Let $a, b$ be integers such that $b$ is not a square. Show that the field extension $\Bbb Q \subseteq \Bbb Q(\sqrt {a + \sqrt b})$ is normal if and only if at least one of $a^2 − b$ and $(a^2 − b)/b$ is a square in $\Bbb Q$. [Hint: Consider the normal closure of $\Bbb Q(\sqrt {a + \sqrt b})$ over $\Bbb Q$ as a field extension of $\Bbb Q(\sqrt b)$.]
($\implies$) My try:
Let $a; b$ be integers such that $b$ is not a square. if $a^2 − b$ is a square in $ \Bbb Q$., then the field extension $\Bbb Q \subseteq \Bbb Q(\sqrt {a + \sqrt b})$ is normal.
If $a^2 − b$ is a square in $ \Bbb Q$, it means $\sqrt {a^2 − b} \in \Bbb Q$.
I can write $\sqrt {a^2 − b}=\sqrt {a+\sqrt b}\sqrt {a-\sqrt b}$
then $\sqrt {a − \sqrt b}= \frac{\sqrt {a^2 − b}} {\sqrt {a + \sqrt b}}\in \Bbb Q(\sqrt {a + \sqrt b})$ which is the splitting field of $f(x)=(t^2-a)^2-b$, so this extension is normal.
Similarly I showed that if $(a^2 − b)/b$ is a square, the extension is normal
I am stuck trying to show the converse. How should I do that?
