# splitting primes

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 $$5$$ ?

• Yes${{{}}}{}{}$. Jul 22 '19 at 9:14
• Is the question the last sentence, or are you confused how to prove the statement? Jul 22 '19 at 14:09
• I do not really know how to prove it . I know that I have to consider a prime $a$ over 5 Jul 22 '19 at 14:22

$$\begin{cases} K_1 = \Bbb Q(i) \\ K_2 = \Bbb Q(\sqrt 2) \\ K_3 = \Bbb Q(\sqrt 5) \end{cases}$$
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 $$\mathfrak{p}=\mathcal{O}_{K_2}\cap\mathfrak{P}$$.