$\mathbb{Q}$ adjoining primes and the sum of root of those primes

Question

I have $p$, $q$ as primes, and I want to show that $\mathbb{Q}(\sqrt{p},\sqrt{q})=\mathbb{Q}(\sqrt{p}+\sqrt{q})$.

I was thinking about using inclusion both ways, so what does an element in $\mathbb{Q}(\sqrt{p},\sqrt{q})$ look like? Does it mean that $\sqrt{p}$ and $\sqrt{q}$ satisfy the polynomial $f(x),g(x)\in \mathbb{Q}(x)$? I am sorta stuck on what to do next, and what is the role of the primes?

Answer 1

1. It's clear that $\Bbb Q(\sqrt{p}+\sqrt{q}) \subseteq \Bbb Q(\sqrt{p},\sqrt{q})$. We need to prove the other way. $$\sqrt{p}+\sqrt{q} \in \Bbb Q(\sqrt{p}+\sqrt{q})$$ $$\implies \sqrt{pq} \in \Bbb Q(\sqrt{p}+\sqrt{q})$$ $$\implies q\sqrt{p}+p\sqrt{q} \in \Bbb Q(\sqrt{p}+\sqrt{q})$$ $$\implies \sqrt{q} \in \Bbb Q(\sqrt{p}+\sqrt{q})$$ and we are done.

2. An element in $\Bbb Q(\sqrt{p},\sqrt{q})$ looks like $(a+b\sqrt p +c\sqrt q+d \sqrt pq)$.

3. The polymonial they satisfy is $(x^2-p)(x^2-q) \in \Bbb Q[x]$.

Answer 2

One simple way: $\rm\ F = \Bbb Q(\sqrt{p}+\sqrt{q}) \supseteq \Bbb Q(\sqrt{p},\sqrt{q})\,$ (reverse is clear), since $\rm\,F\,$ contains not only $\, u = \sqrt{p}+\sqrt{q}\, $ but also $\,v = \sqrt{p}-\sqrt{q} = (p-q)/(\sqrt{p}+\sqrt{q}), \, $ so $\,\sqrt{p},\sqrt{q} = (u\pm v)/2 \in\rm F.\, $ QED $ $ For much further discussion see here.