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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?

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  • 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
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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.

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  • $\begingroup$ 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? $\endgroup$ – Ege Erdil 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

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