I am struggling to understand the behavior of inertia groups and ramification:

Let $L/K$ and $N/K$ be finite galois extensions of number fields and $L\cap N = K$. And let $\mathfrak{P}_{LN}$ be a prime ideal in $\mathcal{O}_{LN}$ (the integral closure of a dedekind domain $o_K \subset K$ in $LN$) and $\mathfrak{P}_L := \mathfrak{P}_{LN} \cap L, \mathfrak{P}_N := \mathfrak{P}_{LN} \cap N$ and $\mathfrak{p} := \mathfrak{P}_{LN} \cap K$ be the corresponding prime ideals below. For the grades of ramifications we get: $e(\mathfrak{P}_{LN} \mid \mathfrak{p}) = e(\mathfrak{P}_{LN} \mid \mathfrak{P}_{L})\cdot e(\mathfrak{P}_{L} \mid \mathfrak{p})$. Now my question is, does $e(\mathfrak{P}_{LN} \mid \mathfrak{P}_{L}) = e(\mathfrak{P}_{N} \mid \mathfrak{p})$ hold?

I think it does, by the following argument: $\mathfrak{P}_{N}^{e(\mathfrak{P}_{N} \mid \mathfrak{p})}=\mathfrak{p}\mathcal{O}_N = \mathfrak{P}_L \mathcal{O}_{LN} \cap N = \mathfrak{P}_{LN}^{e(\mathfrak{P}_{LN} \mid \mathfrak{P}_{L})} \cap N = \mathfrak{P}_N^{e(\mathfrak{P}_{LN} \mid \mathfrak{P}_{L})}$

Yet I end up with the following contradiction:

Suppose $K = \mathbb{Q}$ and $[L:K] = [N:K] = l$ where $l$ is prime. Let $\mathfrak{p} \neq (l)$ be ramified in both $L/K$ and $N/K$ (i.e. $\mathfrak{p}\mathcal{O}_L = \mathfrak{P}_L^l$ and $\mathfrak{p}\mathcal{O}_N = \mathfrak{P}_N^l$). If the above holds, then $I_{\mathfrak{P}_{LN}}(LN/K) \cong C_l \times C_l$. However, as $(\mathfrak{p}, (l) ) = 1$, $\mathfrak{p}$ is only tamely ramified in $LN/K$. Therefore $I_{\mathfrak{P}_{LN}}(LN/K)$ is cyclic, which is obviously not true for $C_l \times C_l$.

Where am I messing up? Thank you for your help!


1 Answer 1


No, it doesn't necessarily hold. For example, take $ L = \mathbf Q(\sqrt{3}) $, $ N = \mathbf Q(\sqrt{6}) $ and $ K = \mathbf Q $, and look at the behavior of the prime $ \mathfrak p = 3 $. (This is exactly the setting you outline in your second argument.)

I am not very sure what you are doing in your first argument; if you try to justify your manipulations explicitly, you will see that they don't work. Taking powers of ideals doesn't behave so regularly under taking intersections, in other words, you don't generally have $ \mathfrak q^n \cap \mathcal O_K = (\mathfrak q \cap \mathcal O_K)^n $; which invalidates the final equality in your argument, for example.

  • $\begingroup$ Thank you! May I ask a follow up question: Is there something like an inequality, e.g. $e(\mathfrak{P}_{LN} \mid \mathfrak{p}) = e(\mathfrak{P}_{LN} \mid \mathfrak{P}_{L})\cdot e(\mathfrak{P}_{L} \mid \mathfrak{p}) \leq e(\mathfrak{P}_{N} \mid \mathfrak{p})\cdot e(\mathfrak{P}_{L} \mid \mathfrak{p})$. I.e. if a prime is neither ramified in $L/K$ nor in $N/K$, does this insure that it is unramified in $LN/K$? $\endgroup$
    – Philipp
    Mar 15, 2017 at 16:43
  • $\begingroup$ It is true that compositums of unramified extensions are unramified, yes. $\endgroup$
    – Ege Erdil
    Mar 15, 2017 at 16:45

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