Irreducibility

From the definition of irreducible polynomial it is clear that reducibility depends on the field 𝕂. For example the polynomial x² − 2 (with coefficients in ℚ, hence in ℝ as well) is reducible in ℝ, because x² − 2 = (x + √2) (x − √2) but not in ℚ.

We are interested in determining when a polynomial is irreducible or reducible over 𝕂, when this field could represent ℚ, ℝ or ℂ.

Theorem 4.18. (Fundamental Theorem of algebra) A polynomial f(x) in ℂ[x] of degree n ≥ 1 admits at least one root in ℂ.

We postpone the proof to a later moment when we'll developed the appropriate mathematical tools.

Corollary 4.19. Every polynomial f(x) in ℂ[x] of degree n has in ℂ exaclty n roots.

Proof. Let α₁ ∈ ℂ, a root of f(x) (which exists owing to the fundamental theorem), by Ruffini's theorem

with q(x) having coefficients in ℂ. Then q(x), for the fundamental theorem of algebra, admits a root α₂ ∈ ℂ, etc. The repetition of this argument yields to

if n = ∂f(x) = n.□

Thus any polynomial with coefficients in C[x] can be factorized on ℂ in term of linear factors.

To sum up:

Irreducible polynomials in ℂ[x] are precisely the polynomials of degree 1.

Irreducible polynomials on ℝ

Let f(x) a polynomial in ℝ[x], with degree n > 1. We can consider f(x) as a polynomial with complex coefficients. Let α ∈ ℂ a root of f(x), (α exists for the fundamental theorem of algebra). As regards a polynomial over ℝ, notice that, if α ∈ ℂ is one of its roots and is not real, then its conjugate ᾱ is a root as well; To show this fact let f(x) = ∑ⁿ_i=0 a_k x^k then

if we take the complex conjugate of both sides of the equation and noting that real numbers are conjugates of themselves

thus ᾱ is as well a root of f(x). We this premise let's suppose f(x) (with degree n > 1) to be irreducible on ℝ. Then f has not real roots. Its decomposition in linear factors on ℂ is

with ∂q(x) = n = 2t. We have ∀i = 1, .., t

Thus f(x) is written as a product of t polynomials of degree 2, with real coefficients. Under the hyphothesis of f(x) irreducible on ℝ, there can be only one factor hence f(x) is a second degree polynomial without real roots. (a second degree polynomial with having discriminant Δ < 0). Since second degree polynomial without roots is irreducible we can sump by saying

the irreducible polynomials in ℝ[x] are exactly the polynomials of degree one and the ones of degree two with Δ < 0.

Remark 4.20. For second degree polynomial onnly the implication no roots ⇒ irreducible hols true. For instance, the polynomial x⁴ + 8x² + 15 with coefficients in ℝ with no real roots may nevertheless be reducible as follows

but when f(x) ∈ 𝕂[x] is a polynomial of degree 2 or 3 with no roots, then f(x) is irreducible in 𝕂[x]. Indeed, if f(x) had a factor isation, at least one of its factors would have degree 1, so it would have a root in 𝕂.