# On Iwasawa theory of elliptic curves in $\mathrm{PGL}_2(\mathbb{Z}_p)$-extension

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 $$K/\mathbb{Q}$$?
• 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. Nov 22, 2023 at 14:25
• @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 $y^2=x^3-x$.) Nov 24, 2023 at 7:46
• ... 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. Nov 24, 2023 at 7:49

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.)

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