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

Question

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?

Answer 1

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.

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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.