Extension $L/K$ with specific decomposition properties

Find an extension $$L/K$$ of number fields with Galois group $$G$$ and respective rings of integers $$O_L$$ and $$O_K$$ for each of the following requirements:

1. The decomposition group $$G_q$$ of some prime ideal $$q$$ of $$O_L$$ over $$p = q \cap O_K$$ is not a normal subgroup of $$G$$.

2. $$G=I_q\times I_{q'}$$ is the direct product of two nontrivial inertia subgroups $$I_q$$ and $$I_{q'}$$, where $$q, q'$$ are prime ideals of $$O_L$$

3. The inertia group of $$I_q$$ is not cyclic for a prime ideal $$q$$ of $$O_L$$.

The only examples I know how to work with (like $$\mathbb{Q}(i)$$, $$\mathbb{Q}(\sqrt[3]{2})$$ or simple cyclotomic extensions) apperently are not enough for this exercise. Is there some strategy to find these examples?

• Hi -- I really like @Starfall's answer below; but thought I should point out that (1) can indeed be answered when $K=\mathbb{Q}$ say, at least for the normal closure $L$ of $\mathbb{Q}(\sqrt[3]{2})$ (because the Galois group is $S_3$ and so has $3$ non-normal cyclic subgroups of order 2) and so for example taking the decomp gp for any of the primes above 5 will give you C2. – GaryMak Oct 12 '18 at 12:27

For (1), let $L/\mathbf Q$ be the splitting field of $X^5 - 4X + 2$ over $\mathbf Q$. This has Galois group $S_5$, and letting $K = \mathbf Q(\alpha)$ where $\alpha$ is a root of $X^5 - 4X + 2$, the prime $13$ factors as $13 = \mathfrak p \mathfrak q \mathfrak r$ in $K$. Since $13$ is not completely split in $K/\mathbf Q$, it is not completely split in $L/\mathbf Q$; and thus it follows that letting $g$ be the number of distinct primes of $L$ lying over $13$, we have $3 \leq g < 120$. We have $120 = efg$, where $ef$ is the order of the decomposition group of any prime lying over $13$, and it follows that $1 < ef \leq 40$. However, the only nontrivial normal subgroup of $S_5$ is $A_5$, which has order $60$. It follows that the decomposition group cannot be normal.
For (2), let $L = \mathbf Q(\sqrt{3}, \sqrt{5})$. Check that $3 = (\sqrt{3})^2$ and $5 = (\sqrt{5})^2$ are prime factorizations in $L/\mathbf Q$, and that the inertia groups of $\sqrt 5$ and $\sqrt 3$ intersect trivially, conclude.
For (3), let $L = \mathbf Q(\sqrt{2}, \sqrt{3})$. Show that $L/\mathbf Q$ is totally ramified at $2$, and that $\textrm{Gal}(L/\mathbf Q) \cong C_2 \times C_2$, which is not cyclic; conclude.
• how do you know that $13=\frak{p}\frak{q}\frak{r}$ in item $1$? In item $2$, I know that $2$ ramifies, since $2|\Delta_{L}$, but in this case I have two possibilities: $(2)=\frak{p}^4$ or $(2)=\frak{p}^2\frak{q}^2$. How do I know that the first is the right one? – rmdmc89 Feb 14 '17 at 14:09
• @AguirreK Item $1$ uses Dedekind's factorization criterion - the discriminant of the polynomial $X^5 - 4X + 2$ is coprime to $13$, thus the splitting of this polynomial modulo $13$ determines the splitting of $13$ in $L$. For (3), note that the inertia field of $\mathfrak p$ has to be unramified at $2$ over $\mathbf Q$. How many subextensions of $L/\mathbf Q$ are there with that property? – Ege Erdil Feb 14 '17 at 14:26
• why does the inertia field have to be unramified at $2$? And how do I check that? – rmdmc89 Feb 14 '17 at 18:45