Eisenstein's Criterion
📂Abstract AlgebraEisenstein's Criterion
Theorem
f(x)=anxn+⋯+a0∈Z[x], if it satisfies the following conditions for the primes p∈Z and k=0,1,2,⋯,n−1, then f(x) is irreducible over Q.
- (i): an≡0(modp)
- (ii): ak≡0(modp)
- (iii): a0≡0(modp2)
Description
It has significance as a very simple judgment method for integer polynomials of the form f(x)=axn+b. It can be usefully discussed in relation to algebraic numbers as a judgment method for Q.
Example
Show that f(x)=25x5−9x4−3x2−12 is irreducible over Q.
Solution
- (i) 25≡0(mod3)
- (ii) −9≡−3≡0(mod3)
- (iii) 12≡0(mod9)
Applying Eisenstein’s criterion to p=3 shows that f(x) is irreducible.
Proof
If f(x)∈Z[x] is said to be f(x)=R(x)S(x) for R(x) of degree r<n and S(x) of degree s<n, then
R(x),S(x)∈Q[x]⟺R(x),S(x)∈Z[x]
is known. Assuming from Z[x] that
f(x)=(brxr+⋯+b0)(csxs+⋯+c0)
satisfies the three conditions (i), (ii), and (iii), condition (iii) means that
b0c0=a0≡0(modp2)
so b0 and c0 are not both
b0≡c0≡p≡0(modp)
Instead, considering the case where one is congruent,
{b0≡0(modp)c0≡0(modp)
then from condition (i),
brcs=an≡0(modp)
must be
{br≡0(modp)⋯(⋆)cs≡0(modp)
Now, if we call the smallest value among k that satisfies ck≡0(modp) as m, then
am=b0cm+b1cm−1+⋯+{bmc0brcm−r,r≥m,r<m
From (⋆), b0≡0(modp) and by the definition of m, cm≡0(modp), therefore,
cm−1≡⋯≡c0≡0(modp)
Thus,
am≡0(modp)
and according to conditions (i) and (ii), m=n must be.
Eventually, it is s≥m=n, which contradicts the premise that s<n.
■