If $E$ is an extension of $F$ in which $f$ is a product of linear factors then no square of these factors divides $f$ in $E [X]$ (that is $f$ is over $E$ ).^[6]

square-free

There exists an extension $E$ of $F$ such that $f$ has $deg(f)$ pairwise distinct roots in $E$ .

[6]

The constant $1$ is a of $f$ and $f'$ .^[7]

polynomial greatest common divisor

The formal derivative $f'$ of $f$ is not the zero polynomial.

[8]

Either the characteristic of $F$ is zero, or the characteristic is $p$ , and $f$ is not of the form $\textstyle \sum _{i=0}^{k}a_{i}X^{pi}.$

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 $F)$ .^[5] Let $f$ in $F [X]$ be an irreducible polynomial and $f'$ its formal derivative. 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 $f (X)= g (X p$ ) for some irreducible polynomial $g$ in $F [X]$ .^[9] 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).^[10]

If the Frobenius endomorphism $x\mapsto x^{p}$ of $F$ is not surjective, there is an element $a\in F$ that is not a $p$ th 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 $\textstyle f(X)=\sum a_{i}X^{ip}$ in $F [X]$ , then the Frobenius endomorphism of $F$ cannot be an automorphism, since, otherwise, we would have $a_{i}=b_{i}^{p}$ for some $b_{i}$ , and the polynomial $f$ would factor as $\textstyle \sum a_{i}X^{ip}=\left(\sum b_{i}X^{i}\right)^{p}.$ ^[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 $f (Y)= Y p - X$ 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. 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 extensions within algebraic extensions[edit]

Let $E\supseteq F$ be an algebraic extension of fields of characteristic $p$ . The separable closure of $F$ in $E$ is $S=\{\alpha \in E\mid \alpha {\text{ is separable over }}F\}.$ For every element $x\in E\setminus S$ there exists a positive integer $k$ such that $x^{p^{k}}\in S,$ 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.^[15]

If $E\supseteq F$ 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 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.^[16] The inseparable degree is 1 in characteristic zero and a power of $p$ in characteristic $p > 0$ .^[17]

On the other hand, an arbitrary algebraic extension $E\supseteq F$ 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, $E\supseteq F$ 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 $E$ .^[18] The known proofs of this equality use the fact that if $K\supseteq F$ is a purely inseparable extension, and if $f$ is a separable irreducible polynomial in $F [X]$ , then $f$ remains irreducible in K[X]^[19]). This equality implies that, if $[E : F]$ is finite, and $U$ is an intermediate field between $F$ and $E$ , then $[E : F] sep = [E : U] sep \cdot[U : F] sep$ .^[20]

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.

$E$ is a separable extension of $F$ ,

$E^{p}$ and $F$ are over $F^{p},$

linearly disjoint

$F^{1/p}\otimes _{F}E$ is ,

reduced

$L\otimes _{F}E$ is reduced for every field extension $L$ of $E$ ,

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 $T$ 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.^[21]

Let $E\supseteq F$ 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 $\otimes _{F}$ denotes the tensor product of fields, $F^{p}$ is the field of the $p$ th powers of the elements of $F$ (for any field $F$ ), and $F^{1/p}$ is the field obtained by adjoining to $F$ the $p$ th root of all its elements (see Separable algebra for details).

Borel, A. Linear algebraic groups, 2nd ed.

P.M. Cohn (2003). Basic algebra

Fried, Michael D.; Jarden, Moshe (2008). Field arithmetic. Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge. Vol. 11 (3rd ed.). . ISBN 978-3-540-77269-9. Zbl 1145.12001.

Springer-Verlag

(1993). Algebra, a graduate course (1st ed.). Brooks/Cole Publishing Company. ISBN 0-534-19002-2.

I. Martin Isaacs

(1972). Fields and rings. Chicago lectures in mathematics (Second ed.). University of Chicago Press. pp. 55–59. ISBN 0-226-42451-0. Zbl 1001.16500.

Kaplansky, Irving

(2002), Algebra, Graduate Texts in Mathematics, vol. 211 (Revised third ed.), New York: Springer-Verlag, ISBN 978-0-387-95385-4, MR 1878556

Lang, Serge

M. Nagata (1985). Commutative field theory: new edition, Shokabo. (Japanese)

[1]

Silverman, Joseph (1993). The Arithmetic of Elliptic Curves. Springer. 0-387-96203-4.

ISBN

, Encyclopedia of Mathematics, EMS Press, 2001 [1994]