# Eisenstein criterion

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

 $f(x)=a_{0}+a_{1}x+a_{2}x^{2}+\ldots+a_{n}x^{n}\;.$

If $R$ has an irreducible element  $p$ such that

 $p\mid a_{m}\qquad 0\leq m\leq n-1$
 $p^{2}\nmid a_{0}$
 $p\nmid a_{n}$

###### Proof.

Suppose

 $f=(b_{0}+\ldots+b_{s}x^{s})(c_{0}+\ldots+c_{t}x^{t})$

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 $a_{k}=b_{0}c_{k}+b_{1}c_{k-1}+\ldots+b_{k}c_{0}$. We also have $p\mid a_{k}$, and $p$ divides every summand except one on the right side, which yields a contradiction   . QED ∎

Title Eisenstein criterion EisensteinCriterion 2013-03-22 12:16:32 2013-03-22 12:16:32 Daume (40) Daume (40) 13 Daume (40) Theorem msc 13A05 Eisenstein irreducibility criterion GausssLemmaII IrreduciblePolynomial2 Monic2 AlternativeProofThatSqrt2IsIrrational