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Show $\sqrt{1+\sqrt{2}}$ is algebraic over $\mathbb{Q}$ with degree $4$

Let $\alpha=\sqrt{1+\sqrt{2}}$, and it is a root of $f(x)=x^4-2x^2+1\in \mathbb{Q}[X]$, so $f(x)$ is irreducible in $\mathbb{Q}[X]$, thus $\alpha$ is algebraic over $\mathbb{Q}$. Clearly, $\alpha$ is not a root of polynomial with degree one. How to show $\sqrt{1+\sqrt{2}}$ has degree $4$? My thought is let $g(x_1)=a_1x^2+b_1x+c_1, q(x)=a_2x^3+b_2x^2+c_2x+d$ and let $g(x_1)=0$, $q(x_2)=0$, but is seems not right. Can anyone give a hit to do it? Thanks

Thomas Andrews
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    If you've proved that $f$ is irreducible, then you're done. – lhf Jun 02 '15 at 19:20
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    If $f(x)$ is irreducible, $\alpha$ has degree $4$. But it is not irreducible ($x^2-1)^2$). Actually there's an error, you should find $,f(x)=x^4-2x^2-1$, and you must prove it is irreducible. – Bernard Jun 02 '15 at 19:23

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As pointed out by Bernard in the comments, $\alpha$ is a root of $f(x) = x^4 - 2x^2 \color{red}{-} 1$, not of $x^4 - 2x^2 + 1 = (x^2 - 1)^2$. Now observe that $f(\pm1) = -2$, so $f(x)$ doesn't have any rational root. This means that $\alpha \notin \Bbb{Q}$ has either degree $2$ or $4$.

On the other hand, if $\alpha$ had degree $2$ we would have $$ \Bbb{Q} = \Bbb{Q}(\alpha^2) = \Bbb{Q}(1 + \sqrt{2}) = \Bbb{Q}(\sqrt{2}) $$ which is absurd, thus $\alpha$ must have degree $4$. In particular, this means that $f(x)$ must be irreducible.

A.P.
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