The Main Theorem of Complex Multiplication

Today we’ll state the Main Theorem of Complex Multiplication. We have our standing assumptions. Let {K} be a quadratic imaginary field and {R_K} the ring of integers. For a prime {\frak{p}} of {K} define {K_\frak{p}} to be the completion at {\frak{p}} and {R_\frak{p}} the ring of integers. For a fractional ideal {\frak{a}} of {K} define {\frak{a}_\frak{p}} to be the fractional ideal {\frak{a}R_\frak{p}}.

We will need a few facts about modules over {R_K}. We know {R_K} is a Dedekind domain, so these will easily follow from the sequence of posts about the structure of modules over Dedekind domains. We will denote the {\frak{p}}-primary part of an {R_K}-module, {M}, to be {M[\frak{p}^\infty]}. This is just the set of elements of the module annihilated by some power of {\frak{p}}.

Back in our posts on this topic we only talked about finitely generated modules, but some things are true in more generality. For example, if {M} is a torsion {R_K}-module, then the natural map {\displaystyle \bigoplus_{\frak{p}}M[\frak{p}^\infty]\rightarrow M} is an isomorphism. This is essentially the part of the theorem that says if we attempt to decompose a module into a projective and torsion part, the torsion part has a nice primary decomposition. We also will need that for any fractional {\frak{a}} the natural map {(K/\frak{a})[\frak{p}^\infty]\rightarrow R_\frak{p}/\frak{a}_p} is an isomorphism. This comes from the inclusion {K\hookrightarrow K_\frak{p}}.

Putting the above to facts together gives us that {\displaystyle K/\frak{a}\simeq \bigoplus_{\frak{p}}K_\frak{p}/\frak{a}_\frak{p}}. Now let {x\in \mathcal{J}_K} be an idele. The fractional ideal generated by this is {\displaystyle (x)=\prod_\frak{p} \frak{p}^{ord_p(x_p)}}. Given any fractional ideal {\frak{a}} of {K} we define {x\frak{a}:=(x)\frak{a}}. Since {\displaystyle K/x\frak{a}\simeq \bigoplus_\frak{p} K_p/x_p\frak{a}_p} we can define multiplication by {x} on all of {K/\frak{a}\rightarrow K/x\frak{a}} piece by piece on the {\frak{p}}-primary parts {(t_\frak{p})\mapsto (x_\frak{p}t_\frak{p})}.

Let {E/\mathbb{C}} be an elliptic curve with CM by {R_K}. Fix some {\sigma\in Aut(\mathbb{C})}. We were calling our surjection from idele class field theory {\lambda: \mathcal{J}_K\rightarrow Gal(K^{ab}/K)}. Choose some {s\in \mathcal{J}_K} with the property that {\lambda(s)=\sigma|_{K^{ab}}}. Let {\frak{a}} be a fractional ideal and fix a complex analytic isomorphism {f:\mathbb{C}/\frak{a}\stackrel{\sim}{\rightarrow} E(\mathbb{C})}.

The Main Theorem of Complex Multiplication tells us that with this setup there is a unique isomorphism {f':\mathbb{C}/s^{-1}\frak{a}\stackrel{\sim}{\rightarrow} E^\sigma(\mathbb{C})} which makes the following diagram commute:

\displaystyle \begin{matrix} \mathbb{C}/\frak{a} & \stackrel{s^{-1}}{\rightarrow} & \mathbb{C}/s^{-1}\frak{a} \\ \downarrow & & \downarrow \\ E(\mathbb{C}) & \stackrel{\sigma}{\rightarrow} & E^\sigma(\mathbb{C}) \end{matrix}

This gives us the following simple way to translate between the analytic action of multiplying by {s^{-1}} and the algebraic action of {\sigma} by {f(t)^{\lambda(s)}=f'(s^{-1}t)}. The proof involves several reductions to a manageable case, and then a lot of work. As usual, the statements are all we are really interested in right now. The next two posts will be some applications of this, so we might prove a few more things there.


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))}.

