# Brauer Groups of Fields

Today we’ll talk about the basic theory of Brauer groups for certain types of fields. If the last post was too poorly written to comprehend, the only thing that will be used from it is that for fields we can refer to “the” Brauer group without any ambiguity because the cohomological definition and the Azumaya (central, simple) algebra definition are canonically isomorphic in this case.

Let’s just work our way from algebraically closed to furthest away from being algebraically closed. Thus, suppose ${K}$ is an algebraically closed field. The two ways to think about ${Br(K)}$ both tell us quickly that this is ${0}$. Cohomologically this is because ${G_K=1}$, so there are no non-trivial Galois cohomology classes. The slightly more interesting approach is that any central, simple algebra over ${K}$ is already split, i.e. a matrix algebra, so it is the zero class modulo the relation we defined last time.

I’m pretty sure I’ve blogged about this before, but there is a nice set of definitions that measures how “far away” from being algebraically closed you are. A field is called ${C_r}$ if for any ${d,n}$ such that ${n>d^r}$ any homogeneous polynomial (with ${K}$ coefficients) of degree ${d}$ in ${n}$ variables has a non-trivial solution.

Thus the condition ${C_0}$ just says that all polynomials have non-trivial solutions, i.e. ${K}$ is algebraically closed. The condition ${C_1}$ is usually called being quasi-algebraically closed. Examples include, but are not limited to finite fields and function fields of curves over algebraically closed fields. A more complicated example that may come up later is that the maximal, unramified extension of a complete, discretely valued field with perfect residue field is ${C_1}$.

A beautiful result is that if ${K}$ is ${C_1}$, then we still get that ${Br(K)=0}$. One could consider this result “classical” if done properly. First, by Artin-Wedderburn any finite dimensional, central, simple algebra has the form ${M_n(D)}$ where ${D}$ is a finite dimensional division algebra with center ${K}$. If you play around with norms (I swear I did this in a previous post somewhere that I can’t find!) you produce the right degree homogeneous polynomial and use the ${C_1}$ condition to conclude that ${D=K}$. Thus any central, simple algebra is already split giving ${Br(K)=0}$.

We might give up and think the Brauer group of any field is ${0}$, but this is not the case (exercise to test understanding: think of ${\mathbb{R}}$). Let’s move on to the easiest example we can think of for a non-${C_1}$ field: ${\mathbb{Q}_p}$ for some prime ${p}$. The computation we do will be totally general and will actually work to show what ${Br(K)}$ is for any ${K}$ that is complete with respect to some non-archimedean discrete valuation, and hence for ${K}$ a local field.

The trick is to use the valuation ring, ${R=\mathbb{Z}_p}$ to interpolate between the Brauer group of ${K}$ and the Brauer group of ${R/m=\mathbb{F}_p}$, a ${C_1}$ field! Since ${K}$ is the fraction field of ${R}$, the first thing we should check is the Leray spectral sequence at the generic point ${i:Spec(K)\hookrightarrow Spec(R)}$. This is given by ${E_2^{p,q}=H^p(Spec(R), R^qi_*\mathbb{G}_m)\Rightarrow H^{p+q}(G_K, (K^s)^\times)}$.

By Hilbert’s Theorem 90, we have ${R^1i_*\mathbb{G}_m=0}$. Recall that last time we said there is a canonical isomorphism ${Br(R)\rightarrow Br(\mathbb{F}_p)}$ given by specialization. This gives us a short exact sequence from the long exact sequence of low degree terms:

$\displaystyle 0\rightarrow Br(\mathbb{F}_p)\rightarrow Br(\mathbb{Q}_p)\rightarrow Hom(G_{\mathbb{F}_p}, \mathbb{Q}/\mathbb{Z})\rightarrow 0$

Now we use that ${Br(\mathbb{F}_p)=0}$ and ${G_{\mathbb{F}_p}\simeq \widehat{\mathbb{Z}}}$ to get that ${Br(\mathbb{Q}_p)\simeq \mathbb{Q}/\mathbb{Z}}$. As already mentioned, nothing in the above argument was specific to ${\mathbb{Q}_p}$. The same argument shows that any (strict) non-archimedean local field also has Brauer group ${\mathbb{Q}/\mathbb{Z}}$.

To get away from local fields, I’ll just end by pointing out that if you start with some global field ${K}$ you can try to use a local-to-global idea to get information about the global field. From class field theory we get an exact sequence

$\displaystyle 0\rightarrow Br(K)\rightarrow \bigoplus_v Br(K_v)\rightarrow \mathbb{Q}/\mathbb{Z}\rightarrow 0,$

which eventually we may talk about. We know what all the maps are already from this and the previous post. The first is specialization (or corestriction from a few posts ago, or most usually this is called taking invariants). Then the second map is just summing since each term of the direct sum is a ${\mathbb{Q}/\mathbb{Z}}$.

Next time we’ll move on to Brauer groups of curves even though so much more can still be said about fields.

# Complex Multiplication 2

I’ll start with thanking Google/Chrome. I sent in a request to fix the CSS rendering issue that happened with the blog and the next day it was fixed (maybe a coincidence?).

We start with our standing assumptions. Let ${E}$ be an elliptic curve with CM by ${R_K}$ for some quadratic imaginary ${K}$ (with normalized iso ${[\cdot]: R_K\rightarrow End(R)}$). We will need the following fact. Suppose ${E}$ is defined over a number field ${L}$. Let ${\frak{p}}$ be a prime of ${L}$ and ${\overline{E}}$ the reduction of ${E}$ mod ${\frak{p}}$. An element ${\gamma}$ is in the image of the natural map ${End(E)\rightarrow End(\overline{E})}$ if and only if ${\gamma}$ commutes with every element in the image of the map.

The proof is a straightforward follow your nose type argument by cases. Caution: You must consider the case that ${End(\overline{E})}$ is an order in a quaternion algebra because ${\overline{E}}$ is defined over a field of positive characteristic. Recall that last time we said that ${H:=K(j(E))}$ is the Hilbert class field of ${K}$. Thus without loss of generality we may assume from here on that ${E}$ is defined over ${H}$ (because ${j(E)\in H}$).

