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