# Normal extension

In abstract algebra, a normal extension is an algebraic field extension L/K for which every polynomial that is irreducible over K either has no root in L or splits into linear factors in L. Bourbaki calls such an extension a quasi-Galois extension.

## Definition

The algebraic field extension L/K is normal (we also say that L is normal over K) if every irreducible polynomial over K that has at least one root in L splits over L. In other words, if αL, then all conjugates of α over K (i.e., all roots of the minimal polynomial of α over K) belong to L.

## Equivalent properties

The normality of L/K is equivalent to either of the following properties. Let Ka be an algebraic closure of K containing L.

• Every embedding σ of L in Ka that restricts to the identity on K, satisfies σ(L) = L (σ is an automorphism of L over K.)
• Every irreducible polynomial in K[X] that has one root in L, has all of its roots in L, that is, it decomposes into linear factors in L[X]. (One says that the polynomial splits in L.)

If L is a finite extension of K that is separable (for example, this is automatically satisfied if K is finite or has characteristic zero) then the following property is also equivalent:

• There exists an irreducible polynomial whose roots, together with the elements of K, generate L. (One says that L is the splitting field for the polynomial.)

## 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 (i.e., L ⊃ E ⊃ K), then L is a normal extension of E.[citation needed]
• 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.[citation needed]

## Examples and counterexamples

For example, $\mathbb {Q} ({\sqrt {2}})$ is a normal extension of $\mathbb {Q} ,$ since it is a splitting field of $x^{2}-2.$ On the other hand, $\mathbb {Q} ({\sqrt[{3}]{2}})$ is not a normal extension of $\mathbb {Q}$ 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 ${\overline {\mathbb {Q} }}$ of algebraic numbers is the algebraic closure of $\mathbb {Q} ,$ i.e., it contains $\mathbb {Q} ({\sqrt[{3}]{2}}).$ Since,

$\mathbb {Q} ({\sqrt[{3}]{2}})=\left.\left\{a+b{\sqrt[{3}]{2}}+c{\sqrt[{3}]{4}}\in {\overline {\mathbb {Q} }}\,\,\right|\,\,a,b,c\in \mathbb {Q} \right\}$ and, if ω is a primitive cubic root of unity, then the map

${\begin{cases}\sigma :\mathbb {Q} ({\sqrt[{3}]{2}})\longrightarrow {\overline {\mathbb {Q} }}\\a+b{\sqrt[{3}]{2}}+c{\sqrt[{3}]{4}}\longmapsto a+b\omega {\sqrt[{3}]{2}}+c\omega ^{2}{\sqrt[{3}]{4}}\end{cases}}$ is an embedding of $\mathbb {Q} ({\sqrt[{3}]{2}})$ in ${\overline {\mathbb {Q} }}$ whose restriction to $\mathbb {Q}$ is the identity. However, σ is not an automorphism of $\mathbb {Q} ({\sqrt[{3}]{2}})$ .

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

## 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, i.e., 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.