8
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – GaryMak Oct 12 '18 at 12:27
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – rmdmc89 Feb 14 '17 at 14:09
  • $\begingroup$ @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? $\endgroup$ – Starfall Feb 14 '17 at 14:26
  • $\begingroup$ why does the inertia field have to be unramified at $2$? And how do I check that? $\endgroup$ – rmdmc89 Feb 14 '17 at 18:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.