Normal extension

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. 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.

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, any of which can be regarded as a definition of normal extension, are equivalent:
 * Every embedding of L in $$\overline{K}$$ over K induces an automorphism of L.
 * L is the splitting field of a family of polynomials in $$K[X]$$.
 * Every irreducible polynomial of $$K[X]$$ that 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 each splits 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 $$L \to \bar{K}$$ that fix all elements of K have the same image;
 * The group of automorphisms, $$\text{Aut}(L/K),$$ of L that fix all elements of K, acts transitively on the set of homomorphisms $$L \to \bar{K}$$ that fix all elements of 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 $$\Q,$$ and thus it contains $$\Q(\sqrt[3]{2}).$$ Let $$\omega$$ be a primitive cubic root of unity. Then since, $$\Q (\sqrt[3]{2})=\left. \left \{a+b\sqrt[3]{2}+c\sqrt[3]{4}\in\overline{\Q }\,\,\right | \,\,a,b,c\in\Q \right \}$$ 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 $$p$$th 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 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.