Some Fields of Definition Results

Suppose {E/\mathbb{C}} is an elliptic curve. We will need some standard facts. First, if {\sigma: \mathbb{C}\rightarrow \mathbb{C}} is a field automorphism, then {E^\sigma} will denote the elliptic curve formed by acting on the coefficients of a Weierstrass equation by {\sigma}. We have an isomorphism {End(E)\stackrel{\sim}{\rightarrow} End(E^\sigma)} by {f\mapsto f^\sigma}. If {E} has CM by the full ring of integers {R_K} in a quadratic imaginary {K}, then {j(E)\in \overline{\mathbb{Q}}}. This follows by exploiting the fact that {j(E^\sigma)=j(E)^\sigma} to argue that {j(E)} lies in a finite extension of {\mathbb{Q}}.

The natural map {Ell_{\overline{\mathbb{Q}}}(R_K)\rightarrow Ell(R_K)} from isomorphism classes of curves over {\overline{\mathbb{Q}}} with CM by {R_K} to isomorphism classes over {\mathbb{C}} is an isomorphism. This follows easily from the fact that {j(E)\in\overline{\mathbb{Q}}}. It turns out that if we use our normalized isomorphism {[\cdot]: R\rightarrow End(E)}, we get the nice relation {[\alpha]_E^\sigma=[\alpha^\sigma]_{E^\sigma}} for all {\alpha\in R} and {\sigma\in Aut(\mathbb{C})}.

If {E} is defined over {L} and has CM by {R} in {K}, then every endomorphism of {E} is defined over {LK}. It is useful to see why this is. Let {\sigma\in Aut(\mathbb{C})} fix {L}. Then by definition {E^\sigma=E}, so using the normalized {[\cdot]} relation we get that {[\alpha]^\sigma=[\alpha^\sigma]} for all {\alpha\in R}. Thus if {\sigma} also fixes {K}, then {[\alpha]} will be fixed by {\sigma} and hence by Galois descent every endomorphism descends to {LK}.

We can use this result to show that if {E} has CM by the full ring of integers {R_K}, then {[\mathbb{Q}(j(E)): \mathbb{Q}]\leq h_K} the class number of {K}. As an example application, without knowing the Weierstrass equation for our curve we can argue that the unique curve with CM by {\mathbb{Z}[i]} must be defined over {\mathbb{Q}}. This is because {h_K=1}, so the degree of the extension is {1} and hence {j(E)\in\mathbb{Q}}. Of course, we already knew that because we said last time it is given by {y^2=x^3+x}. Likewise, we could make the same argument for the curve with CM by {\mathbb{Z}[\rho]} where {\rho=e^{2\pi i/3}}.

Now we get to our first real relation between CM and class field theory. Suppose {E/\mathbb{C}} has CM by {R_K}. The field {L=K(j(E), E_{tors})} generated by the {j}-invariant and the torsion points is an abelian extension of {K(j(E))}. The proof uses our field of definition result above about the endomorphisms. If {H=K(j(E))}, then we can reduce to showing that each {L_m=H(E[m])} is abelian over {H}. Using the standard Galois representation, we see that {Gal(L_m/H)} injects into {GL_2(\mathbb{Z}/m)}.

Since we have CM floating around we know every endomorphism is already defined over {H}. Using the fact that the image of {Gal(L_m/H)} commutes with {R_K}, we get that {Gal(L_m/H)} injects into {Aut_{R_k/mR_k}(E[m])\simeq (R_K/mR_K)^*} an abelian group and hence is abelian. This result about being abelian I think is an exercise in Silverman’s first book.

Fix a quadratic imaginary field {K}. Let’s return to the idea that we have a simply transitive action of {Cl(R_K)} on {Ell(R_K)} by {\frak{a}\cdot E_{\Lambda}=E_{\frak{a}^{-1}\Lambda}}. We also have a nice action of {Gal(\overline{K}/K)} on {Ell(R_K)} given by {\sigma\cdot E=E^{\sigma}}. Fix some {E\in Ell(R_K)}. This gives us a map {F: Gal(\overline{K}/K)\rightarrow Cl(R_K)} by sending {\sigma} to the unique element {\frak{a}} with the property that {\frak{a}\cdot E=E^\sigma}. It turns out that {F} is a group homomorphism that is independent of the choice of {E}. This is actually not so easy to prove, and the underlying philosophy of why it is hard is that the action of {Cl(R_K)} is analytic in nature since it acts on the lattice {\Lambda\subset \mathbb{C}}, whereas the Galois action is purely algebraic. To prove the independence of {E} requires translating between the two. Although it is sort of a fun proof, we’ll skip it in order to get to the meat of CM theory more quickly.

