proof of Eisenstein criterion

Let f(x)R[x] be a polynomialMathworldPlanetmathPlanetmath satisfying Eisenstein’s Criterion with prime p.

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

Let f(x)=i=0naixi, g(x)=j=0bjxj, h(x)=k=0mckxk be the expansions of f(x),g(x), and h(x) respectively.

Let φ:R[x]R/pR[x] be the natural homomorphismMathworldPlanetmathPlanetmath from R[x] to R/pR[x]. Note that since pai for i<n and pan, we have φ(ai)=0 for i<n and φ(ai)=α0


Therefore we have αxn=φ(f(x))=φ(g(x)h(x))=φ(g(x))φ(h(x)) so we must have φ(g(x))=βx and φ(h(x)=γxm for some β,γR/pR and some integers ,m.

Clearly deg(g(x))= and mdeg(h(x))=m, and therefore since m=n=m, we must have = and m=m. Thus φ(g(x))=βx and φ(h(x))=γxm.

If >0, then φ(bi)=0 for i<. In particular, φ(b0)=0, hence pb0. Similarly if m>0, then pc0.

Since f(x)=g(x)h(x), by equating coefficients we see that a0=b0c0.

If >0 and m>0, then pb0 and pc0, which implies that p2a0. But this contradicts our assumptions on f(x), and therefore we must have =0 or m=0, that is, we must have a trivial factorization. Therefore f(x) is irreduciblePlanetmathPlanetmathPlanetmath.

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