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Find an uncomplicated basis for $\mathbb Q(d)$ over $\mathbb Q$ where $d$ is a root of $x^4-14x^2+9$. Then a basis for $\mathbb Q(\sqrt{7+2\sqrt{10}})$ over $\mathbb Q$.

I thought the way to make a basis was to take the root of $d$ (which is $\pm\sqrt{7\pm2\sqrt{10}}$) and then these are the elements of the basis along with 1. So

{$1, \sqrt{7+2\sqrt{10}}, -\sqrt{7+2\sqrt{10}}, \sqrt{7-2\sqrt{10}}, -\sqrt{7-2\sqrt{10}}$} is our basis for $\mathbb Q(d)$.

I don't know how to find a basis of $\sqrt{7+2\sqrt{10}}$ then because I can't make any linear combinations.

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    $\begingroup$ If $d$ is a root of any irreducible quartic, then $\{1,d,d^2,d^3\}$ is a basis for $\mathbb{Q}(d)$ over $\mathbb{Q}$. $\endgroup$ – Barry Cipra Dec 2 '17 at 23:41
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You should note that, by denesting (justified by $7^2-40=3^2$), $$ \sqrt{7+2\sqrt{10}}=\sqrt{\frac{7+3}{2}}+\sqrt{\frac{7-3}{2}}=\sqrt{5}+\sqrt{2} $$ Since $(\sqrt{5}+\sqrt{2})^2=7+2\sqrt{10}$ and $$ (\sqrt{5}+\sqrt{2})^3= 5\sqrt{5}+15\sqrt{2}+6\sqrt{5}+2\sqrt{2}=11\sqrt{5}+17\sqrt{2} $$ we see that $\{1,\sqrt{2},\sqrt{5},\sqrt{10}\}$ is a basis, because these numbers are linear combinations of $\{1,d,d^2,d^3\}$ and conversely.


Note that, if $d$ is a root of an irreducible polynomial $f(x)$, then $\mathbb{Q}(d)$ is not, in general, generated by the roots of $f(x)$. Consider, for example, $f(x)=x^3-2$.

What's true is that, if $d$ is a root and $f(x)$ has degree $n$, then a basis is $\{1,d,d^2,\dots,d^{n-1}\}$.

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The splitting field is $\mathbb{Q}(\sqrt{2},\sqrt{5})$ since $-14 = -2 \cdot (5+2)$ and $9 = (5 - 2)^2$. In general the splitting field of the biquadratic polynomial $x^4 - 2(a+b)x^2 + (a-b)^2$ is $\mathbb{Q}(\sqrt{a},\sqrt{b})$.

So $1, \sqrt{2},\sqrt{5},\sqrt{10}$ is one possible basis.

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