In abstract algebra, a normal extension is an algebraic field extension L/K for which every irreducible polynomial over K which has a root in L, splits into linear factors in L.
These are one of the conditions for algebraic extensions to be a Galois extension.
Bourbaki calls such an extension a quasi-Galois extension.
Definition
Let L/K be an algebraic extension (i.e. L is an algebraic extension of K), such that L\subseteq \overline{K} (i.e. L is contained in an algebraic closure of K).
Then the following conditions, which any of them can be regarded as a definition of normal extension, are equivalent:
Every embedding of L in \overline{K} induces an automorphism of L.
L is the splitting field of a family of polynomials in K\left[X\right].
Every irreducible polynomial of K\left[X\right] which has a root in L splits into linear factors in L.
Other properties
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.
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.
Equivalent Conditions for Normality
Let 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 S \subseteq K[x] of polynomials that simultaneously split over L, such that if K\subseteq F\subsetneq L are fields, then S has a polynomial that does not split in F;
All homomorphisms L \to \bar{K} have the same image;
The group of automorphisms, \text{Aut}(L/K), of L which fixes elements of K, acts transitively on the set of homomorphisms L \to \bar{K}.
Examples and counterexamples
For example, \Q(\sqrt{2}) is a normal extension of \Q, since it is a splitting field of x^2-2.
On the other hand, \Q(\sqrt[3]{2}) is not a normal extension of \Q since the irreducible polynomial x^3-2 has one root in it (namely, \sqrt[3]{2}), but not all of them (it does not have the non-real cubic roots of 2).
Recall that the field \overline{\Q} of algebraic numbers is the algebraic closure of \Q, that is, it contains \Q(\sqrt[3]{2}).
Since,  \Q (\sqrt[3]{2})=\left.
\left \{a+b\sqrt[3]{2}+c\sqrt[3]{4}\in\overline{\Q }\,\,\right | \,\,a,b,c\in\Q \right \}  and, if \omega is a primitive cubic root of unity, then the map \begin{cases} \sigma:\Q (\sqrt[3]{2})\longrightarrow\overline{\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(\sqrt[3]{2}) in \overline{\Q} whose restriction to \Q  is the identity.
However, \sigma is not an automorphism of \Q (\sqrt[3]{2}).
For any prime p, the extension \Q (\sqrt[p]{2}, \zeta_p) is normal of degree p(p-1).
It is a splitting field of x^p - 2.
Here \zeta_p denotes any pth primitive root of unity.
The field \Q (\sqrt[3]{2}, \zeta_3) is the normal closure (see below) of \Q (\sqrt[3]{2}).
Normal closure
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 which is minimal, that is, the only subfield of M which contains L and which 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.
See also
Galois extension
Normal basis
Citations
References
