Let L / K {\displaystyle L/K} be an algebraic extension (i.e., L is an algebraic extension of K), such that L ⊆ K ¯ {\displaystyle L\subseteq {\overline {K}}} (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 L / K {\displaystyle L/K} be algebraic. The field L is a normal extension if and only if any of the equivalent conditions below hold.
For example, Q ( 2 ) {\displaystyle \mathbb {Q} ({\sqrt {2}})} is a normal extension of Q , {\displaystyle \mathbb {Q} ,} since it is a splitting field of x 2 − 2. {\displaystyle x^{2}-2.} On the other hand, Q ( 2 3 ) {\displaystyle \mathbb {Q} ({\sqrt[{3}]{2}})} is not a normal extension of Q {\displaystyle \mathbb {Q} } since the irreducible polynomial x 3 − 2 {\displaystyle x^{3}-2} has one root in it (namely, 2 3 {\displaystyle {\sqrt[{3}]{2}}} ), but not all of them (it does not have the non-real cubic roots of 2). Recall that the field Q ¯ {\displaystyle {\overline {\mathbb {Q} }}} of algebraic numbers is the algebraic closure of Q , {\displaystyle \mathbb {Q} ,} and thus it contains Q ( 2 3 ) . {\displaystyle \mathbb {Q} ({\sqrt[{3}]{2}}).} Let ω {\displaystyle \omega } be a primitive cubic root of unity. Then since, Q ( 2 3 ) = { a + b 2 3 + c 4 3 ∈ Q ¯ | a , b , c ∈ Q } {\displaystyle \mathbb {Q} ({\sqrt[{3}]{2}})=\left.\left\{a+b{\sqrt[{3}]{2}}+c{\sqrt[{3}]{4}}\in {\overline {\mathbb {Q} }}\,\,\right|\,\,a,b,c\in \mathbb {Q} \right\}} the map { σ : Q ( 2 3 ) ⟶ Q ¯ a + b 2 3 + c 4 3 ⟼ a + b ω 2 3 + c ω 2 4 3 {\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 Q ( 2 3 ) {\displaystyle \mathbb {Q} ({\sqrt[{3}]{2}})} in Q ¯ {\displaystyle {\overline {\mathbb {Q} }}} whose restriction to Q {\displaystyle \mathbb {Q} } is the identity. However, σ {\displaystyle \sigma } is not an automorphism of Q ( 2 3 ) . {\displaystyle \mathbb {Q} ({\sqrt[{3}]{2}}).}
For any prime p , {\displaystyle p,} the extension Q ( 2 p , ζ p ) {\displaystyle \mathbb {Q} ({\sqrt[{p}]{2}},\zeta _{p})} is normal of degree p ( p − 1 ) . {\displaystyle p(p-1).} It is a splitting field of x p − 2. {\displaystyle x^{p}-2.} Here ζ p {\displaystyle \zeta _{p}} denotes any p {\displaystyle p} th primitive root of unity. The field Q ( 2 3 , ζ 3 ) {\displaystyle \mathbb {Q} ({\sqrt[{3}]{2}},\zeta _{3})} is the normal closure (see below) of Q ( 2 3 ) . {\displaystyle \mathbb {Q} ({\sqrt[{3}]{2}}).}
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.
Lang 2002, p. 237, Theorem 3.3, NOR 3. - Lang, Serge (2002), Algebra, Graduate Texts in Mathematics, vol. 211 (Revised third ed.), New York: Springer-Verlag, ISBN 978-0-387-95385-4, MR 1878556 https://mathscinet.ams.org/mathscinet-getitem?mr=1878556 ↩
Jacobson 1989, p. 489, Section 8.7. - Jacobson, Nathan (1989), Basic Algebra II (2nd ed.), W. H. Freeman, ISBN 0-7167-1933-9, MR 1009787 https://mathscinet.ams.org/mathscinet-getitem?mr=1009787 ↩
Lang 2002, p. 237, Theorem 3.3. - Lang, Serge (2002), Algebra, Graduate Texts in Mathematics, vol. 211 (Revised third ed.), New York: Springer-Verlag, ISBN 978-0-387-95385-4, MR 1878556 https://mathscinet.ams.org/mathscinet-getitem?mr=1878556 ↩
Lang 2002, p. 238, Theorem 3.4. - Lang, Serge (2002), Algebra, Graduate Texts in Mathematics, vol. 211 (Revised third ed.), New York: Springer-Verlag, ISBN 978-0-387-95385-4, MR 1878556 https://mathscinet.ams.org/mathscinet-getitem?mr=1878556 ↩