Here is something that seemed bizarre to me the first time I saw it. Let ${\frak{p}}$ be a prime of degree ${1}$ of ${K}$ and ${\frak{b}}$ a prime of ${H}$ lying over ${\frak{p}}$. Suppose ${\frak{p}}$ has the property that the natural map ${E\rightarrow \frak{p}\cdot E}$ is an isogeny of degree ${p}$ (${\frak{p}}$ lies over ${p}$) and the reduction ${\overline{E}\rightarrow \overline{\frak{p}\cdot E}}$ is purely inseparable. This happens for all but finitely many ${\frak{p}}$. Then there is an isogeny ${\lambda: E\rightarrow E^{Frob_\frak{p}}}$ lifting the ${p}$-th power Frobenius. This should feel funny because we’re lifting the Frobenius map to characteristic ${0}$, i.e. the following diagram commutes:

$\displaystyle \begin{matrix} E & \rightarrow & E^{Frob_\frak{p}} \\ \downarrow & & \downarrow \\ \overline{E} & \rightarrow & \overline{E}^{(p)} \end{matrix}$

The proof is to use that we already know there is some ${\lambda: E\rightarrow \frak{p}\cdot E}$ whose reduction is purely inseparable of degree ${p}$ and hence factors as ${\overline{E}\rightarrow \overline{E}^{(p)}\rightarrow \overline{E^{Frob_\frak{p}}}=\overline{E}^{(p)}}$ where the second map of degree ${1}$ and hence an automorphism. We’re done if we can check that this automorphism lifts, but we can do this by checking that it commutes with everything in the image of ${End(E^{Frob_\frak{p}})\rightarrow End(\overline{E}^{(p)})}$ by the above lemma. There’s quite a bit of work checking things element-wise to finish this off.

A special case is the following. For all but finitely many primes ${\frak{p}}$ such that ${\psi_{H/K}(\frak{p})=1}$ (i.e. ${\frak{p}=\pi_\frak{p}R_K}$ is principal) the ${\pi_\frak{p}}$ is unique such that the endomorphism ${[\pi_\frak{p}]}$ descends to the ${p}$-the power Frobenius.

We want to now think about generating abelian extensions of ${K}$ using the torsion points of ${E}$. Fix a finite map ${h:E\rightarrow \mathbb{P}^1}$ defined over ${H}$ (called a Weber function for ${E/H}$). These are easy to come by. For example, if ${j(E)\neq 0, 1728}$, then if our model is ${y^2=x^3+Ax+B}$, then ${h(x,y)=x}$ works. Here is another beautiful relation between CM theory and class field theory. Recall ${E[\frak{a}]}$ are the ${\frak{a}}$-torsion points with respect to the normalized map ${[\cdot]}$, i.e. the set of ${P\in E}$ such that ${[\gamma]P=0}$ for all ${\gamma\in\frak{a}}$.

For any integral ideal ${\frak{a}}$ of ${R_K}$ the field ${K(j(E), h(E[\frak{a}]))}$ is the ray class field of ${K}$ of conductor ${\frak{a}}$. An immediate corollary is that ${K^{ab}=K(j(E), h(E_{tors}))}$. Hence we get the amazing fact that the maximal abelian extension of ${K}$ can be made by taking ${E=\mathbb{C}/R_K}$ and adjoining to ${K}$ the numbers ${j(E)}$ and the ${x}$-coordinates of the torsion points of the Weierstrass model (as long as ${j\neq 0, 1728}$). The proof of this is pretty long and involves all the stuff from earlier in the post, so we’ll skip it.

Last time we said that if we don’t just take ${x}$-coordinates and instead adjoin all torsion coordinates we get ${K(j(E), E_{tors})}$ which is an abelian extension of ${H}$. It is interesting that doing this does not in general produce an abelian extension of ${K}$. Another interesting corollary to this is that we get a situation in which we can adjoin all torsion points. Given any ${K}$ with class number ${1}$ we get a chain ${K^{ab}=K(j(E), h(E_{tors}))\subset K(E_{tors})\subset H^{ab}=K^{ab}}$ and hence ${K^{ab}=K(E_{tors})}$.

# Complex Multiplication 1

Today we’ll really get into the class field theory of CM elliptic curves. Let ${K/\mathbb{Q}}$ be a quadratic imaginary field. Let ${E/\mathbb{C}}$ be an elliptic curve with the property that ${End(E)\simeq R_K}$, e.g. take ${E=\mathbb{C}/R_K}$. At the end of the last post we constructed a group homomorphism ${F:Gal(\overline{K}/K)\rightarrow Cl(R_K)}$ which was defined by the property that ${E^\sigma=F(\sigma)\cdot E}$.

Now we know that ${Cl(R_K)}$ is an abelian group, so in fact ${F}$ factors through (a now relabelled) ${F: Gal(K^{ab}/K)\rightarrow Cl(R_K)}$ where ${K^{ab}}$ is the maximal abelian extension of ${K}$. For the rest of the post we’ll sketch why a few things are true:

1) It turns out that ${H=K(j(E))}$ is the Hilbert class field of ${K}$.

2) Last time we stated that ${[\mathbb{Q}(j(E)): \mathbb{Q}]=[K(j(E)): K]=h_K}$, but given the previous statement, the proof is now easy from class field theory.

3) If ${E_1, \ldots, E_h}$ are a complete set of representatives of ${Ell(R_K)}$, then ${j(E_1), \ldots, j(E_h)}$ form a complete set of Galois conjugates for ${j(E)}$.

4) For any prime ${\frak{p}}$ of ${K}$, we have ${j(E)^{Frob_\frak{p}}=j(F(Frob_\frak{p})\cdot E)}$. Extending to all fractional ideals we get ${j(E)^{\psi_{H/K}(\frak{a})}=j(\frak{a}\cdot E)}$.

Let ${L=\overline{K}^{\ker F}}$. One can check directly that ${L=K(j(E))}$. This shows that ${L/K}$ is abelian. Let ${G=Gal(L/K)}$. By class field theory there is some associated conductor ${\frak{m}}$. Considering the composition with the Artin map ${I_\frak{m}\stackrel{\Psi_{L/K}}{\rightarrow} G\stackrel{F}{\rightarrow} CL(R_K)}$ one can see this is just the natural projection map. Now using the injectivity of ${F}$ one can extrapolate that ${\frak{m}=(1)}$ showing ${L/K}$ is unramified and hence contained in the Hilbert class field.

But class field theory tells us that ${I_\frak{m}}$ surjects onto ${Cl(R_K)}$ and hence ${F}$ is an isomorphism which shows ${H=L}$ and hence ${K(j(E))}$ is the Hilbert class field of ${K}$. We’ve already discussed the second item. The third just follows from identifying ${Ell(R_K)}$ with ${\{j(E_1), \ldots, j(E_h)\}}$ and using the fact that ${Ell(R_K)}$ is a ${Cl(R_K)}$-torsor. The fourth just comes from the fact that ${I_\frak{m}}$ consists of all fractional ideals because ${\frak{m}=(1)}$.

I know, this was pretty skimpy on details, but the point of this series should be to see some of the ideas and results from basic CM theory for elliptic curves and not full blown explanations. The results in this post are really cool in my mind because it takes these very classical purely field theory questions and converts them to geometric questions about elliptic curves and vice-versa. Given a quadratic imaginary ${K}$, it is easy to cook up an elliptic curve ${E}$ with CM by ${R_K}$. If you want to know the maximal unramified abelian extension of ${K}$, we now know we need only figure out ${j(E)}$ because ${H=K(j(E))}$.

