Decomposition of unramified primes in normal closure of pure quintic field Let $\Gamma$ = $\mathbb{Q}(\sqrt[5]{n})$ a pure quintic field, $k$ = $\mathbb{Q}(\zeta_5)$ the $5^{th}$ cyclotomic field, then $N$ = $\mathbb{Q}(\sqrt[5]{n}, \zeta_5)$ is the normal closure of $\Gamma$.
Its known that a prime $p \in \mathbb{Z}$ such that $p \equiv -1\,mod(5)$ decompose in $k$ as $p\mathcal{O}_k = \pi_1\pi_2$,  and if $p$ is unramified in $\Gamma$ then we have $p\mathcal{O}_\Gamma = \mathcal{P}_1\mathcal{P}_2\mathcal{P}_3$
I need to find the decomposition of that unramified $p$ in $N$, its seems to be $p\mathcal{O}_N = \mathcal{B}_1\mathcal{B}_2\mathcal{B}_3\mathcal{B}_4\mathcal{B}_5\mathcal{B}_6$ but I know its wrong because $N/\mathbb{Q}$ is Galois of degree 20. So what is the correct decomposition of $p $ ??
 A: Let’s see whether I can get this right. I think the key is to look first at the behavior of $p$ in $k$, not $\Gamma$.
I’ll look at the case you seem to be interested in, $p\equiv9\pmod{10}$. Then we check that the minimal polynomial for $\zeta_5$, namely $X^4+X^3+X^2+X+1$, splits into two quadratics, by seeing that $\Bbb F_{p^2}$ is the first extension of $\Bbb F_p$ that contains all fifth roots of unity. Since the extension $k\supset\Bbb Q$ is abelian, we have the equation $[k:\Bbb Q]=4=efg$, where $e$ is the ramification, $f$ is the residue field, $g$ is the number of primes above $p$.
We know that $e=1$ (because $k$ is ramified only above $5$), and we’ve seen that $f=2$, so that there are two primes of $k$ above $p$, call them $\mathfrak p_1$ and $\mathfrak p_2$.
Now we want to see how each of these two behaves in $N=k(n^{1/5})$. I claim that each $\mathfrak p_i$ splits completely in $N$, because the
minimal polynomial for $n^{1/5}$, namely $X^5-n$, splits completely in $\Bbb F_{p^2}$. Why is that? Let’s look at the multiplicative group $\Bbb F_{p^2}^\times$, cyclic of order $p^2-1=(p-1)(p+1)$. Let $\gamma$ be a generator. Clearly $\gamma^{p+1}\in\Bbb F_p^\times$, because its $(p-1)$-th power is identity. But $\gamma^{p+1}$ is a tenth power of an element pf $\Bbb F_{p^2}$, so that every element of $\Bbb F_p^\times$ is tenth power of en element of the bigger finite field. So $X^5-n$ has a root in $\Bbb F_{p^2}$, and indeed all roots, because the fifth roots of unity are
in $\Bbb F_{p^2}$. This is enough to show that the residue field extension degree of any prime of $N$ that’s above $\mathfrak p_i$ will be $1$. The extension is unramified (you need to assume that $n$ is indivisible by $5$ for this), so $efg=5$ says $g=5$.
Thus in this case that $p\equiv9\pmod{10}$ and $\gcd(5,n)=1$, there are ten primes of $N$ above $p$.
EDIT — Addendum:
The above is rather prolix, even turgid, but I’ve decided to leave it. Let me just say that exactly the same kind of argument shows easily, in case $p$ is unramified in $N$, then no matter the congruence of $p$ modulo $5$, every prime $\mathfrak p|p$ of $k$ splits completely in $N$, so that the splitting is determined entirely by the residue field extension degree $f$ of the prime(s) $\mathfrak p$ of $k$ over $p$. These numbers are: if $p\equiv1\pmod5$ then $f=1$; if $p\equiv2$ or $p\equiv3\pmod5$ then $f=4$; and as above if $p\equiv4\pmod5$, then $f=2$.
This means that if $p\equiv1\pmod5$, then $p$ splits completely in $N$; if $p\equiv2,3$ then $p$ splits into five primes $\mathfrak P$ of $N$, each with $f=4$; and if $p\equiv4$, then $p$ slits into ten primes $\mathfrak P$ with $f=2$.
