Is $Q(2^{\frac14},\sqrt7)/Q(\sqrt2)$ normal? I think it is because $(x^2-\sqrt2)(x^2-7)$ is a polynomial over $Q(\sqrt2)$ and $Q(2^{\frac14},\sqrt7)$ is the splitting field of this over $Q(\sqrt2)$ $\iff$ $Q(2^{\frac14},\sqrt7)$ is normal.
So it is normal? Is this correct?
 A: It is the compositum of two quadratic extensions, which are abelian extensions.
The compositum of two abelian extensions is always abelian, and a fortiori normal.

More explicit answer:
The extension is generated by two elements, $x = \sqrt[4]{2}$ and $y = \sqrt{7}$. So it suffices to show that all conjugates of $x$ and $y$ live in the same extension.
For $x$: the minimal polynomial of $x$ over $\mathbb{Q}(\sqrt{2})$ is $X^2 - \sqrt{2}$, from which it is clear that the only non-trivial conjugate of $x$ is $-x$, which lives in the same field as $x$.
For $y$: exactly the same argument.
So the extension is normal.

This looks like an illustrative example in Galois theory. Note that the extension $\mathbb{Q}(\sqrt[4]{2}, \sqrt{7})/\mathbb{Q}$ is not normal, while both $\mathbb{Q}(\sqrt[4]{2}, \sqrt{7})/\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$ are normal.
It is a counter-example to the assertion that $K/L$ and $L/M$ normal imply $K/M$ normal.
A: Yes it is correct. There is also another (equivalent) definition of normality, which says that an extension $L/K$ is normal if for any extension $M$ of $L$ and any field homomorphism $\gamma: L \rightarrow M$ that fixes $K$ pointwise we have $\gamma(L) = L$. To convince you even more of the fact it is normal, let us prove it using this definition directly. Suppose $f: \mathbb{Q}(\sqrt[4]{2}, \sqrt{7}) \rightarrow \overline{\mathbb{Q}}$ is a field homomorphism that fixes $\mathbb{Q}(\sqrt{2})$. The homomorphism can send $\sqrt{7}$ only to other roots of $X^2-7$, so to $\sqrt{7}$ or to $-\sqrt{7}$. It can send $\sqrt[4]{2}$ only to other roots of $X^4-2=(X^2-\sqrt{2})(X^2+\sqrt{2})$, from which we see $f$ can send $\sqrt[4]{2}$ only to other roots of $X^2-\sqrt{2}$, so only to $\sqrt[4]{2}$ or to $-\sqrt[4]{2}$. From this we see $f(\mathbb{Q}(\sqrt[4]{2}, \sqrt{7}))=\mathbb{Q}(\sqrt[4]{2}, \sqrt{7})$.
