<s>[INST] uncomplicated basis for $\mathbb Q(d)$ over $\mathbb Q$ where $d$ is a root of $x^4-14x^2+9$ 
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.
 [/INST] 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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