classification | msc 13A05 |

Synonym | Eisenstein irreducibility criterion |

Related topic | GausssLemmaII |

Related topic | IrreduciblePolynomial2 |

Related topic | Monic2 |

Related topic | AlternativeProofThatSqrt2IsIrrational |