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I am trying to find the degree of the splitting field of the irreducible polynomial $f(x)=x^4-14x^2+36\in\mathbb Q[x]$.

The roots of $f$ are $\left\{\pm\sqrt{7\pm\sqrt{13}}\right\}$, so the splitting field must be:

$$E=\mathbb Q\left(\sqrt{7+\sqrt{13}},\sqrt{7-\sqrt{13}}\right)$$

If I consider the following chain of extensions

$$\mathbb Q \subseteq \mathbb Q\left(\sqrt{7+\sqrt{13}}\right)\subseteq E$$

The first extension has degree $4$ because $f(x)=Irr\left(\sqrt{7+\sqrt{13}},\,\mathbb Q\right)$, so the splitting field has at least degree $4$ (and cannot be $4$ because this would mean that $$\sqrt{7-\sqrt{13}}\in\mathbb Q\left(\sqrt{7+\sqrt{13}}\right).$$

I know that the degree must divide $n!$, so it can only be $6,8,12$ or $24$. Thus, by the multiplicative formula of degrees, the extension $E\big /\mathbb Q(\sqrt{7+\sqrt{13}})$ can only have degree $2,3$ or $6$.

Is this argument correct? I got stuck here, how could I continue?

EDIT: solution

In fact, $E=\mathbb Q\left(\sqrt{7+\sqrt{13}}\right)$. To see this we show that $\sqrt{7-\sqrt{13}}$ is generated by the field $\mathbb Q\left(\sqrt{7+\sqrt{13}}\right)$.

$$\sqrt{7-\sqrt{13}} = \frac{6}{\sqrt{7+\sqrt{13}}}\in \mathbb Q\left(\sqrt{7+\sqrt{13}}\right)$$

So, the extension has degree four because is a simple extension and $f(x)=Irr\left(\sqrt{7+\sqrt{13}},\,\mathbb Q\right)$.

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Try to figure out the minimal polynomial of $\sqrt{7-\sqrt{13}}$ over $\mathbb{Q}(\sqrt{7+\sqrt{13}})$. If you don't know where to start, factor $f(x)$ in $\mathbb{Q}(\sqrt{7+\sqrt{13}})[x]$. – user27126 Jan 27 '13 at 8:54
mmh, I get $f(x)=(x-\sqrt{7-\sqrt{13}})(x+\sqrt{7-\sqrt{13}})(x^2 + ( \sqrt{7+\sqrt{13}})(\sqrt{7+\sqrt{13}})-14)$, so I have to check that $x^2 + ( \sqrt{7+\sqrt{13}})(\sqrt{7+\sqrt{13}})-14$ is the polynomial that I'm looking for (ie irreducible), right? – synack Jan 27 '13 at 9:16
up vote 3 down vote accepted

I think you will find that $\sqrt{7-\sqrt{13}}$ is in fact in the field generated by $\sqrt{7+\sqrt{13}}$. See what happens when you multiply these two radicals together.

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+1 you're right, thank you – synack Jan 27 '13 at 9:24

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