1. ## Tower Law problem

Show $[\mathbb{Q}(\alpha):\mathbb{Q}]=2$ where $\alpha$ = $\sqrt{4+2\sqrt{3}}$

This is what i have.

$[\mathbb{Q}(\sqrt{4+2\sqrt{3}}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{4+2\sqrt{3}}):\mathbb{Q}\sqrt{3}] [\mathbb{Q}\sqrt{3}:\mathbb{Q}]$

$[\mathbb{Q}(\sqrt{4+2\sqrt{3}}):\mathbb{Q}] = 2*2 = 4$

2. Originally Posted by frankdent1
Show $[\mathbb{Q}(\alpha):\mathbb{Q}]=2$ where $\alpha$ = $\sqrt{4+2\sqrt{3}}$

This is what i have.

$[\mathbb{Q}(\sqrt{4+2\sqrt{3}}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{4+2\sqrt{3}}):\mathbb{Q}\sqrt{3}] [\mathbb{Q}\sqrt{3}:\mathbb{Q}]$

$[\mathbb{Q}(\sqrt{4+2\sqrt{3}}):\mathbb{Q}] = 2*2 = 4$
Hint: $\sqrt{4+2\sqrt{3}}=\sqrt{3}+ 1.$