Editing Normal extension (section)

== Examples and counterexamples ==

For example, <math>\Q(\sqrt{2})</math> is a normal extension of <math>\Q,</math> since it is a splitting field of <math>x^2-2.</math> On the other hand, <math>\Q(\sqrt[3]{2})</math> is not a normal extension of <math>\Q</math> since the irreducible polynomial <math>x^3-2</math> has one root in it (namely, <math>\sqrt[3]{2}</math>), but not all of them (it does not have the non-real cubic roots of&nbsp;2). Recall that the field <math>\overline{\Q}</math> of [[algebraic number]]s is the algebraic closure of <math>\Q,</math> and thus it contains <math>\Q(\sqrt[3]{2}).</math> Let <math>\omega</math> be a primitive cubic root of unity. Then since, 
<math display="block">\Q (\sqrt[3]{2})=\left. \left \{a+b\sqrt[3]{2}+c\sqrt[3]{4}\in\overline{\Q }\,\,\right | \,\,a,b,c\in\Q \right \}</math> 
the map
<math display="block">\begin{cases} \sigma:\Q (\sqrt[3]{2})\longrightarrow\overline{\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}</math>
is an embedding of <math>\Q(\sqrt[3]{2})</math> in <math>\overline{\Q}</math> whose restriction to <math>\Q </math> is the identity. However, <math>\sigma</math> is not an automorphism of <math>\Q (\sqrt[3]{2}).</math>

For any prime <math>p,</math> the extension <math>\Q (\sqrt[p]{2}, \zeta_p)</math> is normal of degree <math>p(p-1).</math> It is a splitting field of <math>x^p - 2.</math> Here <math>\zeta_p</math> denotes any <math>p</math>th [[primitive root of unity]]. The field <math>\Q (\sqrt[3]{2}, \zeta_3)</math> is the normal closure (see below) of <math>\Q (\sqrt[3]{2}).</math>