Normal extension

In abstract algebra, an algebraic field extension L/K is said to be normal if 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 α ∈ L, 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.
• 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.

Examples and Counterexamples

For example, ${\displaystyle \mathbb {Q} ({\sqrt {2}})}$ is a normal extension of ${\displaystyle \mathbb {Q} ,}$ since it is a splitting field of ${\displaystyle x^{2}-2.}$ On the other hand, ${\displaystyle \mathbb {Q} ({\sqrt[{3}]{2}})}$ is not a normal extension of ${\displaystyle \mathbb {Q} }$ since the irreducible polynomial ${\displaystyle x^{3}-2}$ has one root in it (namely, ${\displaystyle {\sqrt[{3}]{2}}}$), but not all of them (it does not have the non-real cubic roots of 2). Recall that the field ${\displaystyle {\overline {\mathbb {Q} }}}$ of algebraic numbers is the algebraic closure of ${\displaystyle \mathbb {Q} ,}$ i.e., it contains ${\displaystyle \mathbb {Q} ({\sqrt[{3}]{2}}).}$ Since,

${\displaystyle \mathbb {Q} ({\sqrt[{3}]{2}})=\left.\left\{a+b{\sqrt[{3}]{2}}+c{\sqrt[{3}]{4}}\in {\overline {\mathbb {Q} }}\right|a,b,c\in \mathbb {Q} \right\}}$

and, if ω is a primitive cubic root of unity, then the map

${\displaystyle {\begin{cases}\sigma :\mathbb {Q} ({\sqrt[{3}]{2}})\longrightarrow {\overline {\mathbb {Q} }}\\a+b{\sqrt[{3}]{2}}+c{\sqrt[{3}]{4}}\longmapsto a+b\omega {\sqrt[{3}]{2}}+c\omega ^{2}{\sqrt[{3}]{4}}\end{cases}}}$

is an embedding of ${\displaystyle \mathbb {Q} ({\sqrt[{3}]{2}})}$ in ${\displaystyle {\overline {\mathbb {Q} }}}$ whose restriction to ${\displaystyle \mathbb {Q} }$ is the identity. However, σ is not an automorphism of ${\displaystyle \mathbb {Q} ({\sqrt[{3}]{2}})}$.

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

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.