Show that if $\sqrt{a}+\sqrt{b}\neq 0$, then $\mathbb{Q}(\sqrt{a}+\sqrt{b}) = \mathbb{Q}(\sqrt{a},\sqrt{b}), \forall a,b\in\mathbb{Q}$

Question

Show that if $\sqrt{a}+\sqrt{b}\neq 0$, then $\mathbb{Q}(\sqrt{a}+\sqrt{b}) = \mathbb{Q}(\sqrt{a},\sqrt{b}), \forall a,b\in\mathbb{Q}$

It has something to do with $\sqrt{a} = -\sqrt{b}\implies a^2 = b^2$

The inclusion $\mathbb{Q}(\sqrt{a}+\sqrt{b})\subseteq \mathbb{Q}(\sqrt{a},\sqrt{b})$ is obvious

Now take $x\in \mathbb{Q}(\sqrt{a},\sqrt{b})$, that is: $x = c + d\sqrt{a} + e\sqrt{b} + f\sqrt{ab}$. I need to show that this is in $\mathbb{Q}(\sqrt{a}+\sqrt{b})$, that is, $x = g + h\sqrt{a} + i\sqrt{b}$, but I can't get that $\sqrt{ab}$ removed

Answer 1

Let $\newcommand{\la}{\lambda}\la=\sqrt a+\sqrt b$. Then $$(\la-\sqrt a)^2=b$$ that is $$\la^2-2\la\sqrt a+a-b=0$$ or $$\sqrt a=\frac{\la^2+a-b}{2\la}\in\Bbb Q(\la).$$ (NB $\la\ne0$.) Once we have this, $\sqrt b$, $\sqrt{ab}\in\Bbb Q(\la)$ follow easily

Answer 2

To show that $\Bbb{Q}(\sqrt{a},\sqrt{b})\subseteq\Bbb{Q}(\sqrt{a}+\sqrt{b})$, consider $$(\sqrt{a}+\sqrt{b})^{-1}=\frac{1}{a-b}(\sqrt{a}-\sqrt{b})\ .$$ Can you take it from there?