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In [[abstract algebra|algebra]], a '''purely inseparable extension''' of
==Purely inseparable extensions==
An [[algebraic extension]] <math>E\supseteq F</math> is a ''purely inseparable extension'' if and only if for every <math>\alpha\in E\setminus F</math>, the [[Minimal polynomial (field theory)|minimal polynomial]] of <math>\alpha</math> over ''F'' is ''not'' a [[separable polynomial]].<ref name="Isaacs298">Isaacs, p. 298</ref> If ''F'' is any field, the trivial extension <math>F\supseteq F</math> 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 <math>E\supseteq F</math> is an algebraic extension with (non-zero) prime characteristic ''p'', then the following are equivalent:<ref>Isaacs, Theorem 19.10, p. 298</ref>
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==Galois correspondence for purely inseparable extensions==
{{harvs|txt|last=Jacobson|year1=1937|year2=1944}} introduced a variation of [[Galois theory]] for purely inseparable extensions of exponent 1, where the Galois groups of field automorphisms in Galois theory are replaced by [[restricted Lie algebra]]s of derivations. The simplest case is for finite index purely inseparable extensions ''K''⊆''L'' of exponent at most 1 (meaning that the ''p''th power of every element of ''L'' is in ''K''). In this case the Lie algebra of ''K''-derivations of ''L'' is a restricted Lie algebra that is also a vector space of dimension ''n'' over ''L'', where [''L'':''K''] = ''p''<sup>''n''</sup>, and the intermediate fields in ''L'' containing ''K'' correspond to the restricted Lie subalgebras of this Lie algebra that are vector spaces over ''L''. Although the Lie algebra of derivations is a vector space over ''L'', it is not in general a Lie algebra over ''L'', but is a Lie algebra over ''K'' of dimension ''n''[''L'':''K''] = ''np''<sup>''n''</sup>.
A purely inseparable extension is called a '''modular extension''' if it is a tensor product of simple extensions, so in particular every extension of exponent 1 is modular, but there are non-modular extensions of exponent 2 {{harv|Weisfeld|1965}}.
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