6
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I want to find the degree of $\mathbb{Q}(\sqrt{3+2\sqrt{2}})$ over $\mathbb{Q}$. I observe that $3+2\sqrt{2}=2+2\sqrt{2}+1=(\sqrt{2}+1)^2$ so $$ \mathbb{Q}(\sqrt{3+2\sqrt{2}})=\mathbb{Q}(\sqrt{2}+1)=\mathbb{Q}(\sqrt{2}) $$ so the degree is 2.

Is there a more mechanical way to show this without noticing the factorization?

3 Answers 3