## Eisenstein’s Irreducibility Criterion

TheoremLet

\[f(x) = a_0 + a_1 x + ... + a_n x^n\]

be a polynomial with integer coefficients. Suppose a prime \[p\]

divides each of \[a_0, a_1, ..., a_{n-1}\]

(every coefficient except coefficient of \[x^n\]

), and that \[p^2\]

does not divide \[a_0\]

. Then \[f(x)\]

has no factors with integer coefficients.Proof

Suppose

\[f = g h\]

for polynomials \[g, h\]

with integer coefficients. Look at this factorization modulo \[p\]

: we get \[f(x) = a_n x^n\]

, so \[g(x) = b_d x^d\]

, \[h(x) = c_e x^e\]

for some constants \[b_d, c_e\]

and for some integers \[d, e\]

with \[d e = n\]

. This implies the constant term of \[g(x)\]

is a multiple of \[p\]

, and similarly for the constant term of \[h(x)\]

, hence \[p^2\]

divides the constant term of \[f(x)\]

, a contradiction.Example

Take

\[f(x)=x^3+4x^2+10x+14\]

2 divides every coefficient except the coefficient of

\[x^3\]

and \[2^2=4\]

does not divide the constant term 14. Therefore \[f(x)\]

does not factorise.