If $p$, $q$ are two distinct prime numbers in $\Bbb N$ then $\Bbb Q(\sqrt p, \sqrt q)) = \Bbb Q(\sqrt p + \sqrt q)$.

Question

If $p$, $q$ are two distinct prime numbers in $\Bbb N$ then

$$\Bbb Q(\sqrt p, \sqrt q) = \Bbb Q(\sqrt p + \sqrt q)$$

How to prove that the first is included in the second, and why is it ligical to assume the equality

Answer 1

Let $a=\sqrt p$ and $b=\sqrt q$.

Let $c=(a^2+3b^2)a+(3a^2+b^2)b.$ Then $c=a^3+3a^2b+3ab^2+b^3=(a+b)^3\in \Bbb Q(a+b).$

Let $d=(a^2+3b^2)(a+b).$ Then $d=(p+3q)(a+b)\in \Bbb Q(a+b).$

So $2(p-q)b=2(a^2-b^2)b=c-d\in \Bbb Q(a+b).$

So $\sqrt q=b\in \Bbb Q(a+b)=\Bbb Q(\sqrt p +\sqrt q).$

The rest is up to you.