f can be written as a product of irreducible factors

We show by induction on deg(f) that f can be written as a product of irreducible factors.

• The case deg(f) = 1.
  Then f itself is obviously irreducible and we are in the case s = 1 and f = p₁.
• The case deg(f) > 1.
  The induction hypothesis says that every polynomial of degree smaller than deg(f) can be written as a product of irreducible factors. If f is irreducible, we are done. If not, then f has a divisor g satisfying 0 < deg(g) < deg(f). The induction hypothesis implies that g and f/g can both be written as a product of irreducible factors:

  g = p₁ ··· p_r,

  f/g = p_{r + 1} ··· p_s.

  Since f = g · f/g, we find

  f = p₁ ··· p_r·p_{r + 1} ··· p_s.

Uniqueness

We prove the unicity of the factorization. Again we use induction. Put n = deg(f). The case n = 1 is simple and left to the reader. Now suppose that n > 1, and suppose that uniqueness has been proved for polynomials of degree < n.

If f = p₁ ··· p_r and f = q₁ ··· q_s are two possible ways of writing f as a product of irreducible factors, then it follows that p₁ | p₁ ·· · p_r = q₁ ··· q_s. From Corollary we conclude that there exists an index k {1, ..., s} with p₁ | q_k. But then p ₁ = q_k because q_k is irreducible. Apply the induction hypothesis to the polynomial f/p₁ with the two ways of writing it as a product of irreducible factors:

These factorizations of f/p₁ are equal (up to the order of the factors and multiplications by constants), and thus the factorizations of f are equal (up to the order of the factors and multiplications by constants).