This is less of a direct question and more of an argument that I've been worried about for a while and want to check (apologies for the length and if my writing is unclear).

Suppose I have an elliptic curve $E / \mathbb{Q}$ given by an equation of the form $y^{2} = x(x - 1)(x - \lambda)$ with $\lambda \in \mathbb{Q}$. For any rational prime $p$, let $v_{p}$ denote the $p$-adic valuation on $\mathbb{Q}$. Let $k_{p}$ denote the fraction field of the strict Henselization of $\mathbb{Z}_{(p)}$; we fix an embedding into $\bar{\mathbb{Q}}$.

Suppose that $m := v_{p}(\lambda) > 0$ and $m' := v_{p}(\lambda') > 0$ for (distinct) rational primes $p$ and $p'$. Now the theory of Tate curves tells us that if $p \neq 2$ or $m \geq 5$, then the $\ell$-adic image of $\mathrm{Gal}(\bar{\mathbb{Q}} / k_{p}(\mu_{\ell})$ is cyclically generated by a power of a transvection in $\mathrm{SL}(T_{\ell}(E))$. More precisely, if $p \neq 2$ (resp. if $p = 2$ and $m \geq 5$), it is generated by the $s := \ell^{v_{\ell}(2m)}$-th power (resp. the $s := \ell^{v_{\ell}(2m - 8)}$-th power) of the transvection with respect to an element $a \in T_{\ell}(E)$. Of course, the analogous statement for $p'$ also holds (call the analogous element $a' \in T_{\ell}(E)$ and power $s'$). This follows from the fact that if $j$ is the $j$-invariant of $E$, then $v_{p}(j) = -2m$ (resp. $v_{p}(j) = -2m + 8$) if $p \neq 2$ (resp. if $p = 2$), and analogously for $p'$. (See Exercise 5.13(b) of Silverman's Advanced Topics, or \SA.1.5 of Serre's Abelian \ell-adic representations.)

Let's assume (*) that $\{a, a'\}$ are always independent in $T_{\ell}(E)$. In fact, I've proven this to be the case when $\ell = 2$ and $p, p' \neq 2$, and believe it's probably true in general. Let's also assume (**) that if $\tilde{p} \neq p'$ is another prime such that $v_{\tilde{p}}(\lambda - 1) > 0$, then the $\ell$-adic images of $\mathrm{Gal}(\bar{\mathrm{Q}} / k_{p'}(\mu_{\ell}))$ and $\mathrm{Gal}(\bar{\mathrm{Q}} / k_{\tilde{p}}(\mu_{\ell}))$ are generated by powers of the same transvection.

Then, by considering $k_{p}, k_{p'}$ as subextensions of $\bar{\mathbb{Q}} / \mathbb{Q}$, we see that the $\ell$-adic image of the absolute Galois group of $\mathbb{Q}$ contains the powers of transvections given above. Since $a$ and $a'$ are independent, it's not hard to show that these two powers of transvections generate a subgroup $G$ of $\mathrm{SL}(T_{\ell}(E))$ containing the principal congruence subgroup $\Gamma(\ell^{s + s'})$; it's also easy to see by reducing mod $\ell^{s + s'}$ that it can't contain any larger principal congruence subgroup. Since $\mathbb{Q}(\mu_{\ell})$ is the intersection of the extensions $k_{p}(\mu_{\ell})$ for all primes $p$ and we only need to consider primes of bad reduction by Neron-Ogg-Shafarevich, it follows from our assumption (**) that the $\ell$-adic Galois image of $\mathrm{Gal}(\bar{\mathrm{Q}} / \mathrm{Q}(\mu_{\ell}))$ actually coincides with $G$.

Now choose primes $\ell$ and $p \neq \ell, 2$ and consider the infinite subset of fibers of the Legendre family given by $\lambda = p^{\ell^{n}}$ for $n = 1, 2, 3, ...$. By the above arguments, for each $n$, the $\ell$-adic Galois image will contain $\Gamma(\ell^{s + s'})$ but no larger principal congruence subgroup, where $s = n$ and $s' > 0$ depends on $n$. Thus, there is no $N$ such that the image of Galois contains $\Gamma(\ell^{N})$ for all fibers of the Legendre family.

