Let f ∈ ℚ[X] irreducible, then f doesn’t have multiple roots in ℂ

To prove the assertion suppose f=a₀+a₁X+…+anXⁿ and let α be a multiple root of f. Consider g=an⁻¹f . Then g is irreducible, monic and has α has a multiple root (which make, by the way, g the minimal polynomial over ℚ of α) . Then α has to be a root of g’ (the derivative of g). This implies that g|g’ and so the degree of g is less or equal to g’. If g’≠0 this is not possible so g’=0. Suppose g=b₀+b₁X+…+bmXᵐ. We will prove that g=b₀. We have g’=0 so expanding the expression of g’ every coefficient has to be 0 so that b1=0,b2=0,…bm-1=0,bm=0 and so g=b0=an⁻¹a0. This implies f=a0 so f is constant, contradiction. ■

Note that the same proof can be done if f is a polynomial in K where K is a subfield of ℂ.