Irreducible Polynomials over Finite Fields

As we will see, modular arithmetic aids in testing the irreducibility of polynomials and even in completely factoring polynomials in Z[x]. If we expect a polynomial f(x) is irreducible, for example, it is not unreasonable to try to find a prime p such that f(x) is irreducible modulo p. If we can find such a prime p and p does not divide the leading coefficient of f(x), then f(x) is irreducible in Z[x] (see the example on page 4). It is the case that there exist polynomials which are irreducible in Z[x] but are reducible modulo every prime (see Exercise (4.1)), but as it turns out, one can show that such polynomials are rare and verifying that a polynomial f(x) is irreducible by trying to find a prime p for which f(x) is irreducible modulo p will almost always work rather quickly (see Chapter 5). This is already strong motivation for looking into the idea of using modular arithmetic, but in this chapter, we plan to explore other aspects of modular arithmetic as well. We begin with a definition. Let p be a prime, and let f(x) ∈ Z[x]. Suppose further that f(x) 6≡ 0 (mod p). We say that u(x) ≡ v(x) (mod p, f(x)), where u(x) and v(x) are in Z[x], if there exist g(x) and h(x) in Z[x] such that u(x) = v(x) + f(x)g(x) + ph(x). (In other words, u(x) ≡ v(x) (mod p, f(x)) if u(x)− v(x) is in the ideal generated by p and f(x) in the ring Z[x].) One easily checks that if u(x) ≡ v(x) (mod p, f(x)) and v(x) ≡ w(x) (mod p, f(x)), then u(x) ≡ w(x) (mod p, f(x)). Suppose that u1(x) ≡ v1(x) (mod p, f(x)) and u2(x) ≡ v2(x) (mod p, f(x)). Then u1(x)± u2(x) ≡ v1(x)± v2(x) (mod p, f(x)).