Separable Extension - Separable and Inseparable Polynomials

Separable and Inseparable Polynomials

A polynomial f in F is a separable polynomial if and only if every irreducible factor of f in F has distinct roots. The separability of a polynomial depends on the field in which its coefficients are considered to lie; for instance, if g is an inseparable polynomial in F, and one considers a splitting field, E, for g over F, g is necessarily separable in E since an arbitrary irreducible factor of g in E is linear and hence has distinct roots. Despite this, a separable polynomial h in F must necessarily be separable over every extension field of F.

Let f in F be an irreducible polynomial and f' its formal derivative. Then the following are equivalent conditions for f to be separable; that is, to have distinct roots:

• If and, then does not divide f in E.
• There exists such that f has deg(f) roots in K.
• f and f' do not have a common root in any extension field of F.
• f' is not the zero polynomial.

By the last condition above, if an irreducible polynomial does not have distinct roots, its derivative must be zero. Since the formal derivative of a positive degree polynomial can be zero only if the field has prime characteristic, for an irreducible polynomial to not have distinct roots its coefficients must lie in a field of prime characteristic. More generally, if an irreducible (non-zero) polynomial f in F does not have distinct roots, not only must the characteristic of F be a (non-zero) prime number p, but also f(X)=g(Xp) for some irreducible polynomial g in F. By repeated application of this property, it follows that in fact, for a non-negative integer n and some separable irreducible polynomial g in F (where F is assumed to have prime characteristic p).

By the property noted in the above paragraph, if f is an irreducible (non-zero) polynomial with coefficients in the field F of prime characteristic p, and does not have distinct roots, it is possible to write f(X)=g(Xp). Furthermore, if, and if the Frobenius endomorphism of F is an automorphism, g may be written as, and in particular, ; a contradiction of the irreducibility of f. Therefore, if F possesses an inseparable irreducible (non-zero) polynomial, then the Frobenius endomorphism of F cannot be an automorphism (where F is assumed to have prime characteristic p).

If K is a finite field of prime characteristic p, and if X is an indeterminant, then the field of rational functions over K, K(X), is necessarily imperfect. Furthermore, the polynomial f(Y)=Yp−X is inseparable. (To see this, note that there is some extension field in which f has a root ; necessarily, in E. Therefore, working over E, (the final equality in the sequence follows from freshman's dream), and f does not have distinct roots.) More generally, if F is any field of (non-zero) prime characteristic for which the Frobenius endomorphism is not an automorphism, F possesses an inseparable algebraic extension.

A field F is perfect if and only if all of its algebraic extensions are separable (in fact, all algebraic extensions of F are separable if and only if all finite degree extensions of F are separable). By the argument outlined in the above paragraphs, it follows that F is perfect if and only if F has characteristic zero, or F has (non-zero) prime characteristic p and the Frobenius endomorphism of F is an automorphism.