advanced28 min studyLesson 155 in Mathematics

Separability: count distinct conjugates, not repeated copies

Detect repeated roots with a formal derivative and explain how positive characteristic permits algebraic degree without distinct conjugates.

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What you’ll understand

  • Apply the gcd criterion for a polynomial to have distinct roots.
  • Distinguish an inseparable polynomial from an inseparable algebraic element.
  • Construct a purely inseparable extension over a rational function field.

A polynomial of degree d has d roots when counted with multiplicity in a splitting field, but those need not be d different elements. This distinction changes the available symmetries. If an algebraic element has a degree-three minimal polynomial with only one distinct root, there are not three possible images for that element under embeddings. Separability identifies the case where degree really does count distinct conjugates. Characteristic zero guarantees it for irreducible polynomials; positive characteristic requires a closer look.

Differentiate a polynomial formally, not analytically

The formal derivative applies the rule that the derivative of Xⁿ is nX raised to n − 1, with coefficients interpreted in the field. No limit or metric is required. If α is a repeated root, write the polynomial with a factor (X − α)² and differentiate: both the polynomial and its derivative vanish at α. Conversely, if the polynomial and derivative share a root in a splitting field, the corresponding linear factor occurs with multiplicity at least two.

Consequently a nonzero polynomial is separable, meaning all its roots are distinct, exactly when its greatest common divisor with its derivative is one, normalized to be monic. This criterion can be checked over the original field using the Euclidean algorithm; explicit roots are unnecessary. For an irreducible polynomial, a nontrivial common divisor must be the entire polynomial. Since a nonzero derivative has smaller degree, an irreducible polynomial is inseparable precisely when its derivative is zero.

f has distinct roots⟺gcd⁡(f,f′)=1(0≠f∈F[X])f\text{ has distinct roots}\quad\Longleftrightarrow\quad\gcd(f,f')=1\qquad(0\ne f\in F[X])
Repeated roots are detected algebraically by a common factor with the formal derivative, in any characteristic.↔ Scroll wide equations horizontally.

Positive characteristic can erase the derivative

In characteristic p, the coefficient p equals zero. Thus the derivative of Xᵖ − t is zero, even though the polynomial is nonconstant. Take F = Fₚ(t), the rational function field in a transcendental indeterminate t. The polynomial Xᵖ − t is irreducible: Eisenstein at t applies in Fₚ[t][X], followed by Gauss's lemma to pass to its fraction field. A root α therefore has degree p over F.

Nevertheless the binomial identity in characteristic p gives Xᵖ − t = (X − α)ᵖ in the field containing α. There is just one distinct root. Every F-embedding must send α to a root of its minimal polynomial, so there is only one such embedding into an algebraic closure. This is a purely inseparable extension. It demonstrates that degree counts linear coordinates even when repeated roots prevent those coordinates from corresponding to different conjugate choices.

char⁡F=p,αp=t⟹Xp−t=(X−α)p,[F(α):F]=p(F=Fp(t))\operatorname{char}F=p,\quad\alpha^p=t\quad\Longrightarrow\quad X^p-t=(X-\alpha)^p,\qquad [F(\alpha):F]=p\quad(F=\mathbb F_p(t))
The displayed degree uses irreducibility over the rational function field; the factorization then shows a single root of multiplicity p.↔ Scroll wide equations horizontally.

Check the minimal polynomial, and know the perfect cases

An algebraic element is separable over F when its minimal polynomial over F has distinct roots. A field extension is separable when every element is separable. A polynomial with repeated factors can be inseparable as a polynomial even when all of its individual roots are separable elements. For instance, (X − 1)² over Q has a repeated root, but the minimal polynomial of its root 1 is X − 1, which is separable. Never use a redundant equation to diagnose an element.

Every irreducible polynomial in characteristic zero has nonzero derivative and hence is separable. Finite fields also have only separable irreducible polynomials: their Frobenius map a ↦ aᵖ is injective, therefore surjective on the finite set. If an irreducible nonconstant polynomial had zero derivative, every exponent would be a multiple of p and every coefficient would be a pth power. The polynomial would itself be a pth power, contradicting irreducibility. Thus positive characteristic permits inseparability without forcing it.

Compare two cubics in characteristic three

Over F = F₃(t), compare the root multiplicities of f(X) = X³ − t and g(X) = X³ − X − t. Is irreducibility needed to apply the distinct-root test?

  1. Compute formal derivatives in characteristic three. The coefficient 3 vanishes, so f′ = 0, whereas g′ = −1 is a nonzero constant.

  2. Since gcd(f, 0) is f, the polynomial f has repeated roots. In fact, for a root α with α³ = t, the factorization is f = (X − α)³, giving one distinct root.

  3. The gcd of g with the nonzero constant −1 is one. Thus g has three distinct roots in its splitting field. This conclusion uses no irreducibility assumption; the gcd criterion applies to every nonzero polynomial.

The result

The first cubic has one root with multiplicity three; the second has three distinct roots. Irreducibility is only needed for the separate minimal-polynomial degree claim.

Explore the model

Interactive lab

Interactive lab · function

A zero slope is weaker than a repeated-root test

Set a to a positive value and move x to zero in the lab's y = ax² family. Predict both height and tangent slope there, then check the readouts. This real quadratic has a repeated root at zero. The lab cannot switch characteristic or compute polynomial gcds, and a zero derivative at one point alone does not establish a root unless the function value is also zero.

A zero slope is weaker than a repeated-root testThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA repeated root is shared by a polynomial and its derivative; characteristic can change the derivative itself.

Check your understanding

Over a field of characteristic zero, which statement is correct?

Try it yourself

Over F₂, show that X³ + X + 1 is irreducible and separable. Explain why the derivative being nonconstant causes no problem.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

A cubic over a field is reducible only if it has a root. Test 0 and 1, then inspect common roots with X² + 1.

Compare with the solution

The values at 0 and 1 are both 1 in F₂, so the cubic has no root and is irreducible. Its derivative is X² + 1 = (X + 1)². The only root of that derivative is 1, which is not a root of the cubic. Their gcd is therefore one and the cubic is separable. The criterion asks for a coprime derivative, not a constant derivative.

Take it with you

  • Separability of an element is tested on its minimal polynomial.
  • A polynomial has distinct roots exactly when it is coprime to its derivative.
  • Characteristic zero and finite fields are perfect, while rational function fields in positive characteristic can be imperfect.

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