# example of normal extension

Let $F=\mathbb{Q}(\sqrt{2})$. Then the extension $F/\mathbb{Q}$ is normal because $F$ is clearly the splitting field^{} of the polynomial^{} $f(x)={x}^{2}-2$. Furthermore $F/\mathbb{Q}$ is a Galois extension^{} with $\mathrm{Gal}(F/\mathbb{Q})\cong \mathbb{Z}/2\mathbb{Z}$.

Now, let ${2}^{1/4}$ denote the positive real fourth root of $2$ and define $K=F({2}^{1/4})$. Then the extension $K/F$ is normal because $K$ is the splitting field of $k(x)={x}^{2}-\sqrt{2}$, and as before $K/F$ is a Galois extension with $\mathrm{Gal}(K/F)\cong \mathbb{Z}/2\mathbb{Z}$.

However, the extension $K/\mathbb{Q}$ is neither normal nor Galois. Indeed, the polynomial $g(x)={x}^{4}-2$ has one root in $K$ (actually two), namely ${2}^{1/4}$, and yet $g(x)$ does not split in $K$ into linear factors.

$$g(x)={x}^{4}-2=({x}^{2}-\sqrt{2})\cdot ({x}^{2}+\sqrt{2})=(x-{2}^{1/4})\cdot (x+{2}^{1/4})\cdot ({x}^{2}+\sqrt{2})$$ |

The Galois closure of $K$ over $\mathbb{Q}$ is $L=\mathbb{Q}({2}^{1/4},i)$.

Title | example of normal extension^{} |
---|---|

Canonical name | ExampleOfNormalExtension |

Date of creation | 2013-03-22 14:30:46 |

Last modified on | 2013-03-22 14:30:46 |

Owner | alozano (2414) |

Last modified by | alozano (2414) |

Numerical id | 4 |

Author | alozano (2414) |

Entry type | Example |

Classification | msc 12F10 |

Related topic | GaloisExtension |

Related topic | CompositumOfAGaloisExtensionAndAnotherExtensionIsGalois |

Related topic | NormalIsNotTransitive |

Related topic | GaloisIsNotTransitive |