4
$\begingroup$

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 :)

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.