I follow the notation of Georges Gras: Class Field Theory, some of which I recall for convenience; feel free to skip the following lines if you are familiar with the notation. Let $K$ be a number field, $T$ a finite set of finite places of $K$, and $S=S_0\cup S_\infty$ a finite set of respectively finite and infinite places, with $S_0$ disjoint from $T$. Let $\mathfrak m=\prod_{v\in T}\mathfrak p_v^{m_v}$ ($m_v\ge0$) be a modulus built from $T$.

He defines the generalized class group $\mathcal C\ell^S_\mathfrak{m}$ of $K$ as $$\mathcal C\ell^S_\mathfrak{m}=I_T/P_{T,\mathfrak m, \Delta_\infty}\cdot\langle S_0\rangle$$ where $\Delta_\infty=\text{Pl}_\infty\setminus S_\infty$; $I_T$ are the fractional ideals prime to $T$; $P_{T,\mathfrak m, \Delta_\infty}$ is the group of principal ideals $(x)$ where $v(x)=0$ for all $v\in T$, $x\equiv1\pmod{\mathfrak m}$ and positive on $\Delta_\infty$; and where $\langle S_0\rangle$ is the free group on $S_0$.

Later Gras states a result saying that $$|\mathcal C\ell^S_\mathfrak{m}|=|\mathcal C\ell^S|\dfrac{\varphi(\mathfrak m)}{(E^S:E^S_\mathfrak{m})}$$ where $\varphi$ is the generalized Euler function; $E^S$ the $S$-units and $E^S_\mathfrak m$ the $S$-units congruent to 1 modulo $\mathfrak m$.

I guess I simply do not understand everything going on... If I let $K=\mathbb Q(\sqrt{-5})$, $S=\text{Pl}_\infty$ (or $S=\varnothing$, don't think it matters?) $\mathfrak m=(2,1+\sqrt{-5})$ and $T$ the place corresponding to $\mathfrak m$, I would suppose that $\mathcal C\ell^S_\mathfrak{m}$ was trivial? Since any fractional ideal prime to $\mathfrak m$ is already principal? But I get that $\varphi(\mathfrak m)=1$, that $E^S=\{\pm1\}=E^S_\mathfrak{m}$ and $|\mathcal C\ell^S|=2$ as being the ordinary class group of $K$. What's the stupid mistake(s)? Thanks in advance :)

  • $\begingroup$ Just to be sure, you meant $\mathcal C\ell^S_\mathfrak{m}=I_T/(P_{T,\mathfrak m, \Delta_\infty}\cdot\langle S_0\rangle)$ ? $\endgroup$ – mercio Oct 20 '14 at 12:16
  • $\begingroup$ why would any fractional ideal prime to $\mathfrak m$ be principal ?? You do know that the class group of $K$ is nontrivial, so there are non-principal ideals ? $\endgroup$ – mercio Oct 20 '14 at 12:27
  • $\begingroup$ @mercio Yes, I mean $I_T/(P_{T,\mathfrak{m},\Delta_\infty}\cdot\langle S_0\rangle)$. When it comes to $I_T$, I just thought that, loosely speaking, since $\mathfrak m$ represents the nontrivial class in the class group of $\mathbb Q(\sqrt{-5})$, we "remove" all elements of this class when going to $I_T$. But yeah, I know that sounds fishy... $\endgroup$ – user183865 Oct 20 '14 at 19:59

In Gras's book, $\Delta_{\infty}$ is always a set of real Archimedean places (that is, your definition should have read $\Delta_{\infty} = \mathrm{Pl}_{\infty}^r \setminus S_{\infty}$ -- you're missing the superscript $r$. So in your example, $\Delta_{\infty}$ is empty and we can ignore it:

$$\mathrm{Cl}_{\mathfrak{m}}^S = I_T/P_{T,\mathfrak{m}}.$$

To answer your question about whether it makes a difference if $S$ is empty or not: in your example, no. Including the Archimedean place in $S$ has no effect on the definition of the generalized class group, and it does not affect the definition of $S$-units since the restriction added by Archimedean places is vacuous when those places are complex.

In your example, $\mathrm{Cl}^S = \mathrm{Cl}$, $\mathrm{Cl}_{\mathfrak{m}}^S = \mathrm{Cl}_{\mathfrak{m}}$, and your computations of $|\mathrm{Cl}|$, $\phi(\mathfrak{m})$ and $(E^S \colon E_{\mathfrak{m}}^S)$ are correct. So everything comes down to seeing why $|\mathrm{Cl}_{\mathfrak{m}}| = |\mathrm{Cl}|$.

The key fact that your post indicates you are missing is that if $T$ is any finite set of prime ideals in $K$ and $P_T$ is the group of principal ideals coprime to the ideals in $T$, then

$$I_T/P_T \cong I/P.$$ In particular, this proves that it is not true that every ideal coprime to $\mathfrak{m}$ is principal.

To get a map $I_T/P_T \rightarrow I/P$, we take an ideal coprime to $T$ and form a class in $I/P$. If we change our ideal by a principal ideal in $P_T$, we still get the same class in $I/P$, so the map is well-defined. It is immediate that the kernel is trivial. But the map is also surjective. A simple reason (which is much more high-powered than it needs to be) is that there are infinitely many prime ideals in any ideal class of any number field, so given a class $\mathfrak{c}$ in $I/P$, pick a prime ideal in the class outside of the (finite!) list of primes in $T$ and its class in $I_T/P_T$ will map to $\mathfrak{c}$.

Since $P_{T, \mathfrak{m}} \subseteq P_T$, there is a surjection $I_{T}/P_{T, \mathfrak{m}} \rightarrow I_T/P_T$. Composing with an isomorphism $I_T/P_T \cong \mathrm{Cl}$ gives us a surjection $\pi \colon \mathrm{Cl}_{\mathfrak{m}} \rightarrow \mathrm{Cl}$. Thus, $|\mathrm{Cl}_{\mathfrak{m}}| \geq |\mathrm{Cl}|$ will always hold. In your example, we have equality because $P_{T, \mathfrak{m}} = P_T$ (if a principal ideal is not divisible my $\mathfrak{m}$, then there exists a generator that is congruent to $1$ modulo $\mathfrak{m}$ -- in fact, every generator is).


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