# Is the "order of the decomposition group of an infinite prime" equal to $1$? $|D_{\mathfrak{P}}|=1$? (About infinite primes and decomposition groups)

Sorry for this very easy question. I am reading Janusz's book "Algebraic Number Fields".

I am in section 2 of the fifth chapter of Janusz's book "Algebraic Number Fields", exactly in Proposition 2.3. There it speaks about "decomposition groups of infinite primes", which I don't know them well. I have some doubts about this concept. Can we say something special about some relations between $$|Gal (L/K)|$$ and $$|D_{\mathfrak{P}}|$$?

In Chapter III it is written that:

"If $$\mathfrak{P}_i$$ is an infinite prime corresponding to an embedding $$\tau_i$$ of $$L$$ into either $$\mathbb{R}$$ or $$\mathbb{C}$$, then $$\sigma(\mathfrak{P}_i)$$ is the prime corresponding to the embedding $$\tau_i\sigma^{-1}$$. The inverse is necessary so that the rule al $$\sigma_1(\sigma_2(\mathfrak{P}_i))=(\sigma_1\sigma_2)(\mathfrak{P}_i)$$ is valid." (I can see this because we have $$(\tau_i\sigma_2^{-1})\sigma_1^{-1}=(\tau_i)(\sigma_1\sigma_2)^{-1}$$)

(Q.1): I think the decomposition group has just one element in the case of infinite primes ($$|D_{\mathfrak{P}}|=1$$). Am I mistaken? Simply because for any fixed $$\tau \in Gal(L/K)$$ the equation $$\sigma\tau=\tau$$ (or the equation $$\tau\sigma ^{-1}=\tau$$) has only one solution, i.e $$\sigma=$$identity.

(Q.2): I think the author could not have written the previous paragraph and written this instead: "If $$\mathfrak{P}_i$$ is an infinite prime corresponding to an embedding $$\tau_i$$ of $$L$$ into either $$\mathbb{R}$$ or $$\mathbb{C}$$, then $$\sigma(\mathfrak{P}_i)$$ is the prime corresponding to the embedding $$\sigma\tau_i$$." Why did he define the action in that way?

Also, it is written that:

"The Galois group $$G$$ permutes the $$\mathfrak{P}_i$$ transitively." I can understand this, but it doesn't coincide with the case of finite primes. In the case of infinite primes, the number of primes is $$g=|Gal(L/K)|$$, and if the ramification indexes or inertia degrees are greater than $$1$$, then the analogous formula $$\sum_{i=1}^{g}e_if_i=|Gal(L/K)|$$ does not hold. So I can conclude that in the case of infinite primes we can not expect an analogous formula.

(Q.3): Is my above conclusion right?

Especially I am looking for these examples (just because of my curiosity):

(Q.4): An explicit extension $$L/K$$, and a prime $$\mathfrak{p}$$ of $$\mathcal{O}_K$$, for which the number of primes $$\mathfrak{P}_i$$'s lying above $$\mathfrak{p}$$ are strictly less than $$|Gal(L/K)|$$.

and

(Q.5): An explicit extension $$L/K$$ where $$K$$ is a quadratic extension, and a prime $$\mathfrak{P}$$ of $$\mathcal{O}_L$$, for which the size of the decomposition group $$D_{\mathfrak{P}}$$ is greater than 2.

(Q.6): Can we say something about the ramification index $$e_{\mathfrak{P}|\mathfrak{p}}$$ and the inertia degree $$f_{\mathfrak{P}|\mathfrak{p}}$$? I mean I guess there should be some bounds, and at least in the case of the complex embeddings, I think the theory speaks about something like $$e_{\mathfrak{P}|\mathfrak{p}}=1$$.

Any other clarifying explanation is welcome.

I think the confusion here mainly comes from the following misconception: For a number field $$K$$, infinite primes are not the same as embeddings $$K\hookrightarrow\mathbb{C}$$. They are rather equivalence classes of archimedean valuations on $$K$$. As Janusz proves, any such valuation is indeed induced by an embedding of $$K$$. However, if the embedding $$\sigma$$ is complex, then the conjugate embedding $$\overline{\sigma}$$ induces an equivalent valuation, i.e. the embeddings $$\sigma$$ and $$\overline{\sigma}$$ correspond to the same infinite prime! Thus, you have one infinite prime for each real embedding and one infinite prime for each conjugate pair of complex embeddings (note that indeed Janusz writes $$\frak{P}_\text{1}, \dots, \frak{P}_\text{r}$$ and $$\frak{P}_\text{r+1}, \dots, \frak{P}_\text{r+s}$$ in Prop. 2.3 for the infinite primes indicating that, indeed, you do not have $$r+2s=n$$ infinite primes, but only $$r+s$$).

Now for a relative extension $$L/K$$, let $$\sigma$$ be a real embedding of $$K$$ and let $$\tau$$ be a complex embedding of $$L$$ extending $$\sigma$$. Then the conjugate embedding $$\overline{\tau}$$ also extends $$\sigma$$! However, as we just saw, both $$\tau$$ and $$\overline{\tau}$$ correspond to the same infinite prime $$\frak{P}$$, which is why one can say that $$\frak{P}$$ has ramification index $$2$$ over the infinite prime $$\frak{p}$$ of $$K$$ corresponding to $$\sigma$$, i.e. $$e_{\frak{P}|\frak{p}}=2$$. If $$\sigma$$ is complex, this cannot occur. Here is another way to think about it: As $$\tau=\mathrm{id}\circ\tau$$ and $$\overline{\tau}=(\overline{\tau}\circ\tau^{-1})\circ\tau$$ are complex conjugate embeddings, they correspond to the same infinite prime $$\frak{P}$$ of $$L$$ and hence both $$\mathrm{id}$$ and $$\overline{\tau}\circ\tau^{-1}\neq\mathrm{id}$$ are in $$D_\frak{P}$$, i.e. $$|D_\frak{P}\text{|=2}$$.

