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Let $E$ be an elliptic curve over the rationals $\mathbb{Q}$. We consider the Galois representation attached to $E$ by acting on its $p$-adic Tate module $T_p(E)$, $$ \rho_E: G_{K} \rightarrow \mathrm{Aut}(T_p(E)) \cong \mathrm{GL}_2(\mathbb{Z}_p). $$ Then it cuts out a field $M := \overline{\mathbb{Q}}^{\ker(\rho_E)}$, whose Galois group over $\mathbb{Q}$ is isomorphic to the image of $\rho_E$. A theorem of Serre tells us that if $E$ has no CM, then $\mathrm{im}(\rho_E)$ is a finite index subgroup of $\mathrm{GL}_2(\mathbb{Z}_p)$ and strictly equals to $\mathrm{GL}_2(\mathbb{Z}_p)$ for all but finitely many primes $p$. So for simplicity, we take sufficiently large odd prime $p$ to guarantee $\rho_E$ has full image.

Further $\widetilde{\rho_E}: G_{K} \rightarrow \mathrm{PGL}_2(\mathbb{Z}_p)$. Then similarly consider $K := \overline{\mathbb{Q}}^{\ker(\widetilde{\rho_E})}$. We assume $E$ has no CM and $p$ sufficiently large so that $\mathrm{Gal}(K/\mathbb{Q}) \cong \mathrm{PGL}_2(\mathbb{Z}_p)$.

Ultimate Goal: The group of torsion points $E(K)_{\mathrm{tors}}$ is a finite group.

A crucial step in proving this (from my naive thought) is to ensure that

Claim 1: there exist infinitely many primes "good primes" $q$ such that the residue field for any prime of $K$ lying above $q$ is finite.

This post is asking how to prove (or disprove) this boldface claim 1?


My "attempt" is to imitate the proof of the following claim in the anticyclotomic $\mathbb{Z}_p$-extension case.

Claim 2: Let $F$ be an imaginary quadratic field and $F_{\infty}^{\mathrm{ac}}$ is its anticyclotomic $\mathbb{Z}_p$-extension. Then for any prime of $F$ which is inert in $F/\mathbb{Q}$ must split completely in $F_{\infty}^{\mathrm{ac}}/F$.

Then since there are infinitely many inert primes in $F$ (by density arguments), we have infinitely many primes that "has finite residue field in $F_{\infty}^{\mathrm{ac}}$".

The proof of Claim 2 may date back to Iwasawa in the second section of his article "On the $\mu$-invariants of $\mathbb{Z}_{\ell}$-extensions (1973)". A sketch of proof goes like this:

Let $q$ be a prime coprime to $p$ that is inert in $F$, write $\mathfrak{q}$ for the unique prime of $F$ lying above $q$, then $\mathfrak{q}$ is unramified in the $n$-th layer $F_{n}^{\mathrm{ac}}$. Let $\mathfrak{q}_n$ be a prime of $F_{n}^{\mathrm{ac}}$ lying above $\mathfrak{q}$ and $Z_n$ be the decomposition group of it for the Galois extension $F_{n}^{\mathrm{ac}}/\mathbb{Q}$. Then since $q$ is unramified in $F_{n}^{\mathrm{ac}}$ and is inert in $F$, $Z_n$ is a cyclic group of $G_n := \mathrm{Gal}(F_{n}^{\mathrm{ac}}/\mathbb{Q})$ such that $G_n = Z_n H_n$, where $H_n := \mathrm{Gal}(F_{n}^{\mathrm{ac}}/F)$. As $G_n$ is a dihedral group of order $2p^n$, it follows that $Z_n$ is a cyclic group of order two satisfying $Z_n \cap H_n = 1$. However, $Z_n \cap H_n$ is nothing but the decomposition group of $\mathfrak{q}_n$ for the extension $F_{n}^{\mathrm{ac}}/F$, hence $\mathfrak{q}$ splits completely in $F_{n}^{\mathrm{ac}}$.

Then to imitate the proof above, we need to know the parallel facts on general $\mathrm{PGL}_2(\mathbb{Z}_p)$-extensions $K/\mathbb{Q}$, for example:

  1. Specify all intermediate fields of the $\mathrm{PGL}_2(\mathbb{Z}_p)$-extension $K/\mathbb{Q}$. This is equivalent to classifying all closed subgroups of $\mathrm{PGL}_2(\mathbb{Z}_p)$ and figuring out which of them are open.

Maybe we can try some congruence subgroups? Let $\Gamma_n$ be the group of matrices in $\mathrm{GL}_2(\mathbb{Z}_p)$ that are congruence to identity matrix modulo $p^n$, and let $\overline{\Gamma_n}$ be its image of $\mathrm{PGL}_2(\mathbb{Z}_p)$. Do these groups work and are these all such subgroups? (I got hints from here.)

