# 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 $\operatorname{Gal}(F/\mathbb{Q})\cong\mathbb{Z}/2\mathbb{Z}$.

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 $\operatorname{Gal}(K/F)\cong\mathbb{Z}/2\mathbb{Z}$.

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 ExampleOfNormalExtension 2013-03-22 14:30:46 2013-03-22 14:30:46 alozano (2414) alozano (2414) 4 alozano (2414) Example msc 12F10 GaloisExtension CompositumOfAGaloisExtensionAndAnotherExtensionIsGalois NormalIsNotTransitive GaloisIsNotTransitive