Separable & Inseparable Extensions

Repeated roots, the derivative test, and perfect fields.

The idea

A polynomial of degree $n$ has $n$ roots in a splitting field only if you count them with multiplicity. Repeated roots are a real nuisance rather than a bookkeeping detail: they cut down the number of distinct roots, so an extension ends up with fewer automorphisms than its degree, and the correspondence between subgroups and subfields that Galois theory rests on stops being a bijection. So we want to detect repeated roots — and finding the roots first is no help, since the roots are exactly what we do not have.

The formal derivative does it without them. Define $\frac{d}{dx}$ on polynomials by declaring $\frac{d}{dx} x^{n} = n x^{n-1}$ and extending linearly: no limits, no analysis, and it makes sense over any field. If $f = (x - \alpha)^{2} g$, the product rule gives $f' = 2(x - \alpha) g + (x - \alpha)^{2} g',$ so $(x - \alpha)$ divides $f'$ as well, while a root appearing only once in $f$ does not survive into $f'$. Repeated roots are therefore precisely the common factors of $f$ and $f'$, and $\gcd(f, f')$ settles the matter by a computation that never leaves the base field. A polynomial with no repeated roots is called separable.

The surprise is that in characteristic $p$ the derivative can vanish identically, and the test then reports that every root repeats. That is not a failure of the test — such polynomials genuinely exist, and the fields over which they never arise are singled out as perfect.

Ways to work on it

• Walkthrough. The derivative test for repeated roots and what makes a field perfect.
• Practice. Count repeated roots of a polynomial using (f, f').
• Hardest. The inseparable archetype x^p - t over F_p(t).

Not sure where to start? Take the ten-question placement test.