Factorization of primes in normal closure of Quartic Field Motivation for the question comes from Marcus' book on Number Fields (exercise 13, Chapter 4).

Let $K= \mathbb{Q}[\sqrt[4]{m}, i]$ where $i=\sqrt{-1}$, $m\in \mathbb{Z}$ and $m$ is not a square. Suppose $p$ is an odd prime not dividing $m$. Prove that $p$ is unramified in $K$.

I think that I should use following Theorem:

For $K=\mathbb{Q}[\alpha]$,  where $\alpha \in \mathcal{O}_K$ and $f$ be any monic polynomial over $\mathbb{Z}$ such that $f(\alpha)=0$. If $p$ is a prime such that $p$ doesn't divide norm of $f'(\alpha)$, then $p$ is unramified in $K$.

So, I put $\alpha =\sqrt[4]{m}+i$ (as we do for biquadratic fields) and then I calculate $f(x)$ by simplifying $x=\sqrt[4]{m}+i$ to get following degree 8 equation: $f(x)= (x^4-2x^2-2x+1-m)^2+16x^2(x^2-1)^2$. But I don't know how am I supposed to calculate norm of its derivative evaluated at $\alpha$.
 A: A good strategy for working with complicated fields is to split them into fields you can deal with.
Suppose that an odd prime $p$ is ramified. Then either the ramification occurs in the extension $K/\mathbb Q(\sqrt[4]m)$, or it occurs in the extension $\mathbb Q(\sqrt[4]m)/\mathbb Q$ (or both). These two subextensions are much easier to handle.
You can use your theorem to show that $p$ ramifies in $\mathbb Q(\sqrt[4]m)$ if and only if $p\mid m$.
We can tackle the other extension as follows. Suppose that $\mathfrak p$ is a prime of $\mathbb Q(\sqrt[4]m)$ which ramifies in $L$ - say 
$$\mathfrak p\mathcal O_L = \mathfrak P^2$$
Suppose moreover that $\mathfrak p$ lies above $p\in\mathbb Z$.
Then 


*

*$\mathfrak P\cap\mathbb Z[i]$ must be the unique prime of $\mathbb Z[i]$ lying over $p$.

*We must have
$$p\mathbb Z[i] = (\mathfrak P\cap\mathbb Z[i])^2.$$


In particular, we've shown that if $\mathfrak p\subset \mathbb Q(\sqrt[4]m)$ ramifies in $L$, then $\mathfrak p\cap \mathbb Z$ must ramify in $\mathbb Z[i]$. Hence $p$ must be $2$.
More generally, the following theorem is extremely useful. It is not too hard to prove.

Theorem: If $p$ is unramified in two fields $L$ and $K$, then $p$ is unramified in the compositum $L\cdot K$.

The proof above basically proves this theorem in the case where $L=\mathbb Q(\sqrt[4]m)$ and $K = \mathbb Q(i)$.
