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I was reading the proof of the following fact from Silverman's Advanced Topics in the Arithmetic of Elliptic Curves (p. 122).

The Hilbert class field of a quadratic imaginary field $K$ with ring of integers $\mathcal{O}_K$ is $K(j(E))$ for any elliptic curve $E$ with complex multiplication by $\mathcal{O}_K$.

I'll briefly describe his proof to get to the point where I don't follow his argument. He shows that there is a homomorphism $F : \mathrm{Gal}(\bar{K}/K) \to Cl(\mathcal{O}_K)$ defined by $E^{\sigma} = F(\sigma) \cdot E$. The kernel of this homomorphism is the subgroup $\mathrm{Gal}(\bar{K}/K(j(E)))$. So $F$ factors as an injective map $F : \mathrm{Gal}(K(j(E))) \to Cl(\mathcal{O}_K)$. Henceforth I'll write $L$ for $K(j(E))$. We assume that the conductor of $L/K$ is $\mathfrak{c}_{L/K}$. Silverman wants to show that the conductor is $(1)$. So that will prove that $L$ is a subfield of the Hilbert class field, after which some further work will give the reverse inclusion and we'll be done. In order to show that $\mathfrak{c}_{L/K} = (1)$, Silverman goes as follows. The composition of the Artin symbol with $F$, $${I}_{\mathfrak{c}_{L/K}} \xrightarrow{(\cdot , L/K)} \mathrm{Gal}(L/K) \xrightarrow{F} Cl(\mathcal{O}_K)$$ is simply the projection. In particular, since $F$ is injective, $((\alpha), L/K) = 1$ for any $\alpha \in K^{\times}$, $\alpha$ coprime to $\mathfrak{c}_{L/K}$. From here (quoting Silverman),

But the conductor of $L/K$ is the smallest integral ideal $\mathfrak{c}$ with the property that $$\alpha \equiv 1 \ (\textrm{mod }\mathfrak{c}) \implies ((\alpha), L/K) = 1.$$ It follows that $\mathfrak{c}_{L/K} = (1)$.

It is here that I can't follow his argument. Firstly, I think he means "largest integral ideal". But that's a minor issue. To be precise, I don't understand why having all principal ideals coprime to $\mathfrak{c}_{L/K}$ in the kernel of the Artin symbol should imply that $\mathfrak{c}_{L/K} = (1)$. I know that in general the kernel of the Artin symbol is $(N_{L/K}I_L \cdot {P}_{\mathfrak{c}_{L/K}}) \cap {I}_{\mathfrak{c}_{L/K}}$ where $I_L$ is the group of fractional ideals of $L$ and ${P}_{\mathfrak{c}_{L/K}}$ is the group of ideals $(\alpha), \alpha \in K^{\times}$ such that $\alpha \equiv 1 \textrm{ mod }\mathfrak{c}_{L/K}$.

Let me clarify that I don't have any background in class field theory. I only know the main statements of CFT as outlined for example in $\textrm{Ch. II}, \S 3$ of Silverman's ATAEC, or this handout by Bjorn Poonen. For me the definition of Hilbert class field is that it is the maximal unramified abelian extension or equivalently the ray class field modulo $(1)$.

As an aside, I got to know from this answer of Franz Lemmermeyer that there are at least 3 equivalent definitions of Hilbert class field. One of them is:

Let $H/K$ be a field extension. It is called a Weber-Hilbert class field if the prime ideals of $K$ that split completely in $H$ are precisely the principal prime ideals.

So, in order that an abelian extension $L$ of $K$ be a subfield of the Hilbert class field of $K$ it is necessary that all principal primes split completely in $L$. In our situation we only know that the principal prime ideals which are coprime to the supposed conductor $\mathrm{c}_{L/K}$ split completely in $L$. So our observations do not discard the possibility that there might be principal primes in $K$ which are ramified in $L$.

I would very much appreciate if someone could spell out in more details Silverman's argument. Thanks in advance!

