[1] There is also a more general definition that applies when E is not necessarily algebraic over F. An extension that is not separable is said to be inseparable.
Nevertheless, the concept of separability is important, as the existence of inseparable extensions is the main obstacle for extending many theorems proved in characteristic zero to non-zero characteristic.
For example, the fundamental theorem of Galois theory is a theorem about normal extensions, which remains true in non-zero characteristic only if the extensions are also assumed to be separable.
of fields of non-zero characteristic p is a purely inseparable extension if and only if for every
over F is not a separable polynomial, or, equivalently, for every element x of E, there is a positive integer k such that
[4] The simplest nontrivial example of a (purely) inseparable extension is
This is the case if and only if the greatest common divisor of the polynomial and its derivative is not a constant.
Thus for testing if a polynomial is square-free, it is not necessary to consider explicitly any field extension nor to compute the roots.
In this context, the case of irreducible polynomials requires some care.
A priori, it may seem that being divisible by a square is impossible for an irreducible polynomial, which has no non-constant divisor except itself.
However, irreducibility depends on the ambient field, and a polynomial may be irreducible over F and reducible over some extension of F. Similarly, divisibility by a square depends on the ambient field.
If an irreducible polynomial f over F is divisible by a square over some field extension, then (by the discussion above) the greatest common divisor of f and its derivative f′ is not constant.
Note that the coefficients of f′ belong to the same field as those of f, and the greatest common divisor of two polynomials is independent of the ambient field, so the greatest common divisor of f and f′ has coefficients in F. Since f is irreducible in F, this greatest common divisor is necessarily f itself.
Because the degree of f′ is strictly less than the degree of f, it follows that the derivative of f is zero, which implies that the characteristic of the field is a prime number p, and f may be written A polynomial such as this one, whose formal derivative is zero, is said to be inseparable.
An irreducible polynomial f in F[X] is separable if and only if it has distinct roots in any extension of F (that is if it may be factored in distinct linear factors over an algebraic closure of F).
Then the following are equivalent conditions for the irreducible polynomial f to be separable: 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 be separable, its coefficients must lie in a field of prime characteristic.
More generally, an irreducible (non-zero) polynomial f in F[X] is not separable, if and only if the characteristic of F is a (non-zero) prime number p, and f(X)=g(Xp) for some irreducible polynomial g in F[X].
for a non-negative integer n and some separable irreducible polynomial g in F[X] (where F is assumed to have prime characteristic p).
that is not a pth power of an element of F. In this case, the polynomial
Conversely, if there exists an inseparable irreducible (non-zero) polynomial
[11] If K is a finite field of prime characteristic p, and if X is an indeterminate, then the field of rational functions over K, K(X), is necessarily imperfect, and the polynomial f(Y)=Yp−X is inseparable (its formal derivative in Y is 0).
[1] 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.
[12] A field F is perfect if and only if all irreducible polynomials are separable.
be an algebraic extension of fields of characteristic p. The separable closure of F in E is
may not possess an intermediate extension K that is purely inseparable over F and over which E is separable.
is a finite degree normal extension (in this case, K is the fixed field of the Galois group of E over F).
Suppose that such an intermediate extension does exist, and [E : F] is finite, then [S : F] = [E : K], where S is the separable closure of F in E.[18] The known proofs of this equality use the fact that if
Separability problems may arise when dealing with transcendental extensions.
is the field obtained by adjoining to F the pth root of all its elements (see Separable algebra for details).
the E-vector space of the F-linear derivations of E, one has and the equality holds if and only if E is separable over F (here "tr.deg" denotes the transcendence degree).