Ramification of $5$ in $\mathbb{Q}( \sqrt[5]{n})$

Question

I need to study the ramification of $5$ in $K=\mathbb{Q}(\sqrt[5]{n})$. I know that $5$ ramifies in $\mathbb{Q}(\sqrt[5]{n})$ because $5$ divide the discriminant, my question is about the possible forms of $5\mathcal{O}_{K}$.

Answer 1

I will show that there is only one prime lying over $5$ in $\mathcal{O}_K$.

I will use the following proposition, which may be found in Neukirch's Algebraic Number Theory (prop. 8.2):

Proposition: Suppose the extension $L/K$ is generated by the zero $\alpha$ of the irreducible polynomial $f\in K\left[ X\right]$. Then the valuations $w_1,\dots,w_r$ of $L$ extending a valuation $v$ of $K$ correspond $1-1$ to the irreducible factors $f_1,\dots,f_r$ in the decomposition of $f$ over the completion $K_v$.

Using this proposition, it is enough to show that $X^5-n$ has a unique irreducible factor in $\mathbf{Q}_5$. If $n=0$, this is evident.

Otherwise, one may show that $X^5-n$ is irreducible. First note that $X^n-5$ has no quadratic factor in $\mathbf{Q}_5$: indeed, let $\alpha\in \overline{\mathbf{Q}_5}$ be a root of such a quadratic factor $Q$ so that $Q$ has roots $\alpha$ and $\alpha\zeta$ for some nontrivial fifth root of unity $\zeta$; then $\alpha^2\zeta$ and $\alpha(1+\zeta)$ are in $\mathbf{Q}_5$ (these are the coefficients of $Q$); from this, one deduces that $\alpha^2$ is in $\mathbf{Q}_5$, hence that $\zeta$ is in $\mathbf{Q}_5$, which leads to a contradiction. The same kind of reasoning shows that $X^5-n$ does not split, that it has no cubic factor with two linear factors nor any factor of degree $4$ (in all those cases, the contradiction is always the same: $\mathbf{Q}_5$ would contain a fifth primitve root of unity).

Thus, $X^5-n$ has a unique irreducible factor in $\mathbf{Q}_5$ which means that $5$ is totally ramified in $\mathcal{O}_K$.

Answer 2

my question is about the field $K = \mathbb{Q}(\sqrt[5]{n})$, $5$ is ramified in $K$ and we now that in the decomposition of $5\mathcal{O}_K$ the ramification index of ideals must be superior that $1$, so what is this decomposition.