Purely inseparable extension

In algebra, a Purely inseparable extension of fields is an extension k⊆K of fields of characteristic p>0 such that every element of K is a root of an equation of the form x^q = a, with q a power of p and a in k.

Purely inseparable extensions

An algebraic extension $E\supseteq F$ is a purely inseparable extension if and only if for every $\alpha \in E\setminus F$, the minimal polynomial of $\alpha$ over F is not a separable polynomial. If F is any field, the trivial extension $F\supseteq F$ is purely inseparable; for the field F to possess a non-trivial purely inseparable extension, it must be imperfect as outlined in the above section.

Several equivalent and more concrete definitions for the notion of a purely inseparable extension are known. If $E\supseteq F$ is an algebraic extension with (non-zero) prime characteristic p, then the following are equivalent:

1. E is purely inseparable over F.

2. For each element $\alpha \in E$, there exists $n\geq 0$ such that $\alpha ^{p^{n}}\in F$.

3. Each element of E has minimal polynomial over F of the form $X^{p^{n}}-a$ for some integer $n\geq 0$ and some element $a\in F$.

It follows from the above equivalent characterizations that if $E=F[\alpha ]$ (for F a field of prime characteristic) such that $\alpha ^{p^{n}}\in F$ for some integer $n\geq 0$, then E is purely inseparable over F. (To see this, note that the set of all x such that $x^{p^{n}}\in F$ for some $n\geq 0$ forms a field; since this field contains both $\alpha$ and F, it must be E, and by condition 2 above, $E\supseteq F$ must be purely inseparable.)

If F is an imperfect field of prime characteristic p, choose $a\in F$ such that a is not a pth power in F, and let f(X)=X^p−a. Then f has no root in F, and so if E is a splitting field for f over F, it is possible to choose $\alpha$ with $f(\alpha )=0$. In particular, $\alpha ^{p}=a$ and by the property stated in the paragraph directly above, it follows that $F[\alpha ]\supseteq F$ is a non-trivial purely inseparable extension (in fact, $E=F[\alpha ]$, and so $E\supseteq F$ is automatically a purely inseparable extension).

Purely inseparable extensions do occur naturally; for example, they occur in algebraic geometry over fields of prime characteristic. If K is a field of characteristic p, and if V is an algebraic variety over K of dimension greater than zero, the function field K(V) is a purely inseparable extension over the subfield K(V)^p of pth powers (this follows from condition 2 above). Such extensions occur in the context of multiplication by p on an elliptic curve over a finite field of characteristic p.

Properties

• If the characteristic of a field F is a (non-zero) prime number p, and if $E\supseteq F$ is a purely inseparable extension, then if $F\subseteq K\subseteq E$, K is purely inseparable over F and E is purely inseparable over K. Furthermore, if [E : F] is finite, then it is a power of p, the characteristic of F.
• Conversely, if $F\subseteq K\subseteq E$ is such that $F\subseteq K$ and $K\subseteq E$ are purely inseparable extensions, then E is purely inseparable over F.
• An algebraic extension $E\supseteq F$ is an inseparable extension if and only if there is some $\alpha \in E\setminus F$ such that the minimal polynomial of $\alpha$ over F is not a separable polynomial (i.e., an algebraic extension is inseparable if and only if it is not separable; note, however, that an inseparable extension is not the same thing as a purely inseparable extension). If $E\supseteq F$ is a finite degree non-trivial inseparable extension, then [E : F] is necessarily divisible by the characteristic of F.
• If $E\supseteq F$ is a finite degree normal extension, and if $K={\mbox{Fix}}({\mbox{Gal}}(E/F))$, then K is purely inseparable over F and E is separable over K.