Separable extension

Informal discussion

An arbitrary polynomial f with coefficients in some field F is said to have distinct roots or to be square-free if it has deg(f) roots in some extension field E ⊇ F For instance, the polynomial g(X) = X² – 1 has precisely deg(g) = 2 roots in the complex plane; namely 1 and –1, and hence does have distinct roots. On the other hand, the polynomial h(X) = (X – 2)², which is the square of a non-constant polynomial does not have distinct roots, as its degree is two, and 2 is its only root.

Every polynomial may be factored in linear factors, over an algebraic closure of the field of its coefficients. Therefore, the polynomial does not have distinct roots if and only if it is divisible by the square of a polynomial of positive degree. This 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 ground field, and a polynomial may be irreducible over F and reducible over some extension of F. If an irreducible polynomial f is divisible by a square over some field extension, then the greatest common divisor of f and its derivative is not constant. As its coefficients belong to the same field as those of f, this greatest common divisor is necessarily f itself. This is possible only if the derivative of f is zero, which implies that the characteristic of the field is a prime number p, and f may be written

Such a polynomial is said inseparable. Other irreducible polynomials are said separable. A separable extension is an extension that may be generated by separable elements, that is elements whose minimal polynomials are separable.

Separable and inseparable polynomials

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). Let f in F[X] be an irreducible polynomial and f' its formal derivative. Then the following are equivalent conditions for 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(X^p) for some irreducible polynomial g in F[X]. By repeated application of this property, it follows that in fact, f ( X ) = g ( X p n ) for a non-negative integer n and some separable irreducible polynomial g in F[X] (where F is assumed to have prime characteristic p).

If the Frobenius endomorphism x → x p of F is not surjective, there is an element a ∈ F which is not a pth power of an element of F. In this case, the polynomial X p − a is irreducible and inseparable. Conversely, if there exists an inseparable irreducible (non-zero) polynomial f ( X ) = ∑ a i X i p in F[X], then the Frobenius endomorphism of F cannot be an automorphism, as the image ∑ a i p X i p = ( ∑ a i X i ) p of f(X) by this endomorphism is reducible.

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)=Y^p−X is inseparable (its formal derivative in Y is 0). 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 irreducible polynomials are separable. It follows that F is perfect if and only if either F has characteristic zero, or F has (non-zero) prime characteristic p and the Frobenius endomorphism of F is an automorphism. This includes every finite field.

Separable elements and separable extensions

Let E ⊇ F be a field extension. An element α ∈ E is separable over F if it is algebraic over F, and its minimal polynomial is separable (the minimal polynomial of an element is necessarily irreducible).

If α , β ∈ E are separable over F, then α + β , α β and 1 / α are separable over F.

Thus the set of all elements in E separable over F forms a subfield of E, called the separable closure of F in E.

The separable closure of F in an algebraic closure of F is simply called the separable closure of F. Like the algebraic closure, it is unique up to an isomorphism, and in general, this isomorphism is not unique.

A field extension E ⊇ F is separable, if E is the separable closure of F in E. This is the case if and only if E is generated over F by separable elements.

If E ⊇ L ⊇ F are field extensions, then E is separable over F if and only if E is separable over L, and L is separable over F.

If E ⊇ F is finite extension (that is E is a F-vector space of finite dimension), then the following are equivalent.

1. E is separable over F.
2. E = F ( a 1 , … , a r ) where a 1 , … , a r are separable elements of E.
3. E = F ( a ) where a is a separable element of E.
4. If K is an algebraic closure of F, then there are exactly [ E : F ] field homomorphisms of E into K which fix F.
5. For any normal extension K of F which contains E, then there are exactly [ E : F ] field homomorphisms of E into K which fix F.

The equivalence of 3. and 1. is known as the primitive element theorem or Artin's theorem on primitive elements. Properties 4. and 5. are the basis of Galois theory, and, in particular, of the fundamental theorem of Galois theory.

Separable extensions within algebraic extensions

Let E ⊇ F be an algebraic extension of fields of characteristic p. The separable closure of F in E is S = { α ∈ E | α  is separable over  F } . For every element x ∈ E ∖ S there exists a positive integer k such that x p k ∈ S , and thus E is a purely inseparable extension of S. It follows that S is the unique intermediate field that is separable over F and over which E is purely inseparable.

If E ⊇ F is a finite extension, its degree [E : F] is the product of the degrees [S : F] and [E : S]. The former, often denoted [E : F]_sep is often referred to as the separable part of [E : F], or as the separable degree of E/F; the latter is referred to as the inseparable part of the degree or the inseparable degree. The inseparable degree is 1 in characteristic zero and a power of p in characteristic p > 0.

On the other hand, an arbitrary algebraic extension E ⊇ F may not possess an intermediate extension K that is purely inseparable over F and over which E is separable. However, such an intermediate extension may exist if, for example, E ⊇ F 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. The known proofs of this equality use the fact that if K ⊇ F is a purely inseparable extension, and if f is a separable irreducible polynomial in F[X], then f remains irreducible in K[X]). This equality implies that, if [E : F] is finite, and U is an intermediate field between F and E, then [E : F]_sep = [E : U]_sep⋅[U : F]_sep.

The separable closure F^sep of a field F is the separable closure of F in an algebraic closure of F. It is the maximal Galois extension of F. By definition, F is perfect if and only if its separable and algebraic closures coincide (in particular, the notion of a separable closure is only interesting for imperfect fields).

Separability of transcendental extensions

It may occur that separability problems arise when dealing with transcendental extensions. This is typically the case for algebraic geometry over a field of prime characteristic, where the function field of an algebraic variety has a transcendence degree over the ground field that is equal to the dimension of the variety.

For defining the separability of a transcendental extension, it is natural to use the fact that every field extension is an algebraic extension of a purely transcendental extension. This leads to the following definition.

A separating transcendence basis of an extension E ⊇ F is a transcendence basis T of E such that E is a separable (algebraic extension of F(T). A finitely generated field extension is separable if and only it has a separating transcendence basis; an extension that is not finitely generated is called separable if every finitely generated subextension has a separating transcendence basis.

Let E ⊇ F be a field extension of characteristic exponent p (that is p = 1 in characteristic zero and, otherwise, p is the characteristic). The following properties are equivalent:

where ⊗ F denotes the tensor product of fields, F p is the field of the pth powers of the elements of F (for any field F), and F 1 / p is the field obtained by adjoining to F the pth root of all its elements (see Separable algebra for details).

Differential criteria

The separability can be studied with the aid of derivations. Let E be a finitely generated field extension of a field F. Denoting Der F ⁡ ( E , E ) the F-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).

In particular, if E / F is an algebraic extension, then Der F ⁡ ( E , E ) = 0 if and only if E / F is separable.

Let D 1 , … , D m be a basis of Der F ⁡ ( E , E ) and a 1 , … , a m ∈ E . Then E is separable algebraic over F ( a 1 , … , a m ) if and only if the matrix D i ( a j ) is invertible. In particular, when m = t r . d e g F ⁡ E , this matrix is invertible if and only if { a 1 , … , a m } is a separating transcendence basis.