Following this train of thought, one also gets a fundamental equality for infinite primes since any embedding of $$K$$ can be extended to an embedding of $$L$$ in $$[L:K]$$ ways. Identifying complex conjugate embeddings with the same infinite prime as explained above and setting all inertia degrees to one, one indeed obtains $$[L:K]=\sum_{\frak{P}|\frak{p}} e_{\frak{P}|\frak{p}}f_{\frak{P}|\frak{p}}.$$ In particular, note that the number of infinite primes above the infinite primes $$\frak{p}$$ corresponding to the embedding $$\sigma$$ of $$K$$ is not, as you said, $$n:=|\mathrm{Gal}(L/K)|$$, but rather $$r+s$$, where $$r$$ is the number of real embeddings of $$L$$ extending $$\sigma$$ and $$2s$$ is the number of complex embeddings of $$L$$ extending $$\sigma$$.

In general, we can only have $$e_{\frak{P}|\frak{p}}=1$$ or $$e_{\frak{P}|\frak{p}}=2$$ for infinite primes $$\frak{P}|\frak{p}$$ and the ramification index can only be $$2$$ in the case outlined above, i.e. when $$\frak{p}$$ is real, but $$\frak{P}$$ is complex. The inertia degree $$f_{\frak{P}|\frak{p}}$$ is always $$1$$. (However, note that there are apparently authors who do this the other way around, i.e. they interpret the extension of a real infinite prime by a complex infinite prime as inertia instead of ramification and correspondingly, they always set $$e_{\frak{P}|\frak{p}}=1$$, while $$f_{\frak{P}|\frak{p}}$$ is either $$1$$ or $$2$$.)

As a concrete example, consider $$K=\mathbb{Q}$$ and $$L=\mathbb{Q}(\sqrt{2})$$. Then $$L$$ has two real infinite primes corresponding to $$\sqrt{2}\mapsto\sqrt{2}$$ and $$\sqrt{2}\mapsto-\sqrt{2}$$, which induce inequivalent archimedean valuations. Hence, they both have ramification index and inertia degree $$1$$ and you get the fundamental equality $$1\cdot 1+1\cdot 1=2$$.

If you consider instead $$L=\mathbb{Q}(\sqrt{-2})$$, then the (now complex!) embeddings $$\sqrt{-2}\mapsto \sqrt{-2}$$ and $$\sqrt{-2}\mapsto -\sqrt{-2}$$ are complex conjugates and hence induce the same valuation on $$L$$. Thus, $$L$$ has only one infinite prime, but with ramification index $$2$$ over the infinite prime of $$\mathbb{Q}$$ (this is precisely a case where a complex infinite prime lies over a real infinite prime as explained above) and the fundamental equality becomes $$2\cdot 1=2$$.

EDIT: Maybe one could go about Q2 in the following way. From the definition of ray class groups, we know that $$a\equiv b\pmod{\frak{P}}$$ for a real infinite prime $$\frak{P}$$ corresponding to the embedding $$\sigma$$ should mean $$\sigma(a/b)>0$$. If $$\tau$$ is another real embedding, then surely we would like to have the property that $$a\equiv b\pmod{\frak{P}}$$ implies $$\tau(a)\equiv \tau(b)\pmod{\tau(\frak{P})}$$ (similarly to the case of finite primes). If $$\tau(\frak{P})$$ corresponded to $$\tau\circ\sigma$$, then $$(\tau\circ\sigma)(\tau(a)/\tau(b))=\tau\sigma\tau(a/b)$$ and there is no reason why there should be any condition of the sign of this quantity. However, defining $$\tau(\frak{P})$$ to correspond to the embedding $$\sigma\circ\tau^{-1}$$, we find $$(\sigma\circ\tau^{-1})(\tau(a)/\tau(b))=\sigma(a/b)>0$$ so that, with this definition, the property $$a\equiv b\pmod{\frak{P}}$$ indeed implies $$\tau(a)\equiv \tau(b)\pmod{\tau(\frak{P})}$$.

• Thanks for your very complete answer, and it solved my 1st question, and 3rd, 4th, 5th, and 6th question. I have to think more on my 2nd question, to realize why would we have a problem if the author had defined the group action in that way (in Q.2)? Also, I found: In the case of ramified infinite prime the decomposition group is precisely what you've written, i.e. $D_{\mathfrak{P}}=\{\mathrm{id}, \overline{\tau}\circ\tau^{-1}\}$, and since it is a subgroup of $Gal(L/K)$, we can conclude that the order of $\overline{\tau}\circ\tau^{-1}$ is equal to $2$, which seems surprising to me! Apr 18, 2022 at 10:56
• Well, as $\overline{\tau}\circ\tau^{-1}$ is just complex conjugation, it shouldn't be so surprising after all that is has order $2$, right? ;) Apr 18, 2022 at 11:17
• You are right! I didn't notice that "$\overline{\tau}\circ\tau^{-1}$" is just the complex multiplication! So for now, only my 2nd question remains, which I need to think about more to understand why we could not consider the group action directly. Thanks for your kindness and your help. Apr 18, 2022 at 11:30
• "However, note that there are apparently authors who do this the other way around" - Yes, that caused me a lot of confusion. See also math.stackexchange.com/questions/769178/…. Jun 20, 2022 at 18:48