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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?

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    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 $\Bbb Q$. – 2017-01-05
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    @AdamHughes : yes maybe. But how do you understand "in the first case the inertia in the compositum has order 2" in the answer? – 2017-01-05
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    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? – 2017-01-05
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    Dear @Mathmo123, thank you for your comment. You can read [here](https://i.stack.imgur.com/LtvUa.jpg) the original version in Ribenboim's book. This is really confusing... I don't know what I've made wrong, this is probably silly…! – 2017-01-05
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    @Mathmo123 : so do you think I applied the theorem in a wrong way, or stated it incorrectly (I don't think so, however) ? Thank you in advance :-) – 2017-01-05
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    @Watson do you have the next page? The proof is cut off – 2017-01-05
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    @Mathmo123 : thank you for your reply. Do you have access to page 264 [here](https://books.google.com/books?id=u5443xdaNZcC&pg=PA264#v=onepage&q&f=false)? Otherwise [here](https://i.stack.imgur.com/CeOdq.png) is a copy. – 2017-01-05
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    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. – 2017-01-05
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    @JyrkiLahtonen : thank you for your comment. I read (quickly) the proof of Prop. E, and noticed no flaw in the reasoning. But I agree that it seems very strange when Ribenboim is writing on page 264 (see the link above) : $$\dfrac{Z_{P}(L/K)}{T_P(L/K)} \cong \dfrac{Z_{P_F}(F/K)}{T_{P_F}(F/K)} \times \dfrac{Z_{P_{F'}}(F'/K)}{T_{P_{F'}}(F'/K)}$$ where $Z$ denotes the decomposition group and $T$ the inertia group [cont.] – 2017-01-06
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    [cont.] (NB : it is actually written $\dfrac{Z_{P_F}(L/K)}{T_P(L/K)} \cong \cdots$, but I think that this is a typo), because as you said $Z_P/T_P$ is cyclic of order $f_P(L/K)$, while the direct product may not be cyclic. This is true is the degree $[F':K]$ and $[F:K]$ are coprime, but our condition $F \cap F' = K$ is weaker! So I'm still really confused. – 2017-01-06
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    At the very end of the proof (p. 264), he's writing $$ f_P(L/K) = f_P(L/F) \cdot f_P(L/F'), $$ but I don't know how he's using Theorem 1 here. For me $Z(F/K) / T(F/K)$ has order $f(F/K)$, not $f(L/F)$ (similarly for $F'$ instead of $F$). That's also confusing. – 2017-01-06
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    If I'm not mistaken, $p=3$ is inert in $F =\Bbb Q(\sqrt 5)$ and in $F' =\Bbb Q(\sqrt{13})$ (the discriminants are resp. $5$ and $13$, not quadratic residues mod $p$, $p>2$). Then for any prime $P$ of $L=FF'=\Bbb Q(\sqrt 5, \sqrt{13})$ above $p$, $Z_P(L/\Bbb Q)/T_P(L/\Bbb Q)$ should be $\Bbb Z/2 \times \Bbb Z/2$ according to Proposition E (provided that $F \cap F' = \Bbb Q$) ! – 2017-01-06
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    At least we should have an injective group morphism $$D_P(L/K) \to D_{P \cap F}(F/K) \times D_{P \cap F'}(F'/K)$$ (similarly for the $I_P$'s). The surjectivity is equivalent to asking: if $$\sigma(P \cap F)=P \cap F \qquad \text{and} \qquad \sigma(P \cap F')=P \cap F',$$ then do we have $\sigma(P)=P$ ? – 2017-01-06
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    @Mathmo123 : basically the idea is that the isomorphism $$f : Gal(FF'/K) \to Gal(F/K) \times Gal(F'/K)$$ induces an isomorphism $$D(FF'/K) \cong f(D(FF'/K))$$ The claim is that $$f(D(FF'/K)) = D(F/K) \times D(F'/K)$$ The inclusion $\subset$ is clear : the restriction of $\sigma \in D(FF'/K)$ to $F$ lies in $D(F/K)$ (similarly for $F'$). But I'm not sure why $\supset$ holds (this is what I've written in the above comment). – 2017-01-07
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    Let $(\alpha,\alpha') \in D(F/K) \times D(F'/K) \subset Gal(F/K) \times Gal(F'/K).$ Then there is $\sigma \in Gal(FF'/K)$ such that $$f(\sigma) = (\sigma\vert_F,\sigma\vert_{F'}) = (\alpha, \alpha')$$ We want to show that we can actually choose $\sigma$ in $D(FF'/K)$. – 2017-01-07

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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.

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    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](http://math.stackexchange.com/q/485512/11619), where the answer describes the correct version of the false claim. – 2017-01-08
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    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. – 2017-01-08
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    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. – 2017-01-08
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    @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. – 2017-01-11
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    Thank you very much for this information! I'm looking forward for any news. – 2017-01-11