# Local Class Field Theory

Today will probably be our last class field theory post. I want to end with a brief description of local class field theory. Let ${K}$ be a global field, and ${v\in M_K}$ a place. We have our standard inclusion ${i_v: K_v^*\hookrightarrow \mathcal{J}_K}$ by putting the element in the ${v}$ component and ${1}$‘s everywhere else. Suppose ${L/K}$ is abelian. We have the Artin map ${\psi_{L/K}: C_K/N_{L/K}(C_L)\stackrel{\sim}{\rightarrow} G=Gal(L/K)}$.

What we learned two posts ago is that the image upon composing the maps ${K_v\rightarrow \mathcal{J}_K\twoheadrightarrow C_K/N_{L/K}(C_L)\rightarrow G}$ is exactly the decomposition group ${G_w}$ where ${w}$ lies over ${v}$. The image of the units ${\mathcal{O}_v^*}$ under the map is the inertia group ${I_w}$. This gives us two exact sequences that fit together:

$\displaystyle \begin{matrix} 1 & \rightarrow & N_{L_w/K_v}(L_w^*) & \rightarrow & K_v^* & \rightarrow & G_w & \rightarrow & 1 \\ & & \cup & & \cup & & & & \\ 1 & \rightarrow & N_{L_w/K_v}(\mathcal{O}_w^*) & \rightarrow & \mathcal{O}_v^* & \rightarrow & I_w & \rightarrow & 1\end{matrix}$

This gives us a local Artin map ${\psi_{w/v}: K_v^*\rightarrow G_w}$. The main theorem of local class field theory is that the map is surjective with kernel ${N_{L_w/K_v}(L_w^*)}$ and moreover the map can be defined independently of localizing the global fields. Just like global class field theory there is an “existence” part of the theorem as well.
This part says that every finite abelian extension of local fields arises as the localization of an extension of global fields and the local Artin maps ${\psi_{w/v}}$ give a bijection between

$\displaystyle \left\{ \text{finite index open subgroups in} \ K_v^*\right\} \leftrightarrow \left\{\text{finite abelian ext} \ L_w/K_v \right\}$

where the correspondence is ${U \leftrightarrow \ker \psi_{w/v}}$.
Let’s do the simplest example. Let’s think about the quadratic extensions ${\mathbb{Q}_p}$ by looking at the bijection. The standard construction is that ${\mathbb{Q}_p(\sqrt{d})}$ are the quadratic extensions where ${d}$ is not a square, and ${\mathbb{Q}_p(\sqrt{d'})}$ is the same extension if ${d/d'}$ is a square. Thus there is a nice bijection between the quadratic extensions and the non-trivial elements of ${\mathbb{Q}_p^*/(\mathbb{Q}_p^*)^2}$.

Local class field theory tells us that the quadradic extensions of ${\mathbb{Q}_p}$ are in bijection with the open index ${2}$ subgroups ${U\subset \mathbb{Q}_p^*}$ via ${L/\mathbb{Q}_p\mapsto N_{L/\mathbb{Q}_p}(L^*)}$. It turns out that any index ${2}$ subgroup at all must be open because it will contain ${(\mathbb{Z}_p^*)^2}$.

Now it will be useful to think in different terms, which is actually a more standard modern reformulation of class field theory since it generalizes to “higher dimensions.” There is a bijection between the open index ${2}$ subgroups ${U}$ of ${\mathbb{Q}_p^*}$ and surjective characters ${\chi: \mathbb{Q}_p^*\rightarrow \{\pm 1\}}$ just by taking the kernel of the character. Thus we have reformulated the problem of counting these subgroups to counting characters.

The unit groups have the form (if ${p}$ odd) ${\mathbb{Q}_p^*\simeq p^\mathbb{Z}\times \mathbb{Z}_p^*}$ and ${\mathbb{Z}_p^*\simeq \mu_{p-1}\times (1 + p\mathbb{Z}_p)}$ where ${\mu_{p-1}}$ is thought of via the Teichmuller lift. If ${\chi}$ has order ${2}$, then it is trivial on ${(\mathbb{Q}_p^*)^2\simeq p^{2\mathbb{Z}}\times (\mathbb{Z}_p^*)^2\simeq p^{2\mathbb{Z}}\times (\mu_{p-1})^2\times (1+p\mathbb{Z}_p)}$.

Thus we get the result that ${\chi}$ factors through ${p^\mathbb{Z}/p^{2\mathbb{Z}}\times \mu_{p-1}/(\mu_{p-1})^2}$ which is a finite abelian ${2}$-group of order ${4}$. Thus the number of non-trivial characters is ${3}$. This gives us a nice alternate description to the classical Kummer description. It tells us there are exactly ${3}$ quadratic extensions up to isomorphism. These aren’t hard to figure out explicitly either. Just take some ${a\in \mathbb{Z}_p^*}$ which is not a square. The three quadratic extensions are ${\mathbb{Q}_p(\sqrt{p})}$, ${\mathbb{Q}_p(\sqrt{a})}$, and ${\mathbb{Q}_p(\sqrt{ap})}$. A similar description can be computed when ${p=2}$, but you get ${7}$ in that case.

That is all for class field theory for now. We’ll move on to complex multiplication, and I think we’ve done enough of the basics that we can probably do what we need as we need it now.

# Examples with the ideles

Last time was a bit of a mess, so let’s review what happened using some examples. We take ${L=\mathbb{Q}(\zeta_m)}$ and ${K=\mathbb{Q}}$. The conductor is ${\frak{m}=\{m\mathbb{Z}, w_1\}}$ (assume for simplicity that ${m}$ odd). We saw this before when discussing the Artin map because now ${\Psi_{L/K}:Cl_m(\mathbb{Q})\simeq (\mathbb{Z}/m)^*\stackrel{\sim}{\rightarrow} Gal(L/K)}$ where the map is ${a\mapsto (\zeta_m\mapsto \zeta_m^a)}$.

For the new stuff let’s think about the idele class group ${C_K\simeq \mathcal{J}_K/K^*}$. Let ${v}$ be a finite place associated to ${\ell | m}$. This means there is some number ${0\neq a=ord_\ell(\frak{m})}$. Suppose ${w}$ is a place in ${\mathbb{Q}(\zeta_m)}$ over ${\ell}$. This is ramified, so we’ll get something interesting for the inertia. There is a natural embedding ${\mathbb{Z}_\ell^*\hookrightarrow \mathcal{J}_K}$ given by ${\alpha\mapsto (1, \ldots, 1, \alpha, 1, \ldots )}$ (where the non ${1}$ is at the ${\ell}$ spot).

There is also the Artin map that factors through the class group ${\psi_{L/K}: \mathcal{J}_K\rightarrow Cl_m(\mathbb{Q})\simeq (\mathbb{Z}/m)^*}$. The composite ${\mathbb{Z}_\ell^*\rightarrow Cl_m(\mathbb{Q})}$ is the map ${\alpha\mapsto (1, \ldots, \alpha^{-1}, 1, \ldots) \mod \frak{m}}$. This is picking out the inertia group. We get that ${\mathbb{Z}_\ell^*\rightarrow 1\times \cdots \times (\mathbb{Z}/\ell^a)^*\times \cdots \times 1=I_w}$.

