Find the degree and a basis for $\mathbb{Q}(\sqrt{3} + \sqrt{5})$ over $\mathbb{Q}(\sqrt{15})$ Find the degree and a basis for $\mathbb{Q}(\sqrt{3} + \sqrt{5})$ over $\mathbb{Q}(\sqrt{15})$.
My attempt:
I first proved that $\mathbb{Q}(\sqrt{3} + \sqrt{5}) = \mathbb{Q}(\sqrt{3}, \sqrt{5})$,
Then using tower law its fairly straight forward to establish that $[\mathbb{Q}(\sqrt{3} + \sqrt{5}) : \mathbb{Q}(\sqrt{15})] = 2$.
But I am having difficulty determining the basis for this extension.
I can only take a guess that the answer is $\{1, \sqrt{3}\}$ or $\{1, \sqrt{5}\}$.
Is there a better method than just trail and error to determine the basis in this case.
 A: There's a very obvious way to do this when you know a primitive element: clearly, we have that $ \mathbf Q(\sqrt{3} + \sqrt{5}) = \mathbf Q(\sqrt{15})(\sqrt{3} + \sqrt{5}) $, that is, the extension is generated over $ \mathbf Q(\sqrt{15}) $ by the same primitive element. Since we know that the extension degree is $ 2 $, we can easily surmise that a basis for the extension is simply $ \{ 1, \sqrt{3} + \sqrt{5} \} $, which is readily verified. Of course, the extension has many bases...
A: One way to prove this is :
$a=(\sqrt{3}+\sqrt{5})$
$a^2=(\sqrt{3}+\sqrt{5})^2$
Thus $a^2-8+2\sqrt{15}=0$ and the polynomial $f(X)=X^2-8+2\sqrt{15}$ is irreducible over $\mathbb{Q}(\sqrt{15})$ because its roots do not lie in $\mathbb{Q}(\sqrt{15})$.
Thus a basis is the set $\{1,\sqrt{3}+\sqrt{5}\}$
Also $\{1,\sqrt{5}\}$ is indeed a basis for the extension because an element  $x \in \mathbb{Q}(\sqrt{3},\sqrt{5})$ is of the form $x=a+b \sqrt{3} +c\sqrt{5}+d\sqrt{15}=a+\frac{b}{5} \sqrt{15} \sqrt{5} +c \sqrt{5}+d \sqrt{15}=(a+d \sqrt{15})+(c+ \frac{b}{5} \sqrt{15}) \sqrt{5}$.
In the same way you can prove that $\{1,\sqrt{3}\}$ is a basis of the extension you ask in your question.
