This result is very useful for producing examples of
irreducible
polynomials.
Theorem
Let f=a_{n}x^{n} + a_{n1}x^{n1} + ... + a_{0}
be a nonconstant polynomial with integer coefficients and let
p be a prime number.
Suppose that
 p does not divide a_{n}

pa_{n1},...,a_{0}

p^{2} doesn't
divide a_{0}.
Then
f is an irreducible polynomial
in
Q[x].
Examples

x^{6}  30x^{5} + 6x^{4}  18x^{3} +
12x^{2}  6x +12
is irreducible in Q[x] by Eisenstein
with p=3 (note that we can't use p=2 :)

x^{n}  2 is irreducible by Eisenstein with p=2.

Consider f(x)=x^{3} 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 = x^{3} + 3x^{2} 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=a_{n}x^{n} + a_{n1}x^{n1} + ... + a_{0}
be a nonconstant polynomial in R[x]. Let p be a
prime in R.
Suppose that
 p does not divide a_{n}

pa_{n1},...,a_{0}

p^{2} doesn't
divide a_{0}.
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 a_{n} 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=f_{1}f_{2} in
R[x] so that f_{1},f_{2}
both have degree < deg f.
Let's say that f_{1}=c_{0}+ ... c_{r}x^{r}
and
f_{2}=d_{0}+ ... d_{r}x^{r}.
Now a_{0}=c_{0}d_{0} and a_{0}
is divisible by p but not by p^{2}. Thus one of
c_{0} and d_{0} is divisble by p
and the other is not. WLOG pc_{0}.
but doesn't divide d_{0}.
Now p does not divide a_{n}=c_{r}d_{s}
so it doesn't divide c_{r}. Let k be the smallest
integer such that p does not divide c_{k}.
Thus, k>0 and k<=r. Now
a_{k}=c_{0}d_{k}+ ... + c_{k}d_{0}.
We know that pa_{k} and pc_{0},...,c_{k1}
so it follows that pc_{k}d_{0}. But p
doesn't divide either of the two terms in this product and so this contradicts
the primeness of p, completing the proof.