Let ${\pi_\ell}$ be a uniformizer at ${\ell}$. Then ${\psi_{L/K}(1, \ldots , \pi_\ell, 1, \ldots)}$ is an element of the Frobenius coset of ${I_w}$ in ${G_w}$ (the decomposition group). We could pick the ${\pi_\ell=\ell}$. Now since we could multiply through by ${\ell^{-1}}$ and not change the class in ${C_K=\mathcal{J}_K/K^*}$ we can look at ${\displaystyle \psi_{L/K}(\ell^{-1}, \ell^{-1}, \ldots, 1, \ell^{-1}, \ldots)=(\ell, \ldots, \ell, 1, \ell, \ldots)\in (\mathbb{Z}/m)^*\simeq \prod_{q|m}(\mathbb{Z}/q^{ord_q(m)})^*}$. This shows the Frobenius coset can have anything in the ${\ell}$ spot, but must have ${\ell}$ times something in all other spots.

Now let’s think about ${K=\mathbb{Q}}$ in general. It is pretty quick to check that ${\displaystyle \mathcal{J}_\mathbb{Q}=\mathbb{Q}^*\times \prod_{v \ nonarch} \mathbb{Z}_v^*\times \mathbb{R}_+^*}$ and hence ${\displaystyle C_\mathbb{Q}=\prod_{v \ nonarch} \mathbb{Z}_v^*\times \mathbb{R}_+^*}$ and hence since ${D_K=\mathbb{R}_+^*}$ we get that ${\displaystyle C_K/D_K=\prod_{v \ nonarch} \mathbb{Z}_v^*\stackrel{\sim}{\rightarrow}\lim_{\longleftarrow} Cl_m(\mathbb{Q})\simeq Gal(\mathbb{Q}^{ab}/\mathbb{Q})}$. You can follow through the same maps as above to get an element-wise description.

Slightly more fun, we can try again with the function field case ${K=\mathbb{F}_q(t)}$. The set of places here is ${M_k=\{v_\pi : \pi\in\mathbb{F}_q[t] \ \text{irreducible monic non-constant}\}\cup\{v_\infty\}}$ where ${v_\infty(f)=-deg(f)}$. To get a handle on the ideles, we can look at the degree map. We have ${1\rightarrow J^0\rightarrow \mathcal{J}_K\stackrel{deg}{\rightarrow} \mathbb{Z}\rightarrow 0}$ by ${(j_v)\mapsto \sum ord_v(j_v)[k(v): \mathbb{F}_q]}$. This gives a sequence ${1\rightarrow J^0/K^*\rightarrow C_K\stackrel{deg}{\rightarrow} \mathcal{J}_K/J^0\simeq \mathbb{Z}\rightarrow 0}$.

Strangely, the map ${C_K\rightarrow Gal(K^{ab}/K)}$ is not surjective like the number field case. This is because we now produce a diagram

$\displaystyle \begin{matrix} C_K & \twoheadrightarrow & \mathbb{Z} \\ \downarrow & & \downarrow \\ Gal(K^{ab}/K) & \twoheadrightarrow & Gal(\overline{\mathbb{F}_q}K/K)\simeq \widehat{\mathbb{Z}}\end{matrix}$

where the right vertical arrow is just ${1\mapsto Frob}$. You get the Artin map is an injection, but not surjective. You get a dense subgroup, but miss the profinite powers. Now we get a very similar decomposition of the ideles: ${\displaystyle \mathcal{J}_K=K^*\times \prod_{v\neq v_\infty}\mathcal{O}_v^*\times (1+\pi_{v_\infty}\mathcal{O}_{v_\infty})^*}$.

We can calculate some ray class groups using this stuff. Let ${\frak{m}}$ be a conductor for ${K}$. Note this is nothing more than an effective divisor say ${\sum m(v)v}$ where ${m(v)\geq 0}$. We have a the map ${\mathcal{J}_K\rightarrow Cl_m(K)}$ given by ${(j_v)\mapsto div(j\alpha^{-1})}$ (the alpha coming from the approximation theorem argument). We can compute exactly the kernel of this, so ${\displaystyle Cl_m(K)\simeq \frac{\mathcal{J}_K}{K^*\times \prod (1+\pi_v^{m(v)}\mathcal{O}_v)^*}}$. With a bit of work we can put all this together to get the following results.

If we assume that ${ord_{v_\infty}(m)>0}$, then we put together the sequence ${\displaystyle 1\rightarrow \prod (\mathcal{O}_v/\pi_v^{m(v)}\mathcal{O}_v)^*\rightarrow Cl_m(K)\rightarrow \mathbb{Z}\rightarrow 0}$. In the other case ${ord_{v_\infty}(m)=0}$, then we must divide out by the image of the scalars ${\displaystyle 1\rightarrow \frac{\prod (\mathcal{O}_v/\pi_v^{m(v)}\mathcal{O}_v)^*}{\mathbb{F}_q^*} \rightarrow Cl_m(K)\rightarrow \mathbb{Z}\rightarrow 0}$. Thus all this global idele class field theory tells us something about our ray class groups which in turn can be used to tell us about the abelian extensions of our field. Next time we’ll try to wrap this stuff up with some local class field theory.

# Idele Class Field Theory 1

Today we’ll start reformulating class field theory in the idelic setting. Let ${K}$ be a global field (a number field, or a finite separable extension of ${\mathbb{F}_q(t)}$). Let ${M_k}$ be all the places and ${|\cdot |_v}$ the normalized absolute value at ${v\in M_k}$. Define ${K_v}$ be to be the completion of ${K}$ at ${v}$. Let ${\mathcal{O}_v\subset K_v}$ the ring of integers if ${v}$ is non-archimedean and just ${K_v}$ itself otherwise.

We’re going to assume that a first algebraic number theory course covers some of the basics of the ideles. Define ${\mathcal{J}_K}$ to be the ideles of ${K}$. Recall this is by definition ${\mathcal{J}_K=\{(j_v)_{v\in M_k} : j_v\in K_v^* \ \text{and} \ j_v\in \mathcal{O}_v^* \ \text{for all but finitely many}\}}$. This is a multiplicative group, but moreover there is a natural topology on it from the restricted direct product. We’ll recall properties of this as needed. We should quickly remind the reader that this is not the same as the product topology. It is the topology generated by the local basis around ${1}$ of the form where you take finitely many components to be open sets and the rest must be ${\mathcal{O}_v^*}$.

The ring of adeles is exactly analogous. We take ${\mathbb{A}_K=\{(a_v)_{v\in M_k} : a_v\in K_v \ \text{and} \ a_v\in \mathcal{O}_v \ \text{for all but finitely many}\}}$. This is a locally compact topological ring. By construction ${\mathbb{A}_K^*=\mathcal{J}_K}$ as a set. But the subspace topology is weaker than the topology we give to ${\mathcal{J}_K}$. You do recover the right topology by taking the subspace of the product ${\mathbb{A}_K\times \mathbb{A}_K}$ by ${\mathcal{J}_K=\{(x,y) : xy=1\}}$.

