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 $5\mathcal{O}_{K}$.
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In the case where $\mathcal{O}_K=\mathbb{Z}[\sqrt[5]{n}]$, $5$ is totally ramified , since $X^5-n=X^5-n^5=(X-n)^5 \mod 5$. – GreginGre Nov 26 '19 at 23:00
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Do you know the $p$-adic numbers ? For $p\ne 5$ it is easy to understand the extension $\Bbb{Q}_p(n^{1/5})/\Bbb{Q}_p$ thus the primes above $p$ in $K$. For $p=5$ there are few more cases. – reuns Nov 27 '19 at 05:51
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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 $\mathcal{O}_K$.
Gaussian
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$K=\Bbb{Q}(26^{1/5})$ then the splitting field of $x^5-26\in \Bbb{Q}_5[x]$ is the same as $x^5-1$ thus $5 O_K= I^4 J$. Your argument about the quadratic factor shows that for $b\in 1,2,3,4$ then $x^5-(1+5b)\in \Bbb{Q}_5[x]$ has no root thus it is irreducible, since $Gal(\overline{\Bbb{Q}}_5/\Bbb{Q}_5)$ acts by multiplication by $\zeta_5$ the extension is totally ramified thus $5O_K =I^5$, finally since $(1+5^2c)^{5}\in \Bbb{Q}_5$ then $n=5^l a^5 (1+5^2c)^5 (1+5b),5\nmid a$ and the only additional case to consider is $5\nmid l$ where it is again totally ramified. – reuns Nov 27 '19 at 06:39
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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.
Fouad El
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I do not see how you could conclude with only these informations. You need to say something more: my answer should help for this. – Gaussian Nov 27 '19 at 14:49
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your answer not clear my friend in my case, I dont understand what you mean by $Q_5$ and what is the relation between the root of $X^5-n$ and the ramification of $5$ ??? – Fouad El Nov 27 '19 at 22:05
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$\mathbf{Q}_5$ is the field of $5$-adic numbers. The relation between the irreducible factors of $X^5-n$ in $\mathbf{Q}_p$ and the primes lying over $5$ in $K$ is given by that proposition of Neukirch. Here, I show that $X^5-n$ is irreducible in $\mathbf{Q}_5$ so that there is only one primes lying over $5$ in $\mathcal{O}_K$. This means that $5=ef$ where $e$ is the ramification index of $5$ and $f$ the degree of the residual extension: since you know that $e>1$, you get that $e=5$ which means that $5$ totally ramifies. – Gaussian Nov 27 '19 at 22:12
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now I understand your proof, but in my research I don't use the field of $5$-adic numbers. is there any other proof of this ramification of 5 ?? thanks a lot – Fouad El Nov 28 '19 at 17:19