Separable extension explained

E/F

is called a separable extension if for every

\alpha\inE

, 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).[1] 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.[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

of fields of non-zero characteristic is a purely inseparable extension if and only if for every

\alpha\inE\setminusF

, the minimal polynomial of

\alpha

over is not a separable polynomial, or, equivalently, for every element of, there is a positive integer such that
pk
x

\inF

.[4]

The simplest nontrivial example of a (purely) inseparable extension is

E=Fp(x)\supseteq

p)
F=F
p(x
, fields of rational functions in the indeterminate x with coefficients in the finite field

Fp=Z/(p)

. The element

x\inE

has minimal polynomial

f(X)=Xp-xp\inF[X]

, having

f'(X)=0

and a p-fold multiple root, as

f(X)=(X-x)p\inE[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

Gal(E/F)

is trivial.

Informal discussion

E\supseteqF

. 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)=

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.

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 .[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

)

for a non-negative integer and some separable irreducible polynomial in (where is assumed to have prime characteristic p).[10]

x\mapstoxp

of is not surjective, there is an element

a\inF

that is not a th power of an element of . In this case, the polynomial

Xp-a

is irreducible and inseparable. Conversely, if there exists an inseparable irreducible (non-zero) polynomial

stylef(X)=\sum

ip
a
iX
in, then the Frobenius endomorphism of cannot be an automorphism, since, otherwise, we would have

ai=b

p
i
for some

bi

, and the polynomial would factor as

style\sum

ip
a
iX

=\left(\sum

i
b
iX

\right)p.

[11]

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.

Separable elements and separable extensions

Let

E\supseteqF

be a field extension. An element

\alpha\inE

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\inE

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 .[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

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\supseteqL\supseteqF

are field extensions, then is separable over if and only if is separable over and is separable over .[14]

If

E\supseteqF

is a finite extension (that is is a -vector space of finite dimension), then the following are equivalent.
  1. is separable over .

E=F(a1,\ldots,ar)

where

a1,\ldots,ar

are separable elements of .

E=F(a)

where is a separable element of .
  1. If is an algebraic closure of, then there are exactly

[E:F]

field homomorphisms of into that fix .
  1. For any normal extension of that contains, then there are exactly

[E:F]

field homomorphisms of into that 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\supseteqF

be an algebraic extension of fields of characteristic . The separable closure of in is

S=\{\alpha\inE\mid\alphaisseparableoverF\}.

For every element

x\inE\setminusS

there exists a positive integer such that
pk
x

\inS,

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.[15]

If

E\supseteqF

is a finite extension, its degree is the product of the degrees and . The former, often denoted, is 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 .[16] The inseparable degree is 1 in characteristic zero and a power of in characteristic .

On the other hand, an arbitrary algebraic extension

E\supseteqF

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\supseteqF

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 .[17] The known proofs of this equality use the fact that if

K\supseteqF

is a purely inseparable extension, and if is a separable irreducible polynomial in, then remains irreducible in K[''X''][18]). This equality implies that, if is finite, and is an intermediate field between and, then .[19]

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\supseteqF

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.[20]

Let

E\supseteqF

be a field extension of characteristic exponent (that is in characteristic zero and, otherwise, is the characteristic). The following properties are equivalent:

Ep

and are linearly disjoint over

Fp,

F1/pFE

is reduced,

LFE

is reduced for every field extension of,where

F

denotes the tensor product of fields,

Fp

is the field of the th powers of the elements of (for any field), and

F1/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

\dimE\operatorname{Der}F(E,E)\ge\operatorname{tr.deg}FE,

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.[21]

Let

D1,\ldots,Dm

be a basis of

\operatorname{Der}F(E,E)

and

a1,\ldots,am\inE

. Then

E

is separable algebraic over

F(a1,\ldots,am)

if and only if the matrix

Di(aj)

is invertible. In particular, when

m=\operatorname{tr.deg}FE

, this matrix is invertible if and only if

\{a1,\ldots,am\}

is a separating transcendence basis.

References

Notes and References

  1. Isaacs, p. 281
  2. Isaacs, Theorem 18.11, p. 281
  3. Isaacs, Theorem 18.13, p. 282
  4. Isaacs, p. 298
  5. Isaacs, p. 280
  6. Isaacs, Lemma 18.7, p. 280
  7. Isaacs, Theorem 19.4, p. 295
  8. Isaacs, Corollary 19.5, p. 296
  9. Isaacs, Corollary 19.6, p. 296
  10. Isaacs, Corollary 19.9, p. 298
  11. Isaacs, Theorem 19.7, p. 297
  12. Isaacs, p. 299
  13. Isaacs, Lemma 19.15, p. 300
  14. Isaacs, Corollary 18.12, p. 281 and Corollary 19.17, p. 301
  15. Isaacs, Theorem 19.14, p. 300
  16. Isaacs, p. 302
  17. Isaacs, Theorem 19.19, p. 302
  18. Isaacs, Lemma 19.20, p. 302
  19. Isaacs, Corollary 19.21, p. 303
  20. Fried & Jarden (2008) p.38
  21. Fried & Jarden (2008) p.49