We define now the idele class group to be ${C_K=\mathcal{J}_K /K^*}$. Fix a conductor ${\frak{m}=\{\frak{m}_f, w_1, \ldots, w_i\}}$. Our first theorem is: there is a surjection ${\lambda_m: C_K\rightarrow Cl_\frak{m}(K)}$ where ${\lambda_m}$ is constructed as follows. Given ${j=(j_v)\in J_K}$ use the approximation theorem to find a non-zero ${\alpha\in K^*}$ with the property that ${ord_v(j_v\cdot \alpha^{-1}-1)\geq ord_v(\frak{m}_f)}$ if ${ord_v(\frak{m}_f)>0}$ and ${w_v(j_v\alpha^{-1})>0}$ where ${w_v}$ ranges through ${w_k}$ for ${k=1, \ldots , i}$. Define ${\lambda_m(j)}$ to be the fractional ideal ${\frak{a}}$ such that ${ord_v(\frak{a})=ord_v(j_v \alpha^{-1})}$ for all non-archimedean ${v}$.

Note that if we do this, at very least we’ve constructed a fractional ideal that is relatively prime to ${\frak{m}}$. Thus we have a candidate map. We should check well-definedness, but that is more tedious than it’s worth. Let’s just take the theorem as true. Out of it we get the nice corollary that there is a surjection ${\displaystyle \lambda: C_K\rightarrow \lim_{\leftarrow_{\frak{m}}}Cl_\frak{m}(K)=Gal(K^{ab}/K)}$ where ${K^{ab}}$ is the maximal abelian extension of ${K}$ unramified outside ${\frak{m}}$. The equality comes from piecing together the Artin maps using class field theory. The kernel is the connected component of the identity of ${C_K}$. This corollary is by no means obvious from the theorem.

The fact that the ${\lambda_m}$ form a compatible system to get a map to the inverse limit is easy because from the construction of the ${\frak{a}}$ that we built, we can just use the same for any compatible system. From a few posts ago we worked out what the kernel of the natural surjection ${Cl_\frak{m}(K)\rightarrow Cl(\mathcal{O}_K)}$ was. Call that kernel ${Q}$. To figure out the kernel of ${\lambda}$ let’s consider the diagram:

$\displaystyle \begin{matrix} & & 1 \\ & & \downarrow \\ U_k & \twoheadrightarrow & Q \\ \cap & & \downarrow \\ \mathcal{J}_K & \stackrel{\lambda_m}{\twoheadrightarrow} & Cl_\frak{m}(K)\\ \parallel & & \downarrow \\ \mathcal{J}_K & \stackrel{\lambda_{\mathcal{O}_K}}{\twoheadrightarrow} & Cl(\mathcal{O}_K)\\ & & \downarrow \\ & & 1 \end{matrix}$

Now notice that going around the square ${\mathcal{J}_K\rightarrow Cl_\frak{m}(K)\rightarrow Cl(\mathcal{O}_K)}$ is just taking ${j}$ and sending it to the ideal we got out of this ${j\alpha^{-1}}$, but then pushing forward to ${Cl(\mathcal{O}_K)}$ we see that the ${\alpha^{-1}}$ is changing things by some principal and hence ${\mathcal{J}_K\rightarrow Cl(\mathcal{O}_K)}$ is the map that builds the ideal from ${j}$ alone. So we want to know the kernel and hence when is making the ideal formed from looking at ${ord_v(j)}$ principal.

Suppose ${j}$ maps to ${0}$, then there is some ${\beta\in K^*}$ so that ${ord_v(j)=ord_v(\mathcal{O}_K\beta)}$ for all ${v}$. Thus ${j\beta^{-1}}$ maps to ${\displaystyle U_k=\prod \mathcal{O}_v^*\times \prod K_v^*}$, the so-called unit ideles. Since this map factors through the idele class group ${C_K}$, we haven’t changed where ${j}$ goes by altering it to ${j\beta^{-1}}$. This shows the ${U_k}$ part of the diagram except for the surjection. To finish the proof of this corollary just involves messing with the ord conditions and using the Mittag-Leffler condition, so we’ll omit it. It is the statements in class field theory together with some motivating examples that are of interest for this blog at this time.

Let’s wrap up today by stating the big theorem. Suppose ${L/K}$ is finite abelian with conductor ${\frak{m}}$ and ${G=Gal(L/K)}$. We get maps ${C_K\twoheadrightarrow Cl_\frak{m}(K)\stackrel{\Psi_{L/K}}{\rightarrow} G}$. Call the composite ${\psi_{L/K}}$. Then

1) ${ker(\psi_{L/K})=N_{L/K}(C_L)}$

2) For all places ${v}$ of ${K}$ we get an inclusion ${K_v^*\hookrightarrow \mathcal{J}_K}$ by ${\alpha\mapsto (1, 1, \ldots, 1, \alpha, 1, \ldots )}$ and ${\pi: \mathcal{J}_K\rightarrow C_K}$ and ${\psi_{L/K}\circ \pi\circ i_v(K_v^*)=G_w}$ where ${G_w}$ is the decomposition group and ${w}$ is any place of ${L}$ over ${v}$.

Now this gives us an early glimpse at why the idele formulation is going to give us more. We’ll get some control on inertia and decomposition groups.

# Functoriality of the Artin Map

Instead of restating the functoriality of the Artin map, let’s just review the statement through an example. We’ll re-use our example from last time. Let ${L}$ be the splitting field of ${x^3-x-1}$ over ${\mathbb{Q}}$. We get a non-abelian Galois group ${H\simeq S_3}$ (to keep notation the same, we called this ${H}$ last time). Take the quadratic subextension ${K=\mathbb{Q}(\sqrt{-23})}$. We have an abelian Galois group ${G\simeq \mathbb{Z}/3}$. We need the abelianization ${H^{ab}\simeq \mathbb{Z}/2}$.

By Galois theory we know ${H^{ab}}$ gives us a field extension ${L'}$ sitting between ${L}$ and ${\mathbb{Q}}$. Class field theory tells us that the conductor ${\frak{m}_{L'}=\{(23), w_1\}}$ because we must pick up all ramification from ${\mathbb{Q}(\sqrt{-23})/\mathbb{Q}}$. The general argument we gave a few posts ago shows us that ${Cl_{\frak{m}_{L'}}(\mathbb{Q})\simeq (\mathbb{Z}/23)^*}$. Now ${K}$ is the Hilbert class field, so ${Cl(\mathcal{O}_K)\stackrel{\sim}{\rightarrow} G}$ via the Artin map. We take ${\frak{m}=(\sqrt{-23})\mathcal{O}_K}$ with no embeddings specified.

