Let f(x)∈Z[x]. If we reduce the coefficents of f(x) modulo p, where p is prime, we get a polynomial f∗(x)∈Fp[x]. Then if f∗(x) is irreducible and has the same degree as f(x), the polynomial f(x) is irreducible.

This is one way to show that a polynomial in Z[x] is irreducible, but it does not always work. There are polynomials which are irreducible in Z[x] yet factor in Fp[x] for every prime p. The only examples I know are x4+1 and x4−10x2+1.

I’d like to see more examples, in particular an infinite family of polynomials like this would be interesting. How does one go about finding them? Has anyone ever attempted classifying all polynomials in Z[x] with this property?

**Answer**

There are two crucial results here.

**Dedekind’s theorem:** Let f be a monic irreducible polynomial over Z of degree n and let p be a prime such that f has distinct roots mod (this is true for precisely the primes not dividing the discriminant). Suppose that the prime factorization of f \bmod p is

f \equiv \prod_{i=1}^k f_i \bmod p.

Then the Galois group G of f contains an element of cycle type (\deg f_1, \deg f_2, …). In particular, if f is irreducible \bmod p, then G contains an n-cycle.

**Frobenius density theorem:** The density of the primes with respect to which the factorization of f \bmod p has the above form is equal to the density of elements of G with the corresponding cycle type. In particular, for every cycle type there is at least one such prime p.

It follows that

f is reducible \bmod p for all p if and only if G does not contain an n-cycle.

The smallest value of n for which this is possible is n = 4, where the Galois group V_4 \cong C_2 \times C_2 has no 4-cycle. Thus to write down a family of examples it suffices to write down a family of irreducible quartics with Galois group V_4. As discussed for example in this math.SE question, if

f(x) = (x^2 – a)^2 – b

is irreducible and a^2 – b is a square, then f has Galois group V_4. In particular, taking b = a^2 – 1 the problem reduces to finding infinitely many a such that

f(x) = x^4 – 2ax^2 + 1

is irreducible. We get your examples by setting a = 0, 5.

By the rational root theorem, the only possible rational roots of f are \pm 1, so by taking a \neq 1 we already guarantee that f has no rational roots. If f splits into two quadratic factors, then they both have constant term \pm 1, so we can write

x^4 – 2ax + 1 = (x^2 – bx \pm 1)(x^2 + bx \pm 1)

for some b. This gives

2a = b^2 \mp 2.

Thus f is irreducible if and only if 2a cannot be written in the above form (and also a \neq 1).

Classifying polynomials f with this property seems quite difficult in general. When n = 4, it turns out that V_4 is in fact the only transitive subgroup of S_4 not containing a 4-cycle, but for higher values of n there should be lots more, and then one has to tell whether a polynomial has one of these as a Galois group…

(Except if n = q is prime; in this case q | |G| so it must have a q-cycle.)

**Attribution***Source : Link , Question Author : spin , Answer Author : Community*