Normal extension explained

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\subseteqK[x]

of polynomials that each splits over L, such that if

K\subseteqF\subsetneqL

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,

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

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])=\left. \left \

the map

\begin \sigma:\Q (\sqrt[3])\longrightarrow\overline\\ a+b\sqrt[3]+c\sqrt[3]\longmapsto a+b\omega\sqrt[3]+c\omega^2\sqrt[3]\end

is an embedding of

\Q(\sqrt[3]{2})

\overline{\Q}

whose restriction to

is the identity. However,

\sigma

is not an automorphism of

\Q(\sqrt[3]{2}).

For any prime

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

th primitive root of unity. The field

\Q(\sqrt[3]{2},\zeta₃₎

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.