Show that $\mathbb{Q}(\sqrt{a} + \sqrt{b}) = \mathbb{Q}(\sqrt{a}, \sqrt{b})$

Question

Let $a$ and $b$ be rational numbers. Show that $\mathbb{Q}(\sqrt{a}, \sqrt{b}) = \mathbb{Q}(\sqrt{a}+ \sqrt{b})$.

I think this question might have been asked before, but Google and the stackexchange search don't show anything.

Answer 1

The inclusion $\Bbb Q(\sqrt a+\sqrt b) \subseteq\Bbb Q(\sqrt a, \sqrt,b)$ is clear.

Conversely, $(\sqrt a+\sqrt b)(\sqrt a-\sqrt b)=a-b$ and so $\sqrt a -\sqrt b=(a-b)/(\sqrt a+\sqrt b)$.

Thus $\sqrt a-\sqrt b\in \Bbb Q(\sqrt a+\sqrt b)$ and hence $(\sqrt a-\sqrt b)+(\sqrt a + \sqrt b) = 2\sqrt a$ in $\Bbb Q(\sqrt a+\sqrt b)$, i.e., $\sqrt a\in\Bbb Q(\sqrt a,\sqrt b)$. Similarly, $\sqrt b\in\Bbb Q(\sqrt a,\sqrt b)$. Equality follows.