This gives us the diagram:

$\displaystyle \begin{matrix} Cl_\frak{m}(K) & \twoheadrightarrow & G & \stackrel{\Psi_{L/K}}{\rightarrow} & \mathbb{Z}/3 & \rightarrow & 1 \\ id \downarrow & & & & \downarrow & & \\ Cl_\frak{m}(K) & \stackrel{N_{K/F}}{\rightarrow} & (\mathbb{Z}/23)^* & \stackrel{\Psi_{L'/F}}{\rightarrow} & \mathbb{Z}/2 & \rightarrow & 1 \end{matrix}$

First, the right vertical arrow is clearly the zero map. The other important part of the diagram is that the norm map is taking a fractional ideal (class) that is relatively prime to ${(\sqrt{-23})\mathcal{O}_K}$ and taking the norm of it which lands you in the units ${(\mathbb{Z}/23)^*\simeq \mathbb{Z}/22}$. Moreover, the map ${\mathbb{Z}/22\rightarrow \mathbb{Z}/2}$ is the unique surjective one and the image of the norm map must land in the kernel of this by exactness. Interestingly, this tells us that the positive generator of ${N_{K/F}(\frak{b})}$ for any ${\frak{b}}$ prime to ${(\sqrt{-23})\mathcal{O}_K}$ is a square mod 23.

Let’s wrap up today by stating another functoriality result. Given the same setup of ${L/K/F}$ where ${G=Gal(L/K)}$ is abelian and ${H=Gal(L/F)}$ is finite possibly non-abelian. Suppose now that ${G\triangleleft H}$ and ${K/F}$ Galois with ${T=Gal(K/F)}$. Now ${T}$ acts on ${G}$ as follows. Let ${t\in T}$. Choose a lift ${h_t\in H}$. The action is given by ${t\cdot g=h_t g h_t^{-1}}$. Call this action ${\sigma_t}$.

We can transfer this Galois action to the ray class group as follows:

$\displaystyle \begin{matrix} Cl_\frak{m}(K) & \stackrel{\Psi}{\rightarrow} & G & \rightarrow & 1 \\ \downarrow & & \downarrow \sigma_t & & \\ Cl_\frak{m}(K) & \stackrel{\Psi}{\rightarrow} & G & \rightarrow & 1 \end{matrix}$

where the vertical arrow is just the natural map on ideals. Commutativity of the diagram just comes from the standard fact that ${hFrob_p(Q)h^{-1}=Frob_p(hQ)}$. There is another functoriality we could do, but it doesn’t seem worth it at this point because it is overly complicated and there isn’t a plan to use it anytime soon.

# Class Field Theory 2

Today we’ll start by sketching an example of a Hilbert class field. Let ${L}$ be the splitting field of ${x^3-x-1}$ over ${\mathbb{Q}}$. Thus ${L=\mathbb{Q}(\alpha_1, \alpha_2, \alpha_3)}$ where ${\alpha_j}$ are the roots of the polynomial. Standard Galois theory shows us that ${Gal(L/\mathbb{Q})\simeq S_3}$. I know, we assumed abelian extensions last time, but we aren’t interested in this extension. It turns out that one can also argue that we have a subextension ${K=\mathbb{Q}(\sqrt{-23})\subset L}$.

To see how useful this class field theory machinery is, we can check with some basic algebraic number theory (involves the Minkowski bound) that ${Cl(\mathcal{O}_K)\simeq \mathbb{Z}/3}$. But above we said that ${Gal(L/K)\simeq \mathbb{Z}/3}$. Thus by the main theorem last time we see that ${L}$ must be the Hilbert class field of ${K}$, and hence ${L}$ is the maximal unramified abelian extension of ${\mathbb{Q}(\sqrt{-23})}$. That’s pretty interesting that it is so small.

This gives us an idea of what these Hilbert class fields look like. Note also that class field theory tells us that if the class number is ${1}$, then there are no unramified abelian extensions, and hence fields like ${\mathbb{Q}(i)}$ or ${\mathbb{Q}(\sqrt{-11})}$ have Hilbert class field equal to themselves. The word “maximal” makes these things sound big, but being unramified is pretty strict and we see that in order to have large Hilbert class fields the class group needs to be large.

Let’s try to go the other direction now. Recall a few posts ago we calculated that ${Cl_m(\mathbb{Q})\simeq (\mathbb{Z}/m)^*}$ when we include the embedding in the conductor. We have an isomorphism ${Cl_m(\mathbb{Q})\simeq (\mathbb{Z}/m)^*\stackrel{\sim}{\rightarrow} Gal(\mathbb{Q}(\zeta_m)/\mathbb{Q})}$, where ${\zeta_m}$ is a primitive ${m}$-th root of unity. We’ll suggestively call the isomorphism ${\Psi([a])=\sigma_a}$ where ${\sigma_a(\zeta_m)=\zeta_m^a}$. This comes from standard theory of cyclotomic fields.

We have that the ring of integers is ${\mathbb{Z}[\zeta_m]}$. But now on the primes, we see that ${\Psi(p\mathbb{Z})}$ is just the Frobenius map of raising to the ${p}$. This shows us that our standard isomorphism is actually the Artin map ${\Psi_{\mathbb{Q}(\zeta_m)/\mathbb{Q}}}$ and hence class field theory tells us that the ray class field of ${\mathbb{Q}}$ of conductor ${\frak{m}=\{(m), \mathbb{Q}\hookrightarrow \mathbb{R}\}}$ is just the cyclotomic extension ${\mathbb{Q}(\zeta_m)}$.

This is an incredibly important example, because now let’s apply the Galois correspondence. Let ${L/\mathbb{Q}}$ be any abelian extension at all. We know that ${L}$ corresponds to some conductor ${\frak{m}_L}$. This must be of the form ${\{\mathbb{Z}m\}}$ or ${\{\mathbb{Z}m, w_1\}}$. Since the first one divides the second, we get two surjections ${Cl_{\{(m), w_1\}}(\mathbb{Q})\twoheadrightarrow Cl_\frak{m}(\mathbb{Q})\twoheadrightarrow Gal(L/\mathbb{Q})}$. But the first term of this by the above is that ${Cl_{\{(m), w_1\}}(\mathbb{Q})\simeq Gal(\mathbb{Q}(\zeta_m)/\mathbb{Q})}$. Therefore, ${L\subset \mathbb{Q}(\zeta_m)}$. Out of class field theory we get the classical result called the Kronecker-Weber theorem that any abelian extension of ${\mathbb{Q}}$ must be contained inside a cyclotomic field.

To finish today let’s talk about one more topic. Suppose we have our abelian extension ${L/K}$ of number fields with Galois group ${G}$. If we have a further subextension ${K/F}$, then we get ${Gal(L/F)=H}$ is possibly non-abelian. But we can take the abelianization ${H^{ab}=H/[H,H]}$ and this corresponds via Galois theory to a maximal abelian subextension say ${L'}$ between ${L}$ and ${F}$, i.e. ${Gal(L'/F)\simeq H^{ab}}$. By the universal property of the abelianization, we have a map ${G\rightarrow H^{ab}}$ since ${G\subset H}$.

