# 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 over 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, $\mathbb {Q} ({\sqrt {2}})$ 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 $\mathbb {Q} ,$ and thus it contains $\mathbb {Q} ({\sqrt[{3}]{2}}).$ Let $\omega$ be a primitive cubic root of unity. Then since, $\mathbb {Q} ({\sqrt[{3}]{2}})=\left.\left\{a+b{\sqrt[{3}]{2}}+c{\sqrt[{3}]{4}}\in {\overline {\mathbb {Q} }}\,\,\right|\,\,a,b,c\in \mathbb {Q} \right\}$ 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, $\sigma$ 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 $x^{p}-2.$ Here $\zeta _{p}$ denotes any $p$th 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 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.