In field theory, a subfield of algebra, an algebraic field extension E\supseteq F is called a separable extension if for every \alpha\in E, the minimal polynomial of \alpha over  is a separable polynomial (i.e., its formal derivative is not the zero polynomial, or equivalently it has no repeated roots in any extension field).Isaacs, p. 281  There is also a more general definition that applies when  is not necessarily algebraic over .
An extension that is not separable is said to be inseparable.
Every algebraic extension of a field of characteristic zero is separable, and every algebraic extension of a finite field is separable.Isaacs, Theorem 18.11, p. 281 It follows that most extensions that are considered in mathematics are separable.
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.Isaacs, Theorem 18.13, p. 282
The opposite concept, a purely inseparable extension, also occurs naturally, as every algebraic extension may be decomposed uniquely as a purely inseparable extension of  a separable extension.
An algebraic extension E\supseteq F of fields of non-zero characteristics  is a purely inseparable extension if and only if for every \alpha\in E\setminus F, the minimal polynomial of \alpha over  is not a separable polynomial, or, equivalently, for every element  of , there is a positive integer  such that x^{p^k} \in F.Isaacs, p. 298
The simplest example of a (purely) inseparable extension is E=\mathbb{F}_p(x)\supset F= \mathbb{F}_p(x^p), fields of rational functions in the indeterminate x with coefficients in the finite field \mathbb{F}_p=\mathbb{Z}/(p).
The element x\in E has minimal polynomial f(X)=X^p -x^p \in F[X], having f'\!(X)
= 0 and a p-fold multiple root, as f(X)=(X-x)^p\in E[X].
This is a simple algebraic extension of degree p, as E = F[x], but it is not a normal extension since the Galois group \text{Gal}(E/F) is trivial.
Informal discussion
An arbitrary polynomial  with coefficients in some field  is said to have distinct roots or to be square-free if it has  roots in some extension field E\supseteq F.
For instance, the polynomial  has precisely  roots in the complex plane; namely  and , and hence does have distinct roots.
On the other hand, the polynomial , which is the square of a non-constant polynomial does not have distinct roots, as its degree is two, and  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 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  and reducible over some extension of .
Similarly, divisibility by a square depends on the ambient field.
If an irreducible polynomial  over  is divisible by a square over some field extension, then (by the discussion above) the greatest common divisor of  and its derivative  is not constant.
Note that the coefficients of  belong to the same field as those of , and the greatest common divisor of two polynomials is independent of the ambient field, so the greatest common divisor of  and  has coefficients in .
Since  is irreducible in , this greatest common divisor is necessarily  itself.
Because the degree of  is strictly less than the degree of , it follows that the derivative of  is zero, which implies that the characteristic of the field is a prime number , and  may be written
f(x)= \sum_{i=0}^ka_ix^{pi}.
A polynomial such as this one, whose formal derivative is zero, is said to be inseparable.
Polynomials that are not inseparable are said to be 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  in  is separable if and only if it has distinct roots in any extension of  (that is if it may be factored in distinct linear factors over an algebraic closure of .Isaacs, p. 280  Let  in  be an irreducible polynomial and  its formal derivative.
Then the following are equivalent conditions for the irreducible polynomial  to be separable:
If  is an extension of  in which  is a product of linear factors then no square of these factors divides  in  (that is  is square-free over ).Isaacs, Lemma 18.7, p. 280
There exists an extension  of  such that  has  pairwise distinct roots in .
The constant  is a polynomial greatest common divisor of  and .Isaacs, Theorem 19.4, p. 295
The formal derivative  of  is not the zero polynomial.Isaacs, Corollary 19.5, p. 296
Either the characteristic of   is zero, or the characteristic is , and  is not of the form \textstyle\sum_{i=0}^k a_iX^{pi}.
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  in  is not separable, if and only if the characteristic of  is a (non-zero) prime number , and ) for some irreducible polynomial  in .Isaacs, Corollary 19.6, p. 296 By repeated application of this property, it follows that in fact, f(X)=g(X^{p^n}) for a non-negative integer  and some separable irreducible polynomial  in  (where  is assumed to have prime characteristic p).Isaacs, Corollary 19.9, p. 298
If the Frobenius endomorphism x\to x^p of  is not surjective, there is an element a\in F which is not a th power of an element of .
In this case, the polynomial X^p-a is irreducible and inseparable.
Conversely, if there exists an inseparable irreducible (non-zero) polynomial \textstyle f(X)=\sum a_iX^{ip} in , then the Frobenius endomorphism of  cannot be an automorphism, since, otherwise, we would have a_i=b_i^p for some b_i, and the polynomial  would factor as \textstyle \sum a_iX^{ip}=\left(\sum b_iX^{i}\right)^p.Isaacs, Theorem 19.7, p. 297
If  is a finite field of prime characteristic p, and if  is an indeterminate, then the field of rational functions over , , is necessarily imperfect, and the polynomial  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.Isaacs, p. 299
A field F is perfect if and only if all irreducible polynomials are separable.