But this contradicts well-known results (of Serre and others) on the uniform boundedness of the indices of $\ell$-adic Galois images of non-CM elliptic curves over a fixed number field. The only assumptions I made were (*) and (**). In the $\ell = 2$ case, like I said, I've proven (*), and so ( **) must be false in general. Or did I make some other unjustified assumption in my argument? I would greatly appreciate any insight on this.

EDIT: Okay, thanks to Horace's answer below some things have become clearer to me, in particular that $\bigcap k_{p}(\mu_{\ell})$ doesn't coincide with $\mathbb{Q}(\mu_{\ell})$ as I had assumed before. Therefore, the $\ell$-adic images of $I_{p} := \mathrm{Gal}(\bar{\mathbb{Q}} / k_{p}(\mu_{\ell}))$ do not necessarily generate the whole image of the absolute Galois group of $\mathbb{Q}(\mu_{\ell})$. Thus, a priori the boundedness results of Serre and others aren't relevant here.

Also, Horace points out that my assumption (** ) isn't really sensible as stated, given that one has to choose a decomposition group above $p$ in order to consider $k_{p}$ as a subfield of $\bar{\mathbb{Q}}$. So suppose we amend it to claim that if $q \neq p'$ is another prime such that $v_{q}(\lambda - 1) > 0$, then the $\ell$-adic image $\rho_{\ell}(I_{q})$ is conjugate via an element of $\rho_{\ell}(I_{p})$ to a subgroup $H$ such that $H$ and $\rho_{\ell}(I_{p'})$ are each generated by powers of the same transvection. (By the way, note that we expect the images of the inertia groups to be generated by powers of transvections, not necessarily transvections themselves, so this kind of statement has nothing to do with whether $v_{q}(j)$ has higher $\ell$-adic valuation than $v_{p'}(j)$, etc.) In other words, the claim is that the subgroup of $T_{\ell}(E)$ generated by $\rho_{\ell}(I_{p})$, $\rho_{\ell}(I_{p'})$, and $\rho_{\ell}(I_{q})$ is the same as the subgroup generated by $\rho_{\ell}(I_{p})$ and $H'$ for some cyclic subgroup $H' \geq \rho_{\ell}(I_{p'})$. I believe I proved this directly for $\ell = 2$ and $p, p', q \neq 2$ by using explicit formulas for $2$-power torsion, but I'm not completely sure since I never wrote it down. Anyway, now it seems this new version of assumption (**) might be true, since it wouldn't contradict Serre's boundedness results. I'm still interested to find out when it is.

Also, thank you Horace for correcting my oversight in not including the prime $\ell$ for Neron-Ogg-Shafarevich.

  • $\begingroup$ In the final paragraph, for some reason my () and (*) are not compiling, even though they did above. Does anyone know how to fix this? $\endgroup$ Nov 26, 2015 at 23:43
  • $\begingroup$ I've forced it to compile with an extra space. Perhaps someone can fix it properly. $\endgroup$
    – Myshkin
    Nov 27, 2015 at 0:01
  • $\begingroup$ Firstly, this question is answered, so if you have a new question, you should ask a new question. But secondly... $\endgroup$ Nov 30, 2015 at 17:53
  • $\begingroup$ I can't believe that you are not thinking about these subgroups up to conjugation, as if you completely ignored my main point. Assume, for example, that $E[\ell]$ gives rise to a surjective Galois representation. Then the three transvections $A$, $B$, and $C$ (such that suitable powers generate the image of inertia at $p$, $p'$ and $q$) can be taken to be any transvections in $\mathrm{GL}_2(\mathbf{Z}_{\ell})$ that you like. So, for some choice, you could take inertia at $p$ and $p'$ to be given by powers of the same transvection, and something completely different at $q$. $\endgroup$ Nov 30, 2015 at 18:00
  • $\begingroup$ The image of inertia at such a prime $p$ is completely determined up to conjugation in $\mathrm{GL}_2(\mathbf{Z}_{\ell})$ by the power of $\ell$ dividing $v_p(j_E)$. $\endgroup$ Nov 30, 2015 at 18:03

