# proof of Eisenstein criterion

Let $f(x)\in R[x]$ be a polynomial^{} satisfying Eisenstein’s Criterion with prime $p$.

Suppose that $f(x)=g(x)h(x)$ with $g(x),h(x)\in F[x]$, where $F$ is the field of fractions of $R$. Gauss’ Lemma II there exist ${g}^{\prime}(x),{h}^{\prime}(x)\in R[x]$ such that $f(x)={g}^{\prime}(x){h}^{\prime}(x)$, i.e. any factorization can be converted to a factorization in $R[x]$.

Let $f(x)={\sum}_{i=0}^{n}{a}_{i}{x}^{i}$, ${g}^{\prime}(x)={\sum}_{j=0}^{\mathrm{\ell}}{b}_{j}{x}^{j}$, ${h}^{\prime}(x)={\sum}_{k=0}^{m}{c}_{k}{x}^{k}$ be the expansions of $f(x),{g}^{\prime}(x)$, and ${h}^{\prime}(x)$ respectively.

Let $\phi :R[x]\to R/pR[x]$ be the natural homomorphism^{} from $R[x]$ to $R/pR[x]$. Note that since $p\mid {a}_{i}$ for $$ and $p\nmid {a}_{n}$, we have $\phi ({a}_{i})=0$ for $$ and $\phi ({a}_{i})=\alpha \ne 0$

$$\phi (f(x))=\phi \left(\sum _{i=0}^{n}{a}_{i}{x}^{i}\right)=\sum _{i=0}^{n}\phi ({a}_{i}){x}^{i}=\phi ({a}_{n}){x}^{n}=\alpha {x}^{n}$$ |

Therefore we have $\alpha {x}^{n}=\phi (f(x))=\phi ({g}^{\prime}(x){h}^{\prime}(x))=\phi ({g}^{\prime}(x))\phi ({h}^{\prime}(x))$ so we must have $\phi ({g}^{\prime}(x))=\beta {x}^{{\mathrm{\ell}}^{\prime}}$ and $\phi ({h}^{\prime}(x)=\gamma {x}^{{m}^{\prime}}$ for some $\beta ,\gamma \in R/pR$ and some integers ${\mathrm{\ell}}^{\prime},{m}^{\prime}$.

Clearly ${\mathrm{\ell}}^{\prime}\le \mathrm{deg}({g}^{\prime}(x))=\mathrm{\ell}$ and ${m}^{\prime}\le \mathrm{deg}({h}^{\prime}(x))=m$, and therefore since ${\mathrm{\ell}}^{\prime}{m}^{\prime}=n=\mathrm{\ell}m$, we must have ${\mathrm{\ell}}^{\prime}=\mathrm{\ell}$ and ${m}^{\prime}=m$. Thus $\phi ({g}^{\prime}(x))=\beta {x}^{\mathrm{\ell}}$ and $\phi ({h}^{\prime}(x))=\gamma {x}^{m}$.

If $\mathrm{\ell}>0$, then $\phi ({b}_{i})=0$ for $$. In particular, $\phi ({b}_{0})=0$, hence $p\mid {b}_{0}$. Similarly if $m>0$, then $p\mid {c}_{0}$.

Since $f(x)={g}^{\prime}(x){h}^{\prime}(x)$, by equating coefficients we see that ${a}_{0}={b}_{0}{c}_{0}$.

If $\mathrm{\ell}>0$ and $m>0$, then $p\mid {b}_{0}$ and $p\mid {c}_{0}$, which implies that ${p}^{2}\mid {a}_{0}$. But this contradicts our assumptions on $f(x)$, and therefore we must have $\mathrm{\ell}=0$ or $m=0$, that is, we must have a trivial factorization. Therefore $f(x)$ is irreducible^{}.

Title | proof of Eisenstein criterion |
---|---|

Canonical name | ProofOfEisensteinCriterion |

Date of creation | 2013-03-22 12:42:11 |

Last modified on | 2013-03-22 12:42:11 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 11 |

Author | rspuzio (6075) |

Entry type | Proof |

Classification | msc 11C08 |

Classification | msc 13F15 |

Related topic | GausssLemmaII |