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Let $K=\mathbb{Q}[i,\sqrt{2},\sqrt{5}]$ which is normal over $\mathbb{Q}$ and has degree 8 .

We also know that the prime $5 $ is ramified in $\mathbb{Q}(\sqrt{5})$ , inert in $\mathbb{Q}(\sqrt{2})$ and splits completely and $\mathbb{Q}(i)$.

Now I have to conclude that K contains at least two primes lying over $5$ each of which has inertia degree and ramification index $2$ .

I do not know what is meant by "lying over 5" . Does it mean that these two primes contain $5$ ?

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    $\begingroup$ Yes${{{}}}{}{}$. $\endgroup$ – Angina Seng Jul 22 '19 at 9:14
  • $\begingroup$ Is the question the last sentence, or are you confused how to prove the statement? $\endgroup$ – awllower Jul 22 '19 at 14:09
  • $\begingroup$ I do not really know how to prove it . I know that I have to consider a prime $a$ over 5 $\endgroup$ – AnabolicHorse Jul 22 '19 at 14:22
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First off let

$$\begin{cases} K_1 = \Bbb Q(i) \\ K_2 = \Bbb Q(\sqrt 2) \\ K_3 = \Bbb Q(\sqrt 5) \end{cases}$$

Now since $(5)$ ramifies in $K_3$ as $\mathfrak{p}^2$ we need only note that $5\mathcal{O}_K=\mathfrak{p}^2\mathcal{O}_K$ so no matter how $\mathfrak{p}$ factors in $\mathcal{O}_K$ all the factors of $(5)$ have a positive power (in fact a multiple of $2$ even!) and therefore $5$ is ramified in $K$.

In addition, since $5$ splits in a subfield as the product of two primes, you know there are $2k$ primes above $5$ for some $k$ since if $\mathfrak{q}_1,\mathfrak{q}_2$ are primes of $K_1$ above $(5)$ and $\mathfrak{Q}$ is a prime above $(5)$ in $K$, we know that $\mathfrak{Q}\cap\mathcal{O}_{K_1}\in\{\mathfrak{q}_1,\mathfrak{q}_2\}$ is a prime of $K_1$ and so can only be one of the two--WLOG say $\mathfrak{q}_1$, but then there must be another prime in $K$ above $\mathfrak{q}_2$.

Finally since there is some inertia in a sub-field, you know that there must be some inertia in the composite field, since if $\mathfrak{P}\big| (5)$ is a prime divisor of $(5)$ in $K$, then $\mathcal{O}_K/\mathfrak{P}$ is a field extension of $\mathcal{O}_{K_2}/\mathfrak{p}$ where $\mathfrak{p}=\mathcal{O}_{K_2}\cap\mathfrak{P}$.

This only leaves you with the case stated in the problem.

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