# Elements of Extension Fields

• May 19th 2011, 11:40 AM
h2osprey
Elements of Extension Fields
Prove that $\displaystyle \sqrt{7}$ is not an element of $\displaystyle \mathbb{Q}(\sqrt{3 + \sqrt{2}}).$
• May 19th 2011, 05:38 PM
NonCommAlg
Quote:

Originally Posted by h2osprey
Prove that $\displaystyle \sqrt{7}$ is not an element of $\displaystyle \mathbb{Q}(\sqrt{3 + \sqrt{2}}).$

here is one way to solve the problem:
let$\displaystyle \sqrt{3+\sqrt{2}}=a$ and $\displaystyle \sqrt{3-\sqrt{2}}=b$. if $\displaystyle \sqrt{7} \in \mathbb{Q}(a)$, then $\displaystyle b \in \mathbb{Q}(a)$ because $\displaystyle ab = \sqrt{7}$. that means $\displaystyle \mathbb{Q}(a)$ is the splitting field of $\displaystyle x^4-6x^2+7$. thus $\displaystyle \mathbb{Q}(a)/\mathbb{Q}$ is Galois and hence $\displaystyle |\text{Gal}(\mathbb{Q}(a)/\mathbb{Q})|=[\mathbb{Q}(a):\mathbb{Q}]=4.$ this is a contradiction because the galois group of $\displaystyle x^4-6x^2+7$ is the dihedral group of order $\displaystyle 8$.