Elliptic Curves over C

Today we’ll start exploring the fascinating topic of complex multiplication. I’ll assume familiarity with basic elliptic curve theory. For today we’ll review some of the basic theory when our elliptic curve {E} is defined over {\mathbb{C}}. In this case, recall that an elliptic curve has complex multiplication by {R} if {End(E)\simeq R} is an order in a quadratic imaginary field {End(E)\otimes \mathbb{Q}}. A quick reminder about the terminology. Recall that {E\simeq \mathbb{C}/\Lambda} for some lattice {\Lambda}. Thus if we take a complex number {\alpha} with the property that {\alpha\Lambda\subset \Lambda}, then “multiplication by {\alpha}” descends to a well-defined map {E\rightarrow E} and hence we get the term “complex multiplication” by {R}.

Define a normalized elliptic curve to be a pair {(E, [\cdot ])} where {[\cdot]: R\stackrel{\sim}{\rightarrow} End(E)} is the isomorphism preserving the invariant differential, i.e. {[\alpha]^*\omega=\alpha \omega}. Now let {K=R\otimes \mathbb{Q}} and let {R_K} be the ring of integers of {K}. One important thing to be able to do is classify the elliptic curves with complex multiplication by {R_K}. Given any non-zero ideal {\frak{a}\subset R_K} we have by our theory of projective modules over Dedekind rings that {\frak{a}} is a lattice in {\mathbb{C}} and hence {E_\frak{a}=\mathbb{C}/\frak{a}} is an elliptic curve.

Moreover, using the fact that {\frak{a}} is an ideal, we get that {E_\frak{a}} has complex multiplication by {R_K}. Clearly altering {\frak{a}} to {c\frak{a}} by a principal ideal will give a homothetic lattice and hence an isomorphic curve which leads to the natural guess that elements of the class group {Cl(R_K)} might give the different isomorphism classes of curves with CM by {R_K}. Let’s denote the set of isomorphism classes of elliptic curves with CM by {R_K} as {Ell(R_K)}. It is easily verified that the natural action {Cl(R_K)\times Ell(R_K)\rightarrow Ell(R_K)} by {\frak{a}\cdot E_\Lambda = E_{\frak{a}^{-1}\Lambda}} is well-defined and simply transitive.

Thus the number of curves over {\mathbb{C}} with CM by {R_K} is the class number of {R_K}. As an example we could ask how many elliptic curves have complex multiplication by the Gaussian integers {\mathbb{Z}[i]= \mathcal{O}_{\mathbb{Q}(i)}}. Since the class number is {1}, there is a unique elliptic curve over {\mathbb{C}} with complex multiplication by {\mathbb{Z}[i]}. This curve is well-known to have Weierstrass equation {y^2=x^3-x}, but from our above theory is also constructed as {\mathbb{C}/(1\cdot\mathbb{Z}+i\cdot\mathbb{Z})}.

Recall that {E[n]} is the group of {n}-torsion points. Given {\frak{a}} an ideal of {R_K} we could also form {E[\frak{a}]=\{P : [\alpha]P=0 \ \forall \alpha\in \frak{a}\}} with respect to the normalized {[\cdot ]}. Now since for any lattice {\Lambda} we have {\Lambda\subset \frak{a}^{-1}\Lambda} we get an isogenty {E_\Lambda\rightarrow \frak{a}\cdot E_{\Lambda}}. One can check that the kernel of this isogeny is exactly {E[\frak{a}]} and that {E[\frak{a}]} is a free {R_K/\frak{a}}-module of rank {1} (less easy).

This allows us to compute that the degree of the isogeny {E\rightarrow \frak{a}\cdot E} is {N_{K/\mathbb{Q}}(\frak{a})} and for any {\alpha\in \frak{a}} the map multiplication by {\alpha} is an isogeny of degree {|N_{K/\mathbb{Q}}(\alpha)|}. This allows us to recover the classical fact that multiplication by {n} is a degree {n^2} isogeny (even an exercise in Hartshorne IV). One interesting thing about CM is that there are endomorphisms that have degree other than a square. Going back to our example we see that multiplication by {1+i} on {y^2=x^3-x} has degree {2} since {N(1+i)=1^2+1^2=2}.

This was a short review and introduction. Next time we’ll move onto some more subtle issues that arise when we don’t work over {\mathbb{C}}.

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.