Inertia group of compositum of number fields – erratum in Ribenboim's book?

I have a question about this thread on MO, in which it is written in the answer:

Here is a simple example: take $K_1=\mathbb{Q}(i)$ and $K_2=\mathbb{Q}(\sqrt{-5})$. Then both $K_1/\mathbb{Q}$ and $K_2/\mathbb{Q}$ are (totally) ramified at $p=2$ and $K_1\cap K_2=\mathbb{Q}$, but $F=K_1K_2$ is not totally ramified at $2$. […]

[We have] $I_1=D_1=G_1$ and $I_2=D_2=G_2$ (in your notation) but […] the inertia in the compositum has order $2$ […].

I don't understand why the inertia in the compositum has order $2$. According to Ribenboim, Classical Theory of Algebraic Numbers, chapter 14, proposition E, p. 263, we have :

Let $K \subset F,F' \subset L=FF'$ be number fields such that $F/K$ and $F'/K$ are Galois. If $F \cap F' = K$ then $$I_P(L/K) \cong I_{P \cap F}(F/K) \times I_{P \cap F'}(F'/K)$$ and $$D_P(L/K) \cong D_{P \cap F}(F/K) \times D_{P \cap F'}(F'/K)$$

where $P$ is a prime of $L$ above a prime $\mathfrak p$ of $K$, $I(\cdot)$ is the inertia group, and $D(\cdot)$ the decomposition group.

Maybe the answer on MO was focused on $K_1K_2/K_2$, because when he is writing "the inertia in the compositum has order $2$", it can't be talking about $I_P(K_1K_2/\Bbb Q)$ which has order $4$ by Ribenboim's theorem. Or am I wrong somewhere?

• I think "this is wrong in general" is a reference to the fact that the Galois group of a compositum is not a product in general (i.e. the trivial intersection property is needed here). "Trivial" here means they intersect in the base field under consideration, not necessarily $\Bbb Q$. – Adam Hughes Jan 5 '17 at 15:40
• @AdamHughes : yes maybe. But how do you understand "in the first case the inertia in the compositum has order 2" in the answer? – Watson Jan 5 '17 at 15:42
• The counterexample is definitely correct - $\mathbb Q(i, \sqrt{-5})$ is not totally ramified over $\mathbb Q$ since it is unramified over $\mathbb Q(\sqrt{-5})$, and hence the inertia group must have order $2$. Are you sure you have written the theorem correctly? – Mathmo123 Jan 5 '17 at 17:45
• Dear @Mathmo123, thank you for your comment. You can read here the original version in Ribenboim's book. This is really confusing... I don't know what I've made wrong, this is probably silly…! – Watson Jan 5 '17 at 18:38
• I think something is wrong in Ribenboim. Ramification degree (as well as inertia degree) is multiplicative in a tower of extensions, but not in a compositum. Also, $D_P(L/K)/I_P(L/K)$ is always cyclic because it is isomorphic to the Galois group of the extension of residue class fields (cyclic of order $f$). However, the boxed claims seem to indicate otherwise. Hope you get this sorted out! I am not fully conversant with decomposition/inertia groups in the first place. It is easy to misinterpret something about them. – Jyrki Lahtonen Jan 5 '17 at 23:13

I think there is the following weakness in Ribenboim's argument.

The context is that $L/K$ is the compositum of two trivially intersecting Galois extensions $F/K$ and $F'/K$. Then we know that $$Gal(L/K)\simeq Gal(F/K)\times Gal(F'/K),$$ where the isomorphism maps an automorphism $\sigma\in Gal(L/K)$ to its restrictions $(\sigma\vert_F,\sigma\vert_{F'})$.

On the first lines of page 264 he explains how the restriction homomorphism maps $D_P(L/K)$ onto $D_{P\cap F}(F/K)$ as well as onto $D_{P\cap F'}(F'/K)$. I think this is all well and correct. However, this does NOT imply that we would have $$D_P(L/K)\simeq D_{P\cap F}(F/K)\times D_{P\cap F'}(F'/K).$$

He then simply states that the argument with inertia groups is done similarly.

Essentially Ribenboim seems to be arguing as if the following "Lemma" would hold.

False claim: Let $G=G_1\times G_2$ be a direct product of groups. Let $p_i:G\to G_i$ be the canonical projection, $i=1,2$. Then for all subgroups $H\le G$ we have $H=p_1(H)\times p_2(H)$.

Counterexample: Let $G_1=G_2=C_2$. Let $H$ be the diagonal subgroup. Then $H$ is also cyclic of order two as are the homomorphic images $p_1(H)$ and $p_2(H)$.

The given counterexample about inertia groups of the rational prime $p=2$ in the case $K=\Bbb{Q}$, $F=\Bbb{Q}(i)$, $F'=\Bbb{Q}(\sqrt{-5})$, $L=\Bbb{Q}(i,\sqrt5)$ then parallels this counterexample precisely. In this case the inertia subgroup of the unique prime ideal above $(2)$ of $L$ is the diagonal subgroup of $Gal(F/K)\times Gal(F'/K)$ under the above identifcations.

I don't have a copy of that book, so I cannot say whether this has dire consequences for the rest of the material. It may well be that a corrected version of the above False Claim saves the day on some occasions. We always have the inclusions $$(H\cap G_1)\times (H\cap G_2)\le H\le p_1(H)\times p_2(H),$$ and it may sometimes be possible to work with the intersections $H\cap G_i$ or combinations like $p_1(H)\times (H\cap G_2)$ instead. I am not sure.

• I am fairly sure that this and related fallacies about subgroups of a direct product have been discussed on our site earlier. A moment's search gave me this, where the answer describes the correct version of the false claim. – Jyrki Lahtonen Jan 8 '17 at 8:52
• Thank you very much for your answer! Do you know what to do in such situations, in order to possibly correct this false claim in a new edition of the book (the copy I have is already the 2nd edition…)? Is it possible (and a good idea) to contact Springer (or someone else) about that, for instance? Apparently, this flaw has no consequences for the rest of the material. – Watson Jan 8 '17 at 9:46
• I don't know what is best. I am considering asking my own teachers (Tauno Metsänkylä in particular) for consultation and/or verification. Ribenboim himself is approaching 90, so contacting him may not be prudent. – Jyrki Lahtonen Jan 8 '17 at 10:53
• @Watson: Prof. Metsänkylä told me that he is regularly in touch with one of Ribenboim's students. He will bring this up in his next e-mail. We may hear if an errata has been published somewhere, or whether a new edition/printing is being planned. I will keep you posted. – Jyrki Lahtonen Jan 11 '17 at 12:34
• Thank you very much for this information! I'm looking forward for any news. – Watson Jan 11 '17 at 13:41