This result is very useful for producing examples of irreducible polynomials.

Theorem Let f=anxn + an-1xn-1 + ... + a0 be a nonconstant polynomial with integer coefficients and let p be a prime number. Suppose that

• p does not divide an
• p|an-1,...,a0
• p2 doesn't divide a0.
Then f is an irreducible polynomial in Q[x].

Examples

• x6 - 30x5 + 6x4 - 18x3 + 12x2 - 6x +12 is irreducible in Q[x] by Eisenstein with p=3 (note that we can't use p=2 :)
• xn - 2 is irreducible by Eisenstein with p=2.
• Consider f(x)=x3- 3x - 1. We can't apply Eisenstein directly but consider f(x+1) (Obviously if f(x+1) is irreducible then so is f(x).) We have f(x+1)=(x+1)3 - 3(x+1) - 1 = x3 + 3x2 -3. By Eisenstein (p=3) we deduce that f(x) is irreducible.
In fact Eisenstein's criterion is a special case of a more general result.

Theorem Let R be a unique factorization domain with field of fractions K. Let f=anxn + an-1xn-1 + ... + a0 be a nonconstant polynomial in R[x]. Let p be a prime in R. Suppose that

• p does not divide an
• p|an-1,...,a0
• p2 doesn't divide a0.
Then f is an irreducible polynomial in K[x].

Proof of Eisenstein's criterion: Firstly, we can assume that f is primitive, for if we write f=ch, with c the content and h primitive then since p doesn't divide an it doesn't divide c. It follows quickly that h also satisfies the conditions of the criterion. Finally, if h is irreducible then so is f.

By Gauss's Lemma, if f fails to be irreducible in K[x] then it has a factorization f=f1f2 in R[x] so that f1,f2 both have degree < deg f. Let's say that f1=c0+ ... crxr and f2=d0+ ... drxr. Now a0=c0d0 and a0 is divisible by p but not by p2. Thus one of c0 and d0 is divisble by p and the other is not. WLOG p|c0. but doesn't divide d0. Now p does not divide an=crds so it doesn't divide cr. Let k be the smallest integer such that p does not divide ck. Thus, k>0 and k<=r. Now ak=c0dk+ ... + ckd0. We know that p|ak and p|c0,...,ck-1 so it follows that p|ckd0. But p doesn't divide either of the two terms in this product and so this contradicts the primeness of p, completing the proof.

Let p be a prime integer, and let C(p,i) denote the binomial coefficient. We can use EIC to prove that the polynomial f(x) = xp-1 + ... + x + 1 is irreducible over Q - i.e. that it has no non-trivial factorisation.

Proof:

Suppose that f(x) is reducible; that is, it can be written as f(x)=r(x)*s(x) such that both r(x) and s(x) have degrees less than f(x). Define f'(x), r'(x) and s'(x) by f'(x) = f(x+1), r'(x) = r(x+1) and s'(x) = s(x+1). It is clear that the degree of these new polynomials are the same as the originals, and furthermore:

```    f'(x) = f(x+1) = r(x+1)*s(x+1) = r'(x)*s'(x)
```
So now f'(x) must also be reducible. Noticing that
```            xp - 1
f(x) = --------
x - 1
```
it follows that:
```                      (x+1)p - 1
f'(x) = f(x+1) = ------------
(x+1) - 1
```
Using the binomial theorem to expand the numerator and simplifying, we get that:
```    f'(x) = C(p,1) + C(p,2)*x + ... + C(p,p-1)*xp-2 + xp-1
```
Now since p divides all those coeffients except for the coefficient of xp-1, and C(p,1) = p, we can use EIC (see Noether's writeup above) to deduce that f'(x) is irreducible, which gives us a contradiction!

Hence f(x) = xp-1 + ... + x + 1 is irreducible over Q.