Normal 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, 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 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 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.

Citations

References