  1. The properties of the Galois groups of intermediate fields are good for us to run a similar argument as the anticyclotomic case. In this process, we hope to get a sufficient condition $(\star)$ for such "good primes" (that satisfying Claim 1).

Follow Greenberg's hint, maybe a common property for finite dihedral groups and $\mathrm{PGL}_2(\mathbb{Z}_p)$ is that $g$ and $g^{-1}$ are conjugate for any group element $g$. Unfortunately, I cannot see how this property is used in the above proof of anticyclotomic case either.

  1. There are infinitely many primes satisfying the sufficient condition $(\star)$ in 2.

Following Greenberg's hint, I guess that the primes $q$ such that $E$ has good supersingular reduction satisfies the abstract sufficient condition $(\star)$. Then a result of N. Elkies shows that there are infinitely many good supersingular primes. Yet I cannot see how such good supersingular properties are used in the abstract discussion of $\mathrm{PGL}_2(\mathbb{Z}_p)$-extensions. Since I am not able to do 2, figuring out 3 is impossible.

So though there is a rough three-step roadmap, I got stuck on each step. So I am here to ask if there is any way out.


Some further remarks:

  • This entire problem is motivated by solving Exercise 1.12 of Ralph Greenberg's IAS/Park City note "Introduction to Iwasawa Theory for Elliptic Curves". The hints above by Greenberg is taken from here.
  • The Claim 2 seems to work for anticyclotomic $\mathbb{Z}_p$-extension $\mathcal{F}_{\infty}^{\mathrm{ac}}$ for any CM field $\mathcal{F}/\mathcal{F}^{+}$.
  • Even if all the problems in this post are solved, we still cannot get the full proof of Greenberg's Exercise 1.12 that $E(K)_{\mathrm{tors}}$ is finite since we haven't touched the CM elliptic curves. An exercise in Silverman's book tells us that in this case, $\rho_E$ has abelian image in $\mathrm{GL}_2(\mathbb{Z}_p)$, and hence $\widetilde{\rho_E}$ has abelian image in $\mathrm{PGL}_2(\mathbb{Z}_p)$. Then in this CM elliptic curve case, it seems that we need a classification of abelian subgroups of $\mathrm{PGL}_2(\mathbb{Z}_p)$ as "Step 0" and try to run the argument above.

Or am I so stupid that missed some easy solution to this exercise? Actually, I feel like I am quite good at making things unnecessarily complicated and as a result, obtaining nothing valuable during my Ph.D. study up to now. Quite frustrated. :(

So sorry for such a long post, and thank you all for commenting and answering! :)


EDIT: Even Further Remarks: I found Greenberg's Exercise 1.16 was focusing on the CM case.

  • In Greenberg's Exercise 1.16, he considered a particular elliptic curve $y^2 = x^3 - x$ with CM $\mathbb{Z}[\sqrt{-1}]$. Let $F=\mathbb{Q}(\sqrt{-1})$. Then he asked the reader to show $K$ contains $F_{\infty}^{\mathrm{ac}}$ and $[K:F_{\infty}^{\mathrm{ac}}] < \infty$. This example (though I am still trying to prove) may inspire us to show that maybe (I am not sure at all) $K$ contains an anticyclotomic extension of finite index and then use the Claim 2 directly to conclude?
  • When searching for references, I found arXiv: 2008.04960, where the authors said in the second paragraph of the introduction that the $\mathrm{PGL}(2)$-extension does not contain the cyclotomic extension. This (though still I am trying to prove it) may further provide some evidence on that certain "anticyclotomic line" appears in $K/\mathbb{Q}$?
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    $\begingroup$ There exists no $\mathbf{Z}_p$ extension (cyclotomic or anticyclotomic or whatever) contained in a $PGL_2$ extension, just by elementary group theory: $\mathbf{Z}_p$ is not a quotient of $PGL_2(\mathbf{Z}_p)$. This is precisely why the PGL2 extensions are studied: because they are the first obvious example of p-adic Lie extensions which are "essentially" nonabelian, having no nontrivial abelian extensions inside them. $\endgroup$ Nov 22, 2023 at 14:25
  • $\begingroup$ @DavidLoeffler Thank you so much for your comment and sorry for the late reply. I managed to see the group theory fact you mentioned. Then inspired by your comment, I realized that I mixed up the CM case and the non-CM case, which seems to be completely different. In the CM case, the image of $\rho_E$ is an abelian subgroup of $\mathrm{PGL}_2(\mathbb{Z}_p)$, which could be quite small, and as Greenberg pointed out, it contains a $F_{\infty}^{\mathrm{ac}}$ "of finite index". (I have somehow managed to show this after struggling for two days, at least for $y^2=x^3-x$.) $\endgroup$
    – Hetong Xu
    Nov 24, 2023 at 7:46
  • $\begingroup$ ... Yet for the non-CM case, the image of $\rho_E$ is (except for finitely many prime $p$) the entire PGL2, which makes the story completely different. I hope I got this right this time, though still I have no idea how to get the ultimate goal proved. $\endgroup$
    – Hetong Xu
    Nov 24, 2023 at 7:49

1 Answer 1

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It seems that I have obtained some positive results, but I couldn't believe my "solution" is correct since I haven't completely followed Greenberg's hints.