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  • $\begingroup$ I don't think there is any argument, Silverman is just using the (deep) theorems of class field theory, but he is also proving that $K(j(E))/K$ is an abelian extension with Galois group $Cl(K)$ and that the Artin map factors through $Cl(K)$ for almost every prime. Your main question is how to prove in an elementary way that $K(j(E))/K$ is unramified. It makes sense because historically class field theory was created to generalize the results obtained from CM elliptic curves. $\endgroup$ – reuns Jan 4 '20 at 16:18
  • $\begingroup$ Yes, precisely. As you said, he is "proving that $K(j(E))/K$ is an abelian extension with Galois group $Cl(K)$" but in order to show this he has to show that the Artin map actually factors through $Cl(K)$ for all prime, instead of almost every, which is where his argument seems opaque to me. $\endgroup$ – standard reduction Jan 4 '20 at 16:25
  • $\begingroup$ To prove that $Gal(K(j(E))/K) = Cl(K)$ there is no need of the Artin map, only to show that $\sigma(j(E)) = j(\sigma(E))= j(E^{c^{-1}})$ where $c\in Cl(K)$ is represented by an ideal $(a,b)$ of $O_K=End(E)$ and $E^{c^{-1}} = E/(\ker(a)\cap \ker(b))$ and that the map $\sigma \to c$ is a group homomorphism $\endgroup$ – reuns Jan 4 '20 at 16:30
  • $\begingroup$ I'm sorry, as I understand his proof, he does need the Artin map to prove that the Galois group is $Cl(K)$. $F : Gal(K(j(E))) \to Cl(K)$ is injective trivially. But to show surjectivity he tries to show that $F \circ (\cdot, K(j(E))/K)$ is the projection. Then once we know that $K(j(E))/K$ is unramified, we conclude $F$ is surjective and $K(j(E))$ is a subfield of the Hilbert class field and hence equal to it, since $F$ is now an isomorphism. $\endgroup$ – standard reduction Jan 4 '20 at 16:37
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I think I have figured out a way to circumvent the problem, without deviating too much from Silverman's exposition. In case my question was not clear, let me state it again. As mentioned in the comments above by user reuns, Silverman proves that $\mathrm{Gal}(L/K) \simeq Cl(K)$ and the Artin map factors through $Cl(K)$. In order to prove that the Galois group is the class group Silverman tries to show that the map $F : \mathrm{Gal}(L/K) \to Cl(K)$ is surjective, by instead showing that $$ {I}_{\mathfrak{c}_{L/K}} \xrightarrow{(\cdot, L/K)} \mathrm{Gal}(L/K) \xrightarrow{F} Cl(K) $$ is surjective. The way he shows this composition is surjective is by first showing that it is simply the projection. Then he somehow concludes that this implies the conductor is $(1)$, which shows that $I_{\mathfrak{c}_{L/K}}$ is the group of fractional ideals in $K$, which in turn shows that the composition is surjective. Since I was not understanding how he concluded the conductor is $(1)$, I wanted to avoid this route. This can be avoided by using Dirichlet's theorem on primes in arithmetic progression. So that proves that $F$ is indeed an isomorphism. Now to prove that $\mathfrak{c}_{L/K} = (1)$ we note the following.

Let $\mathbf{A}^*_K$ be the idele group of $K$. For an arbitrary modulus $\mathfrak{m} = \prod {v}^{e_{v}}$, let $\mathbf{A}^{\mathfrak{m}}_K$ be the group of ideles $(\alpha_v)$ such that $v(\alpha_v - 1) \geq e_v$ and $\alpha_v \in \mathcal{O}_{K,v}^{\times}$ whenever $e_v > 0$. Let $U_{\mathfrak{m}}$ be the group of ideles $(\alpha_v)$ such that $v(\alpha_v - 1) \geq e_v$ whenever $e_v > 0$ and $\alpha_v \in \mathcal{O}_{K,v}^{\times}$ for all $e_v = 0$. For any modulus $\mathfrak{m}$, we have $\mathbf{A}^{\mathfrak{m}}_K/U_{\mathfrak{m}}K^{\times} \simeq I_{\mathfrak{m}}/P_{\mathfrak{m}}$. In particular, $\mathbf{A}^*_K/U_{(1)}K^{\times} \simeq Cl(K)$. Denoting the reciprocity map on the ideles by $[\cdot, K]$, we have a diagram as follows. $\require{AMScd}$ \begin{CD} \mathbf{A}^{\mathfrak{c}_{L/K}}_K @>>> \mathbf{A}^*_K @>{[\cdot, K]}>> \mathrm{Gal}(K^{\mathrm{ab}}/K) \\ @VVV @VVV @VVV \\ I_{\mathfrak{c}_{L/K}} @>>> Cl(K) @>{F^{-1}}>> \mathrm{Gal}(L/K) \end{CD}

A priori we do not know that the right small square is commutative and this is exactly what we need to show $L$ is the Hilbert class field. But we do know that the outer rectangle and the left small square are commutative. This along with the fact that $\mathbf{A}^*_K = \mathbf{A}^{\mathfrak{c}_{L/K}}_K K^{\times}$ gives the required commutativity of the small right square. Thus $L$ is the Hilbert class field of $K$.

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