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Determine the degree of the extension $Q(\sqrt{3+2 \sqrt{2}})$ over Q. I can see that $$3+2 \sqrt{2} = (1+ \sqrt2)(1+ \sqrt2) =(1+ \sqrt2)^2$$ does that mean $$x^2 -(1+ \sqrt2)^2)$$ has a degree $2$. Is this correct

user146269
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  • You have to find an irreducible polynomial $f \in \mathbb{Q}[x]$ such that it has $\sqrt{3+2(\sqrt{2})}$ as a root. Is your polynomial in $\mathbb{Q}[x]$? – user1314 Apr 30 '15 at 16:31
  • yes the polynomial is in $Q[x]$ – user146269 Apr 30 '15 at 16:39
  • what polynomial is of degree $2$? What polynomial is in $\mathbb{Q}[x]$?? Write clearly.. –  Apr 30 '15 at 16:42
  • but $(1+ \sqrt{2})^2 = 3 + 2\sqrt{2}$ and this doesnt lie in $\mathbb{Q}$! – user1314 Apr 30 '15 at 16:42
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    http://math.stackexchange.com/questions/193317/degree-of-the-extension-mathbbq-sqrta-sqrtb-over-mathbbq?rq=1 may be of some use.. –  Apr 30 '15 at 16:50

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Example: Let $p(x) = \sqrt{1+\sqrt{2}}$.

Let $u = \sqrt{1+\sqrt{2}} \Rightarrow u^2 = (1+\sqrt{2})^2 = 2+2\sqrt{2} \Rightarrow (u^2-2)^2 = 8 \Rightarrow u^4-4u^2+4=8 \Rightarrow u^4-4u2-4=0$

Hence $u$ is the root of $x^4-4u^2-4$. Try this techniques with your adjoined root and see if you can say anything about the polynomial that it satisifies over $\mathbb{Q}$.

Mr.Fry
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