# Normal extension

In abstract algebra, an algebraic field extension L/K is said to be normal if L is the splitting field of a family of polynomials in K[X]. Bourbaki calls such an extension a quasi-Galois extension.

## Equivalent properties and examples

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. In other words, 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.)

For example, $\mathbb{Q}(\sqrt{2})$ is a normal extension of $\mathbb{Q}$, since it is a splitting field of x2 − 2. On the other hand, $\mathbb{Q}(\sqrt[3]{2})$ is not a normal extension of $\mathbb{Q}$ since the irreducible polynomial x3 − 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).

The fact that $\mathbb{Q}(\sqrt[3]{2})$ is not a normal extension of $\mathbb{Q}$ can also be seen using the first of the three properties above. The field $\mathbb{A}$ of algebraic numbers is an algebraic closure of $\mathbb{Q}$ containing $\mathbb{Q}(\sqrt[3]{2})$. On the other hand

$\mathbb{Q}(\sqrt[3]{2})=\{a+b\sqrt[3]{2}+c\sqrt[3]{4}\in\mathbb{A}\,|\,a,b,c\in\mathbb{Q}\}$

and, if ω is one of the two non-real cubic roots of 2, then the map

$\begin{array}{rccc}\sigma:&\mathbb{Q}(\sqrt[3]{2})&\longrightarrow&\mathbb{A}\\&a+b\sqrt[3]{2}+c\sqrt[3]{4}&\mapsto&a+b\omega\sqrt[3]{2}+c\omega^2\sqrt[3]{4}\end{array}$

is an embedding of $\mathbb{Q}(\sqrt[3]{2})$ in $\mathbb{A}$ 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 $\mathbb{Q}(\sqrt[3]{2})$.

## 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.

## 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. such that 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.