The point is that ${G}$ corresponds to an abelian extension ${L/K}$ and so class field theory over ${K}$ tells us something about it, and ${H^{ab}}$ corresponds to an abelian extension ${L'/F}$, so class field theory over ${F}$ tells us something about it. We just produced a natural map ${G\rightarrow H^{ab}}$, and hence there should be a corresponding statement in class field theory.

Here’s the theorem. Let ${\frak{m}=\frak{m}_{K/F}}$ so that ${\frak{m}_L=\frak{m}_{L/K}}$ divides ${\frak{m}}$ and ${L/F}$ is unramified outside the places of ${F}$ under ${\frak{m}}$. Then we get a commutative diagram:

$\displaystyle \begin{matrix} Cl_\frak{m}(K) & \twoheadrightarrow & Cl_{\frak{m}_L}(K) & \stackrel{\Psi_{L/K}}{\rightarrow} & G & \rightarrow & 1 \\ id \downarrow & & & & \downarrow & & \\ Cl_\frak{m}(K) & \stackrel{N_{K/F}}{\rightarrow} & Cl_{\frak{m}_{L'}}(F) & \stackrel{\Psi_{L'/F}}{\rightarrow} & H^{ab} & \rightarrow & 1 \end{matrix}$

We could call this a functoriality property of the Artin map. The loose description of this is that inclusions of Galois groups go to the corresponding norm map. We’ll pick up here next time.

# Class Field Theory 1

Today we’ll to get to stating class field theory in terms of ray class groups. Let ${L/K}$ be an extension of number fields. Recall that if ${G=Gal(L/K)}$ is abelian and ${\frak{p}\subset \mathcal{O}_K}$ is a prime unramified in ${\mathcal{O}_L}$, then the Frobenius element, ${\Phi(\frak{q}/\frak{p})=Frob_{L/K}(\frak{q})}$, is really an element and is independent of the ${\frak{q}}$ over ${\frak{p}}$. The independence is just because we are assuming an abelian Galois group, and all the Frobenii are conjugate.

Let ${S}$ be the set of places that ramify in ${L}$. Recall that ${I_S(K)}$ are the fractional ideals relatively prime to ${S}$. Thus we can define a group homomorphism called the Artin map ${\Psi_{L/K}: I_S(K)\rightarrow G}$ by sending a prime to its Frobenius element and then extending.

To have something concrete in your head take the example of ${L=\mathbb{Q}(\sqrt{3})/\mathbb{Q}}$. We have that ${\mathcal{O}_L=\mathbb{Z}[\sqrt{3}]}$, and the discriminant is ${12}$. Take a prime ideal downstairs, ${p\mathbb{Z}}$. There are three possibilities, it ramifies so ${p\mathcal{O}_L=Q^2}$, it splits so ${p\mathcal{O}_L=Q_1Q_2}$, or it is inert in which case ${p\mathcal{O}_L=Q}$.

From the discriminant we know the only primes that ramify are ${p=2, 3}$. To determine whether the other primes split or are inert depends on whether or not ${x^2-3}$ has a root mod ${p}$ by Hensel’s lemma. It turns out ${p}$ splits if ${3}$ is a square mod ${p}$ and is inert if ${3}$ is not a square mod ${p}$. Thus we can read off these two cases from the Legendre symbol.

In the case that it splits, ${p\mathcal{O}_L=Q_1Q_2}$. The decomposition group ${D_{Q_1}=\{id\}}$ because the Galois group acts transitively on ${Q_1}$ and ${Q_2}$, but there is only one non-trivial element which must switch the two primes. Thus only the identity fixes ${Q_1}$ which is the definition ${D_{Q_1}}$. So we get ${\Psi_{L/\mathbb{Q}}(p)=1}$ when ${p}$ splits (i.e. 3 a square mod ${p}$). Likewise, in the inert case we get ${D_{Q}=G}$. Thus the Frobenius must be the non-trivial element of ${G}$ since it generates ${G}$. Thus ${\Psi_{L/\mathbb{Q}}(p)=\sigma}$ (the generator, i.e. raising to the ${p}$ power). This should make us think that quadratic reciprocity will turn out to be a special case of where we’re going.

Now let’s state the main theorem of class field theory (in our special case above). Artin actually proved this version of it. There is a conductor ${\frak{m}=(\frak{m}_f, w_1, \ldots, w_i)}$ depending on ${L/K}$ such that:

a) ${\text{Supp}(\frak{m})=S}$ where the support is the set of places that divide ${\frak{m}_f}$ together with real embeddings ${K\hookrightarrow \mathbb{R}}$.

b) The Artin map ${\Psi_{L/K}: I_S(K)\rightarrow G}$ factors through the ray class group, i.e. ${\Psi_{L/K}: I_S(K)\rightarrow Cl_{\frak{m}}(K)\rightarrow G}$.

c) ${\Psi_{L/K}}$ is surjective.

There is a smallest such ${\frak{m}}$ (under the partial order by dividing on the ideals and inclusion on the embeddings) called the conductor of the extension. There is an existence part of the theorem that gives a Galois correspondence. It says that for each conductor ${\frak{m}}$ and subgroup ${H}$ of ${Cl_\frak{m}(K)}$, there is an abelian extension ${E/K}$ such that ${\frak{m}_{E/K}}$ (the conductor coming from the earlier statement) divides ${\frak{m}}$ and the composition ${Cl_\frak{m}(K)\rightarrow Cl_{\frak{m}_{E/K}}(K)\stackrel{\Psi}{\rightarrow} Gal(E/K)}$ is surjective and has kernel ${H}$.

In particular, you can take ${H=\{e\}}$ and force the maps to be isomorphisms. In that case ${E}$ is called the ray class field of ${K}$ of conductor ${\frak{m}_{E/K}}$. One caution is that even in this nice situation you can’t force ${\frak{m}_{E/K}=\frak{m}}$. For example, take ${\frak{m}}$ to be ${\mathcal{O}_K}$ with no embeddings. The ray class field in this situation is called the Hilbert class field and the ray class group is just the ideal class group.

Thus we get ${Cl(\mathcal{O}_K)\stackrel{\sim}{\rightarrow} Gal(E/K)}$ via the Artin map for the Hilbert class field. Now take ${\frak{m}=\{\mathcal{O}_K, w_1\}}$. Just throw in an embedding. The ideal class group ${Cl(\mathcal{O}_K)}$ is the group of fractional ideals mod the principal ideals. This ${Cl_\frak{m}(K)}$ is the group of fractional ideals mod principal ideals generated by something that is positive under ${w_1}$. These are isomorphic because you can always multiply the generator by ${-1}$. Thus starting with this ${\frak{m}}$ and ${H=\{e\}}$ we get the Hilbert class field again, but ${\frak{m}_{E/K}=\{\mathcal{O}_K\}}$ without the embedding and hence in general we can’t force ${\frak{m}=\frak{m}_{E/K}}$ in general.

