Elements of Extension Fields

**h2osprey** · May 19th 2011, 11:40 AM

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

**NonCommAlg** · May 19th 2011, 05:38 PM

Originally Posted by h2osprey

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

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

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