Throughout, $p$ is an odd prime.

Let's focus on the non-CM case only, and suppose $p$ is sufficiently large so that $\tilde{\rho}_E$ has full image $\mathrm{PGL}_2(\mathbb{Z}_p)$. For "small primes", the image is of finite index in $\mathrm{PGL}_2(\mathbb{Z}_p)$, so it doesn't harm.

Recall that we have the tower of extensions $$ \mathbb{Q} \subseteq K := \overline{\mathbb{Q}}^{\ker \tilde{\rho}_E} \subseteq M := \overline{\mathbb{Q}}^{\ker \rho_E} = \mathbb{Q}(E[p^{\infty}]). $$ Our goal is to understand $K/\mathbb{Q}$. To imitate the anticyclotomic $\mathbb{Z}_p$-case, we observe that there is a short exact sequence of abstract groups $$ 0 \rightarrow \mathrm{PSL}_2(\mathbb{Z}_p) \rightarrow \mathrm{PGL}_2(\mathbb{Z}_p) \xrightarrow{\det} \mathbb{Z}_p^{\times}/(\mathbb{Z}_p^{\times})^{2} = \{1, -1\} \rightarrow 0, $$ and $\mathrm{PGL}_2(\mathbb{Z}_p) = \mathrm{PSL}_2(\mathbb{Z}_p) \rtimes C_2$, where $C_2$ is a cyclic group of order two. So we have a quadratic extension $F/\mathbb{Q}$ such that $\mathrm{Gal}(K/F) \cong \mathrm{PSL}_2(\mathbb{Z}_p)$. Actually from the construction we see that $F$ is nothing but the fixed field of $\overline{\mathbb{Q}}$ by $\ker(\det \circ \tilde{\rho}_E)$.

In the $\mathrm{PSL}_2(\mathbb{Z}_p)$-extension $K/F$, we have many finite subextensions. We consider the quotient $\mathrm{PSL}_2(\mathbb{Z}/p^n)$ of $\mathrm{PSL}_2(\mathbb{Z}_p)$. It gives a finite subextension of $F_n$ of $K/F$ such that $$ \mathrm{Gal}(F_n/K) \cong \mathrm{PSL}_2(\mathbb{Z}/p^n), \quad \mathrm{Gal}(F_n/\mathbb{Q}) \cong \mathrm{PSL}_2(\mathbb{Z}/p^n) \rtimes C_2 \cong \mathrm{PGL}_2(\mathbb{Z}/p^n). \quad (\ast) $$

Pity 1: I still have no idea about the classification of closed/open subgroups of $\mathrm{PSL}_2(\mathbb{Z}_p)$ or $\mathrm{PGL}_2(\mathbb{Z}_p)$. But knowing some of these is sufficient to get this exercise done.

Then the story is similar to the anticyclotomic $\mathbb{Z}_p$-extension. But still, an important property need to be established:

"Property $(\dagger)$": Let $\mathfrak{q}$ be a prime of $F$ not lying above $p$, then $\mathfrak{q}$ is unramified in the $\mathrm{PSL}_2(\mathbb{Z}_p)$-extension $K/F$.

To imitate the $\mathbb{Z}_p$-extension case, it seems that we still need to know the closed subgroups of $\mathrm{PSL}_2(\mathbb{Z}_p)$. But here note that the extension $K/F$ arises from the elliptic curve $E$. We use:

Property (Neron-Ogg-Shafarevich): with notations above, $\mathfrak{q}$ is unramified in $F(E[p^{\infty}])$ if (and only if) $E$ has good reduction at $\mathfrak{q}$.

Then since $K \subseteq M=\mathbb{Q}(E[p^{\infty}]) \subseteq F(E[p^{\infty}])$, we see that at least for prime $\mathfrak{q}$ where $E/F$ has good reduction, $\mathfrak{q}$ is unramified in $K$.

Pity 2: I cannot show the "Property $(\dagger)$" for abstract $\mathrm{PSL}_2(\mathbb{Z}_p)$-extension of number fields.

Now let $q$ be a rational prime that is

  • not equal to $p$, odd
  • inert in $F/\mathbb{Q}$,
  • $E$ has good (not necessarily supersingular) reduction at $q$.

then by [Silverman, AEC, Proposition 5.4(a)], for any prime $\mathfrak{q}$ of $F$ over $q$, $E$ has good reduction at $\mathfrak{q}$, and hence $\mathfrak{q}$ is unramified in $K$. Obviously there are infinitely many such rational primes $q$.

