unramified prime in pure quintic field

Question

Let $\Gamma = \mathbb{Q}(\sqrt[3]n)$ a pure cubic field and $p$ prime in $\mathbb{Z}$. We have the following theorem introduced by Dedekind:

Theorem: (1) If $p$ divides $n$ and $p\neq3$ so $p$ is ramified in $\Gamma$, then $p\mathcal{O}_\Gamma$ = $\mathcal{P}^3$ and $\mathcal{N}(\mathcal{P}) = p$

(2) if $p\nmid n$ and $p\equiv -1\pmod{3}$, so $p$ is unramified in $\Gamma$ then : $p\mathcal{O}_\Gamma$ = $\mathcal{P}\mathcal{P}_1$ and $\mathcal{N}(\mathcal{P}) = p$, $\mathcal{N}(\mathcal{P}_1) = p^2$

(3) if $p\nmid n$ and $p\equiv 1\pmod{3}$, so $p$ is unramified in $\Gamma$ then :

• $p\mathcal{O}_\Gamma$ = $\mathcal{P}\mathcal{P}_1\mathcal{P}_2$ and $\mathcal{N}(\mathcal{P}) = \mathcal{N}(\mathcal{P}_1)=\mathcal{N}(\mathcal{P}_2)$ if $n$ is a cubic residue modulo $p$

• $p\mathcal{O}_\Gamma$ = $\mathcal{P}$ and $\mathcal{N}(\mathcal{P}) = p^3$ if $n$ is not a cubic residue modulo $p$

the decomposition in the pure cubic field is solved by this theorem, my question is about the pure quintic field $\mathbb{Q}(\sqrt[5]n)$, for the first point of the theorem we have the same result for ramified prime, but for the unramified prime I need to now what is their decomposition

Answer 1

You can reason very similarly using the simple result that for any prime $ p $ coprime to the discriminant of $ X^5 - n $ (equal to $ 5n^4 $) you have that the splitting of $ p $ in the ring of integers of $ \mathbf Q(\sqrt[5]{n}) $ is identical to its splitting in the order $ \mathbf Z[\sqrt[5]{n}] \cong \mathbf Z[x]/(x^5 - n) $, which amounts to checking how $ x^5 - n $ factors modulo $ p $. This gives you a very similar set of conditions to the one you have for the cubic case.

If $ p $ is $ 2 $ or $ 3 $ modulo $ 5 $, then fifth roots exist and are unique in both $ \mathbf F_p $ and in $ \mathbf F_{p^2} $, which means $ x^5 - n $ splits as the product of a linear and a quartic factor, corresponding to a splitting of $ (p) = \mathfrak p_1 \mathfrak p_2 $ with $ f_1 = 1, f_2 = 4 $. If $ p $ is $ 4 $ modulo $ 5 $, then fifth roots exist and are unique in $ \mathbf F_p $ but not in $ \mathbf F_{p^2} $, which means $ x^5 - n $ splits as the product of a linear and two quadratic factors, giving a splitting $ (p) = \mathfrak p_1 \mathfrak p_2 \mathfrak p_3 $ with $ f_1 = 1, f_2 = f_3 = 2 $. If $ p $ is $ 1 $ modulo $ 5 $ then depending on whether $ n $ is a fifth power modulo $ p $ or not, $ p $ either splits completely or remains inert.

Answer 2

Suppose $p\nmid 5n$. Then the decomposition of the ideal $(p)$ is governed by the factorisation of $X^5-n$ over the field $\Bbb F_p$ of $p$ elements.

If $5\mid(p-1)$ then $\Bbb F_p$ has five fifth roots of unity. Therefore either $X^5-n$ splits into five linear factors, or is irreducible, according to whether or not $n$ is a quintic residue modulo $p$. So in these two cases, $(p)$ splits into five ideals of norm $p$, or is irreducible.

If $5\nmid(p-1)$ then $X^5-n$ has a unique linear factor over $\Bbb F_p$ so either it splits into a linear times an irreducible quartic, or a linear times two irreducible quadratics. We need to distinguish these cases.

Suppose that $5\nmid(p-1)$. Let $X-a$ be the linear factor of $X^5-n$ over $\Bbb F_p$. The remaining linear factors are $X-\xi^ka$ where $\xi$ is a primitive fifth root of unity and $k\in\{1,2,3,4\}$.

If $p\equiv-1\pmod 5$ then $(X-\xi a)(X-\xi^4 a)$ and $(X-\xi^2 a)(X-\xi^3 a)$ are irreducible quadratics over $\Bbb F_p$. So $X^5-n$ is a linear times two irreducible quadratics over $\Bbb F_p$ and so $(p)$ is a product of a prime ideal of norm $p$ and two of norm $p^2$.

If $p\not\equiv-1\pmod 5$ then the $\xi^k a$ are all conjugates over $\Bbb F_p$. Then $(X-\xi a)(X-\xi^2 a)(X-\xi^3 a)(X-\xi^4 a)$ is irreducible over $\Bbb F_p$. In this case $(p)$ is a product of a prime ideal of norm $p$ and one of norm $p^4$.