E/F
\alpha\inE
\alpha
Every algebraic extension of a field of characteristic zero is separable, and every algebraic extension of a finite field is separable.[2] 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.[3]
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/F
\alpha\inE\setminusF
\alpha
pk | |
x |
\inF
The simplest nontrivial example of a (purely) inseparable extension is
E=Fp(x)\supseteq
p) | |
F=F | |
p(x |
Fp=Z/(p)
x\inE
f(X)=Xp-xp\inF[X]
f'(X)=0
f(X)=(X-x)p\inE[X]
E=F[x]
Gal(E/F)
E\supseteqF
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)=
pi | |
\sum | |
ix |
.
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.
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 .[5] Let in be an irreducible polynomial and its formal derivative. Then the following are equivalent conditions for the irreducible polynomial to be separable:
k | |
style\sum | |
i=0 |
pi | |
a | |
iX |
.
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 .[9] By repeated application of this property, it follows that in fact,
pn | |
f(X)=g(X |
)
x\mapstoxp
a\inF
Xp-a
stylef(X)=\sum
ip | |
a | |
iX |
ai=b
p | |
i |
bi
style\sum
ip | |
a | |
iX |
=\left(\sum
i | |
b | |
iX |
\right)p.
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).[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. 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.
Let
E\supseteqF
\alpha\inE
If
\alpha,\beta\inE
\alpha+\beta
\alpha\beta
1/\alpha
Thus the set of all elements in separable over forms a subfield of, called the separable closure of in .[13]
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\supseteqF
If
E\supseteqL\supseteqF
If
E\supseteqF
E=F(a1,\ldots,ar)
a1,\ldots,ar
E=F(a)
[E:F]
[E:F]
Let
E\supseteqF
S=\{\alpha\inE\mid\alphaisseparableoverF\}.
x\inE\setminusS
pk | |
x |
\inS,
If
E\supseteqF
On the other hand, an arbitrary algebraic extension
E\supseteqF
E\supseteqF
K\supseteqF
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 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\supseteqF
Let
E\supseteqF
Ep
Fp,
F1/p ⊗ FE
L ⊗ FE
⊗ F
Fp
F1/p
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)
\dimE\operatorname{Der}F(E,E)\ge\operatorname{tr.deg}FE,
In particular, if
E/F
\operatorname{Der}F(E,E)=0
E/F
Let
D1,\ldots,Dm
\operatorname{Der}F(E,E)
a1,\ldots,am\inE
E
F(a1,\ldots,am)
Di(aj)
m=\operatorname{tr.deg}FE
\{a1,\ldots,am\}