To end today let’s just say a few more things about the Hilbert class field. It is the maximal abelian unramified extension of ${K}$, say ${L_0}$. The proof that it is unramified is just that ${\text{Supp}(\mathcal{O}_K)=\emptyset}$. Let ${L/K}$ be any unramified abelian extension. Suppose ${L}$ is not contained in ${L_0}$. Then ${LL_0}$ is an unramified extension strictly containing ${L_0}$. Thus by the main theorem of class field theory ${LL_0}$ has a conductor dividing ${\frak{m}}$ and hence equal to ${\mathcal{O}_K}$. Thus ${Cl(\mathcal{O}_K)\simeq Gal(L/K)\rightarrow Gal(LL_0/K)}$ is surjective, a contradiction. This tells us we can take as our definition of the Hilbert class field to be the ray class field of conductor ${\mathcal{O}_K}$ and get the standard definition that it is the maximal unramified abelian extension of ${K}$.

# Ray Class Groups 2

Today let’s relate the ray class groups to the ideal class group. Fix ${K}$ a number field and choose a conductor ${\frak{m}=\{\frak{m}_f, w_1, \ldots, w_i\}}$. There is certainly always a map ${Cl_\frak{m}(K)\rightarrow Cl(\mathcal{O}_K)}$. This is just because given some fractional ideal ${\frak{a}}$ that is relatively prime to ${\frak{m}}$ it is a fractional ideal, so it maps to its class in the ideal class group (well-definedness is just because we mod out by less principal ideals, but they’re all principal).

It turns out that under this map we get an exact sequence

$\displaystyle 1\rightarrow \left(\frac{(\mathcal{O}_K/\frak{m}_f)^*\times \prod_{j=1}^i\{\pm 1\}}{\text{im} \Delta}\right)\rightarrow Cl_\frak{m}(K)\rightarrow Cl(\mathcal{O}_K)\rightarrow 1,$

where the diagonal map ${\Delta: \mathcal{O}_K^* \rightarrow (\mathcal{O}_K/\frak{m}_f)^*\times\prod_{j=1}^i\{\pm 1\}}$ is given by ${\Delta(u)=([u], \text{sgn}(w_i(u)))}$.

This might seem scary, but let’s go back to our example of ${Cl_m(\mathbb{Q})}$. Something fishy was going on with having to pick the positive generator, and in general when we have lots more units floating around the above tells us exactly how that is affecting everything. For one thing we have ${Cl(\mathbb{Z})=1}$, and so this theorem tells us we have an isomorphism from the big quotient to ${Cl_m(\mathbb{Q})}$.

Our units are just ${\{\pm 1\}}$, and our embedding is ${w_1:\mathbb{Q}\hookrightarrow \mathbb{R}}$. Thus our diagonal map is ${1\mapsto ([1], 1)}$ and ${-1\mapsto ([-1], -1)}$. Thus ${\displaystyle Cl_m(\mathbb{Q})\simeq \frac{(\mathbb{Z}/m\mathbb{Z})^*\times\{\pm 1\}}{\text{im}(\Delta)}\simeq (\mathbb{Z}/m\mathbb{Z})^*}$ which is what we got last time.

What this shows is that if ${\frak{m}}$ doesn’t involve the real place, i.e. no embedding is part of the conductor, the map given last time is not well-defined because the two generators of the same fractional ideal ${\frac{a}{b}}$ and ${-\frac{a}{b}}$ must get mapped to the same place. But this exact sequence tells us exactly what gets changed, and we have to mod out by ${\{\pm 1\}}$ (the image of the diagonal) so that they really do go to the same place and hence we get a smaller ray class group ${Cl_\frak{m}(\mathbb{Q})\simeq \frac{(\mathbb{Z}/m\mathbb{Z})^*}{\{\pm 1\}}}$.

Now let’s prove exactness of that sequence in general. We’ll start with surjectivity of the right map. Let ${[\frak{b}]\in Cl(\mathcal{O}_K)}$. We want to find some fractional ideal ${\frak{a}}$ relatively prime to ${\frak{m}}$ such that ${[\frak{a}]=[\frak{b}]}$. We can use the same approximation theorem trick as in this post (see it for more details). We have ${\frak{b}=\prod \frak{p}_i^{e_i}}$, but for any prime appearing that also appears in ${\frak{m}}$ choose some ${a_i}$ with ${ord_{p_i}(a_i)=-e_i}$. Multiplying these together we get some element ${a}$ and ${\frak{a}=\frak{b}\cdot (a)}$ is altered by a principal and hence in the same ideal class, but is by construction relatively prime to ${\frak{m}}$.

Just by definition the kernel of that map is exactly the principal ideals that are relatively prime to ${\frak{m}}$ modulo the principal ideals with the conductor condition, i.e. ${\displaystyle \frac{Prin(\mathcal{O}_K)\cap I_\frak{m}(K)}{Prin_\frak{m}(K)}}$. We’ll establish an isomorphism between this and the first term of the sequence.

Suppose ${\mathcal{O}_K\cdot \beta}$ is a class in this group, so that ${\beta}$ is relatively prime to ${\frak{m}}$. Thus our map will be ${\beta\mapsto \overline{\beta}:=((\beta \mod \frak{p}^{ord_{\frak{p}}(\frak{m}_f)}\mathcal{O}_K), \text{sgn}(w_j(\beta)))}$ in

$\displaystyle \prod_{\frak{p}|\frak{m}_f} \mathcal{O}_{K, \frak{p}}^*/(1+\frak{p}^{ord_{\frak{p}}(\frak{m}_f)}\mathcal{O}_K)^*\times\prod_{j=1}^i\{\pm 1\}$

Now it is not clear this is well-defined because there is ambiguity in the choice of generator ${\beta}$. Thus we must check that ${\mathcal{O}_K \beta=\mathcal{O}_K \beta'}$ (as classes mod the principal ideals generated by elements congruent to ${1\mod \frak{m}}$!) if and only if ${\beta=u\beta'}$ for some ${u\in \mathcal{O}_K^*}$.

Claim: The ideal ${\mathcal{O}_K\beta}$ is in ${Prin_{\frak{m}}(K)}$ if and only if ${\overline{\beta}\in \text{im}(\Delta)}$. If ${\overline{\beta}\in \text{im}(\Delta)}$, then let ${u}$ a unit such that ${\Delta(u)=\overline{\beta}}$. Take ${\beta'=u\beta}$, and you get that ${\beta'\equiv 1\mod \frak{m}}$, thus ${\mathcal{O}_K\beta=\mathcal{O}_K\beta'}$ which is in ${Prin_{\frak{m}}(K)}$.

Conversely, if ${\mathcal{O}_K\beta}$ has the property that ${\mathcal{O}_K\beta=\mathcal{O}_K\beta'}$ where ${\beta'\equiv 1\mod \frak{m}}$, then ${\beta=u\beta'}$ for some unit. But now by definition ${\overline{\beta'}}$ is just ${1}$. Thus ${\overline{\beta}=\Delta(u)}$. This proves the claim.

The proof of this claim actually shows that the map descends to an isomorphism and hence the sequence is exact by identifying the appropriate groups using the Chinese remainder theorem. That seems like enough for today.