In abstract algebra, a normal extension is an algebraic field extension L/K for which every irreducible polynomial over K that has a root in L splits into linear factors in L.[1][2] This is one of the conditions for an algebraic extension to be a Galois extension.
Bourbaki calls such an extension a quasi-Galois extension.
For finite extensions, a normal extension is identical to a splitting field.
be an algebraic extension (i.e., L is an algebraic extension of K), such that
(i.e., L is contained in an algebraic closure of K).
Then the following conditions, any of which can be regarded as a definition of normal extension, are equivalent:[3] Let L be an extension of a field K. Then: Let
The field L is a normal extension if and only if any of the equivalent conditions below hold.
is a normal extension of
since it is a splitting field of
On the other hand,
is not a normal extension of
since the irreducible polynomial
has one root in it (namely,
), but not all of them (it does not have the non-real cubic roots of 2).
Recall that the field
of algebraic numbers is the algebraic closure of
ω
be a primitive cubic root of unity.
the map
{\displaystyle {\begin{cases}\sigma :\mathbb {Q} ({\sqrt[{3}]{2}})\longrightarrow {\overline {\mathbb {Q} }}\\a+b{\sqrt[{3}]{2}}+c{\sqrt[{3}]{4}}\longmapsto a+b\omega {\sqrt[{3}]{2}}+c\omega ^{2}{\sqrt[{3}]{4}}\end{cases}}}
is an embedding of
whose restriction to
is not an automorphism of
For any prime
is normal of degree
It is a splitting field of
th primitive root of unity.
is the normal closure (see below) of
If K is a field and L is an algebraic extension of K, then there is some algebraic extension M of L such that M is a normal extension of K. Furthermore, up to isomorphism there is only one such extension that is minimal, that is, the only subfield of M that contains L and that is a normal extension of K is M itself.
This extension is called the normal closure of the extension L of K. If L is a finite extension of K, then its normal closure is also a finite extension.