Let ${\displaystyle L/K}$  be an algebraic extension (i.e., L is an algebraic extension of K), such that ${\displaystyle L\subseteq {\overline {K}}}$  (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]

• Every embedding of L in ${\displaystyle {\overline {K}}}$  over K induces an automorphism of L.
• L is the splitting field of a family of polynomials in ${\displaystyle K[X]}$ .
• Every irreducible polynomial of ${\displaystyle K[X]}$  that has a root in L splits into linear factors in L.

Let L be an extension of a field K. Then:

• If L is a normal extension of K and if E is an intermediate extension (that is, L ⊇ E ⊇ K), then L is a normal extension of E.^[4]
• If E and F are normal extensions of K contained in L, then the compositum EF and E ∩ F are also normal extensions of K.^[4]

Let ${\displaystyle L/K}$  be algebraic. The field L is a normal extension if and only if any of the equivalent conditions below hold.

• The minimal polynomial over K of every element in L splits in L;
• There is a set ${\displaystyle S\subseteq K[x]}$  of polynomials that each splits over L, such that if ${\displaystyle K\subseteq F\subsetneq L}$  are fields, then S has a polynomial that does not split in F;
• All homomorphisms ${\displaystyle L\to {\bar {K}}}$  that fix all elements of K have the same image;
• The group of automorphisms, ${\displaystyle {\text{Aut}}(L/K),}$  of L that fix all elements of K, acts transitively on the set of homomorphisms ${\displaystyle L\to {\bar {K}}}$  that fix all elements of K.

For example, ${\displaystyle \mathbb {Q} ({\sqrt {2}})}$  is a normal extension of ${\displaystyle \mathbb {Q} ,}$  since it is a splitting field of ${\displaystyle x^{2}-2.}$  On the other hand, ${\displaystyle \mathbb {Q} ({\sqrt[{3}]{2}})}$  is not a normal extension of ${\displaystyle \mathbb {Q} }$  since the irreducible polynomial ${\displaystyle x^{3}-2}$  has one root in it (namely, ${\displaystyle {\sqrt[{3}]{2}}}$ ), but not all of them (it does not have the non-real cubic roots of 2). Recall that the field ${\displaystyle {\overline {\mathbb {Q} }}}$  of algebraic numbers is the algebraic closure of ${\displaystyle \mathbb {Q} ,}$  and thus it contains ${\displaystyle \mathbb {Q} ({\sqrt[{3}]{2}}).}$  Let ${\displaystyle \omega }$  be a primitive cubic root of unity. Then since,

For any prime ${\displaystyle p,}$  the extension ${\displaystyle \mathbb {Q} ({\sqrt[{p}]{2}},\zeta _{p})}$  is normal of degree ${\displaystyle p(p-1).}$  It is a splitting field of ${\displaystyle x^{p}-2.}$  Here ${\displaystyle \zeta _{p}}$  denotes any ${\displaystyle p}$ th primitive root of unity. The field ${\displaystyle \mathbb {Q} ({\sqrt[{3}]{2}},\zeta _{3})}$  is the normal closure (see below) of ${\displaystyle \mathbb {Q} ({\sqrt[{3}]{2}}).}$ 

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.

• Galois extension
• Normal basis

• 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
• Jacobson, Nathan (1989), Basic Algebra II (2nd ed.), W. H. Freeman, ISBN 0-7167-1933-9, MR 1009787