It follows that  is perfect if and only if either  has characteristic zero, or  has (non-zero) prime characteristic  and the Frobenius endomorphism of  is an automorphism.
This includes every finite field.
Separable elements and separable extensions
Let E\supseteq F be a field extension.
An element \alpha\in E is separable over  if it is algebraic over , and its minimal polynomial is separable (the minimal polynomial of an element is necessarily irreducible).
If \alpha,\beta\in E are separable over , then \alpha+\beta, \alpha\beta and 1/\alpha are separable over F.
Thus the set of all elements in  separable over  forms a subfield of , called the separable closure of  in .Isaacs, Lemma 19.15, p. 300
The separable closure of  in an algebraic closure of  is simply called the separable closure of .
Like the algebraic closure, it is unique up to an isomorphism, and in general, this isomorphism is not unique.
A field extension E\supseteq F is separable, if  is the separable closure of  in .
This is the case if and only if  is generated over  by separable elements.
If E\supseteq L\supseteq F are field extensions, then  is separable over  if and only if  is separable over  and  is separable over .Isaacs, Corollary 18.12, p. 281
If E\supseteq F is a finite extension (that is  is a -vector space of finite dimension), then the following are equivalent.
is separable over .
E = F(a_1, \ldots, a_r) where a_1, \ldots, a_r are separable elements of .
E = F(a) where  is a separable element of .
If  is an algebraic closure of , then there are exactly [E : F] field homomorphisms of  into  which fix .
For any normal extension   of  which contains , then there are exactly [E : F] field homomorphisms of  into  which fix .
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\supseteq F be an algebraic extension of fields of characteristic .
The separable closure of  in  is S=\{\alpha\in E|\alpha \text{ is separable over } F\}.
For every element x\in E\setminus S there exists a positive integer  such that x^{p^k}\in S, and thus  is a purely inseparable extension of  .
It follows that   is the unique intermediate field that is separable over  and over which  is purely inseparable.Isaacs, Theorem 19.14, p. 300
If E\supseteq F is a finite extension, its degree   is the product of the degrees  and  .
The former, often denoted  is often referred to as the separable part of  ,  or as the  of ; the latter is referred to as the inseparable part of the degree or the .Isaacs, p. 302 The inseparable degree is 1 in characteristic zero and a power of  in characteristic .
On the other hand, an arbitrary algebraic extension E\supseteq F may not possess an intermediate extension  that is purely inseparable over  and over which  is separable.
However, such an intermediate extension may exist if, for example, E\supseteq F is a finite degree normal extension (in this case,  is the fixed field of the Galois group of  over ).
Suppose that such an intermediate extension does exist, and  is finite, then , where  is the separable closure of  in  .Isaacs, Theorem 19.19, p. 302 The known proofs of this equality use the fact that if K\supseteq F is a purely inseparable extension, and if  is a separable irreducible polynomial in , then  remains irreducible in K[X]Isaacs, Lemma 19.20, p. 302).
This equality implies that, if  is finite, and  is an intermediate field between  and , then .Isaacs, Corollary 19.21, p. 303
The separable closure  of a field  is the separable closure of  in an algebraic closure of .
It is the maximal Galois extension of .
By definition,  is perfect if and only if its separable and algebraic closures coincide.
Separability of transcendental extensions
Separability problems may 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\supseteq F is a transcendence basis   of  such that  is a separable algebraic extension of .
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.Fried & Jarden (2008) p.38
Let E\supseteq F be a field extension of characteristic exponent  (that is  in characteristic zero and, otherwise,  is the characteristic).
The following properties are equivalent:
is a separable extension of ,
E^p and  are linearly disjoint over F^p,
F^{1/p} \otimes_F E is reduced,
L \otimes_F E is reduced for every field extension  of ,
where \otimes_F denotes the tensor product of fields, F^p is the field of the th powers of the elements of  (for any field ), and F^{1/p} is the field obtained by adjoining to  the th root of all its elements (see Separable algebra for details).
Differential criteria
Separability can be studied with the aid of derivations.
Let  be a finitely generated field extension of a field .
Denoting \operatorname{Der}_F(E,E) the -vector space of the -linear derivations of , one has
\dim_E \operatorname{Der}_F(E,E) \ge \operatorname{tr.deg}_F E,
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 \operatorname{Der}_F(E, E) = 0 if and only if E/F is separable.Fried & Jarden (2008) p.49
Let D_1, \ldots, D_m be a basis of \operatorname{Der}_F(E,E) and a_1, \ldots, a_m \in E.
Then E is separable algebraic over F(a_1, \ldots, a_m) if and only if the matrix D_i(a_j) is invertible.
In particular, when m = \operatorname{tr.deg}_F E, this matrix is invertible if and only if \{ a_1, \ldots, a_m \} is a separating transcendence basis.
Notes
References
Borel, A. Linear algebraic groups, 2nd ed.
P.M. Cohn (2003).
Basic algebra
M. Nagata (1985).
Commutative field theory: new edition, Shokabo. (Japanese)
External links
Category:Field extensions
de:Körpererweiterung#Separable Erweiterungen
