Detect repeated roots through derivatives

Try

The polynomial (x−a)² has a repeated root, and its derivative vanishes at that root too. Predict whether this common-root condition characterizes repeated roots for an arbitrary polynomial. Factor out the exact power of x−a before differentiating.

A polynomial is separable when its roots in a splitting field are distinct. An algebraic element is separable when its minimal polynomial is separable. The definition refers to a splitting field, not only roots visible in the base field.

Predict

In characteristic zero, an irreducible positive-degree polynomial has a nonzero derivative of smaller degree. Predict why it cannot share a nonconstant factor with that derivative. Then consider xᵖ−t in characteristic p, whose derivative is zero. The characteristic-zero reasoning has lost its essential step.

Experiment

Compare the derivative mechanisms.

Polynomial and field	Derivative	Consequence
x³−2 over Q	3x²	Coprime to the polynomial
x⁴−x over F₂	−1=1	Every root simple
xᵖ−t over F_p(t)	0	A repeated-root possibility

For f=(x−a)ᵐg with g(a) nonzero, differentiation gives f′=m(x−a)^(m−1)g+(x−a)ᵐg′. If m=1, f′(a)=g(a) is nonzero. If m≥2, both terms vanish at a, even if the integer coefficient m also becomes zero in the field.

Understand

The factor calculation proves that repeated roots are exactly common roots of f and f′. A common root implies their gcd has positive degree, since a Bézout identity for gcd one would evaluate to one equals zero. Conversely a nonconstant gcd has a root in an algebraic closure and therefore supplies a common root. Thus separability is equivalent to gcd(f,f′)=1.

For an irreducible polynomial over a characteristic-zero field, f′ is nonzero and lower degree. Irreducibility forces the gcd to be one. Hence every algebraic extension in characteristic zero is separable. A field over which every irreducible polynomial is separable is called perfect.

Finite fields are perfect too. If an irreducible f over a finite field had repeated roots, irreducibility and the gcd criterion would force f′=0. In characteristic p this means every nonzero term has exponent divisible by p. Frobenius is onto the finite coefficient field, so each coefficient has a pth root. Therefore f is the pth power of a positive-degree polynomial, contradicting irreducibility. Finiteness is used for coefficient roots, not merely for the derivative calculation.

An infinite characteristic-p counterexample is F_p(t). The polynomial xᵖ−t is Eisenstein at the prime element t in the UFD F_p[t], so Gauss’s lemma makes it irreducible over its fraction field. If αᵖ=t in an extension, then xᵖ−t=(x−α)ᵖ. It has degree p but only one distinct root. Thus the extension has degree p while the usual count of distinct conjugates is one.

This example also shows why irreducible and separable are different conditions. Irreducibility concerns factors over the base field; separability concerns multiplicities after all roots become available. A polynomial may be irreducible at home yet acquire a repeated factor in a larger field.

Apply

Compute a formal derivative in F₅, reducing integer coefficients modulo five. Explain why the x⁵ term contributes zero while an x² term generally does not. Then use the gcd criterion instead of concluding that any derivative with a missing term means inseparability.

For a field claimed perfect, identify the actual proof: characteristic zero or coefficientwise Frobenius surjectivity. The statement “all fields are perfect” is disproved by the rational-function example, and the failed hypothesis should be named precisely.

Continue

Normality next asks whether all conjugate roots remain inside the extension. Separability counts distinct embeddings; normality determines whether those embeddings become automorphisms of the same field.

For a coefficient certificate of the same repeated-root condition, continue to discriminants and repeated roots. The derivative product and quotient-algebra determinant provide a second way to check it.