$b$ not a square, $a^2 − b$ a square Why is $\Bbb Q(\sqrt {a + \sqrt b})$ the splitting field of $f(x)=(t^2-a)^2-b$ and a normal extension?

Question

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.

My try: 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 $\sqrt {a − \sqrt b}= \frac{\sqrt {a^2 − b}} {\sqrt {a + \sqrt b}}\in \Bbb Q(\sqrt {a + \sqrt b})$

I have a feeling this is the splitting field of $f(x)=(t^2-a)^2-b$ but I don't know how to justify it.

1 By the definition of splitting field, how do I know $\Bbb Q(\sqrt {a + \sqrt b})$ is the smallest field over which f decomposes into linear factors?

2 To conclude normality I need to show that very irreducible polynomial over $\Bbb Q$ which has a root in $\Bbb Q(\sqrt {a + \sqrt b})$, splits into linear factors in $\Bbb Q(\sqrt {a + \sqrt b})$. How do I show this?

Answer 1

Let $L$ be a field extension of $K$ with $\alpha \in L$. Then we can define $K(\alpha)$ to be the smallest subfield of $L$ which contains $K$ and $\alpha$. Taking this view, it is now trivial why if $f \in K[X]$, $L$ is a field extension of $K$ which contains all the roots of $f$, say $\alpha_1, \dots, \alpha_n$, then $K(\alpha_1, \dots, \alpha_n)$ is the spitting field for $f$.

So taking this view, the answer to your first question is kind of trivial, because the splitting field of $(t^2-a)^2-b$ is just $\mathbb{Q}(\sqrt{a+\sqrt{b}}, \sqrt{a-\sqrt{b}}, -\sqrt{a+\sqrt{b}}, -\sqrt{a-\sqrt{b}})$, which you can see is the same as $\mathbb{Q}(\sqrt{a+\sqrt{b}})$.

If you're thinking of $K(\alpha)$ as the field $K$ with $\alpha$ added in, along with all the other elements required to make a field such as $\alpha^2$, $\alpha + 1$, $\alpha^{-1}$ and so on. Well, this is really the smallest subfield of $L$ containing $K$ and $\alpha$. Intuitively if I add $\alpha$ to $K$ and add just enough elements to make this a field, I should be getting the smallest field which contains both $K$ and $\alpha$. This can be seen with the following:

Exercise: Let $L / K $ be a field extension, $\alpha \in L$ algebraic. Then $K(\alpha)$ is the image of the ring homomorphism $\phi: K[t] \rightarrow L$ defined by $f \mapsto f(\alpha)$.