Then we are finally ready to run the proof of Claim 2 quoted in my question, replacing the intermediate fields "$F_{n}^{\mathrm{ac}}$" there by $F_n$ here. The similar structure of Galois groups in $(\ast)$ helps us to conclude.


Edit: I was a little bit cheating here. Let me write some details here.

Let $\mathfrak{q}_n/\mathfrak{q}/q$ be the primes in the tower $F_n/F/\mathbb{Q}$. Then since $\mathfrak{q}/q$ is inert and $\mathfrak{q}_n/\mathfrak{q}$ is unramified, the decomposition group $Z(\mathfrak{q}_n/q)$ is cyclic of even order. Then by counting $\# \mathrm{PSL}_2(\mathbb{p}^n) = p^{3n-2} (p^2-1)/2$ (citing here), we see that $Z(\mathfrak{q}_n/\mathfrak{q})$ is of order at most $(p^2-1)/2$. (Here we used the assumption that $p$ is odd). Hence the ramification index $e_n$ and $f_n$ satisfies $e_n f_n \leq (p^2-1)/2$, which is bounded independent of $n$. This implies that the residue field of $q$ in $K$ is finite.

Remark: So here is still a little bit difference between the dihedral group case and the $\mathrm{PGL}_2$ case.


I hope that I haven't made mistakes in the proof. Here I would like to add a proof of Prof. David Loeffler's hint that $\mathrm{PGL}_2(\mathbb{Z}_p)$ has no quotient isomorphic to $\mathbb{Z}_p$.

We assume $p \geq 5$. For primes $p=2$ and $p=3$, actually I am not quite certain.

Proof: Suppose $\mathrm{PGL}_2(\mathbb{Z}_p) \twoheadrightarrow \mathbb{Z}_p$, then quotienting out $p\mathbb{Z}_p$, we obtain $\mathrm{PGL}_2(\mathbb{F}_p) \twoheadrightarrow \mathbb{F}_p$. This means (by counting the cardinality of the two groups $\mathrm{PGL}_2(\mathbb{F}_p)$ and $\mathrm{PGL}_2(\mathbb{F}_p)$) that we would have a normal subgroup $H$ of $\mathrm{PGL}_2(\mathbb{F}_p)$ of order $p^2-1$. We show next that this would not happen.

The key is the fact:

Fact of Galois: When $p \geq 5$, $\mathrm{PSL}_2(\mathbb{F}_p)$ is a simple group. (See K. Conrad's note here.)

This is the only place in the proof that we use the assumption $p \geq 5$. Suppose $H$ is a normal subgroup of $\mathrm{PGL}_2(\mathbb{F}_p)$, then by the simplicity result of Galois, $H \cap \mathrm{PSL}_2(\mathbb{F}_p)$ would be either trivial or the entire $\mathrm{PSL}_2(\mathbb{F}_p)$.

  • If $H \cap \mathrm{PSL}_2(\mathbb{F}_p) = \mathrm{PSL}_2(\mathbb{F}_p)$. Then $H \supseteq \mathrm{PSL}_2(\mathbb{F}_p)$. As $\mathrm{PSL}_2(\mathbb{F}_p)$ is of index two in $\mathrm{PGL}_2(\mathbb{F}_p)$, we see that either $H=\mathrm{PSL}_2(\mathbb{F}_p)$ or $H=\mathrm{PGL}_2(\mathbb{F}_p)$.
  • If $H \cap \mathrm{PSL}_2(\mathbb{F}_p) = 1$, then $H \cdot \mathrm{PSL}_2(\mathbb{F}_p) = \mathrm{PGL}_2(\mathbb{F}_p)$. By the second isomorphism theorem, $$ \mathrm{PGL}_2(\mathbb{F}_p)/H \cong H \cdot \mathrm{PSL}_2(\mathbb{F}_p) / H \cong \mathrm{PSL}_2(\mathbb{F}_p) / H \cap \mathrm{PSL}_2(\mathbb{F}_p) \cong \mathrm{PSL}_2(\mathbb{F}_p). $$ So $H$ is a normal subgroup of order $2$, strictly smaller than $p^2-1$.

So we are done.

I doubt if it is correct for $p=2$. For $p=3$, maybe it follows form a more direct computation since $\mathrm{PSL}_2(\mathbb{F}_3)$ is just the alternating group $A_4$.(Again here.)

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  • $\begingroup$ It is very hard to classify open subgroups of $PGL_2$. In the final part, it should also be true for $p=2$. $\endgroup$
    – stupid boy
    Nov 25, 2023 at 19:21

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