1 Answer 1


$\def\Q{\mathbf{Q}}$ $\def\Qbar{\overline{\Q}}$ $\def\Z{\mathbf{Z}}$ $\def\GL{\mathrm{GL}}$

There are many confusions in your post, let me point out just a few

It is dangerous to fix embeddings of $\Qbar$ into $\Qbar_p$ (equivalently, to fix decomposition groups $D_p$ inside the Galois group $G_{\Q}$ for all $p$), and leads you astray on a number of occasions. You say:

Since $\Q(\zeta_{\ell})$ is the intersection of the extensions ... we need only consider primes of bad reduction by Néron-Ogg-Shafarevich, it follows ... that the $\ell$-adic image ... coincides with $G$.

There are at least three things wrong with this paragraph. First, by Néron-Ogg-Shafarevich, one deduces that the ramification on the $\ell$-adic Tate module is unramified outside the primes of bad reduction and the prime $\ell$, which you omitted. Second, a Galois group need not be generated by the decomposition groups at primes of bad reduction; for example, this does not hold for an extension which is unramified everywhere (or which contains any such extension). You can overcome both of these issues by including the prime $\ell$ and working over $\Q$ which has no unramified extensions by Minkowski. However, you still run into trouble, because you fixed a choice of $D_p$, and for this sort of argument you need to consider all conjugates of $D_p$ as well. For example, let $L/\Q$ be the splitting field of

$$x^3 - x - 1 = 0.$$

The Galois group $G$ is $S_3$, and $G$ is ramified only at $23$. If you fix a decomposition group $D \subset G$ at a prime above $p = 23$, you will find that $D = \Z/2\Z$, which doesn't generate $G$. Correspondingly, the fixed field $K = L^{D}$ (which is isomorphic to the cubic field cut out by a root of the equation above) is unramified at one prime above $23$ but ramified at the other prime above $23$.

This leads into the second issue when fixing $D_p$. It's completely non-canonical, and yet both of your assumptions (*) and (**) depend completely on this choice. Let's assume that the $\ell$-adic representation:

$$\rho: G_{\Q} \rightarrow \GL_2(\Z_{\ell})$$

was surjective (it often is). Let's also suppose that $E$ has multiplicative reduction at $p$, and that valuation of $j_E$ is $-1$. Then what does one know about the image of inertia $I_p \subset D_p$ inside the $\ell$-adic Tate module? Well, it will contain a transvection conjugate to

$$\left(\begin{matrix} 1 & u \\ 0 & 1 \end{matrix} \right)$$

for some unit $u \in \Z^{\times}_{\ell}$. However, choosing a different embedding $D_p \subset G_{\Q}$ is equivalent to conjugating this element. And every single transvection $\sigma \in \GL_2(\Z_{\ell})$ which is non-trivial modulo $\ell$ is conjugate to this element. So for example, given another prime $q$ with ordinary reduction and such that $j_E$ has valuation $-1$, the corresponding statement holds. But now * and ** clearly depend on the choice you make. You can choose $D_p$ and $D_{q}$ inside $G_{\Q}$ so that you get exactly the same transvection, or you can choose them such that together they generate $\GL_2(\Z_{\ell})$. So it doesn't make sense to ask whether * or ** holds. This is the main source of your error.

Note that (**) will be false in a stronger sense. You could ask whether the transvection at $q$ is generated by a power of an element in the conjugacy class in $\GL_2(\Z_{\ell})$ of the transvection at $p$. Yet the conjugacy class will be determined by the $\ell$-adic valuation of $v_{p}(j_E)$. So this amounts to saying that $v_q(j_E)$ has the same or higher $\ell$-adic valuation as $v_p(j_E)$. And this is unlikely to be the case, especially of one is choosing $\lambda$ specifically so that $v_p(j_E)$ is highly divisible by powers of $\ell$.

Here is a modification of your thoughts which makes sense. Suppose $E$ is semi-stable at $N$, and we want to force the $\ell$-adic image to be small. One way is to make the $\ell$-adic image small by showing that $E[\ell^m]$ is not ramified at any of the primes dividing $N$. It's hard to eliminate ramification at $\ell$, but if $\ell$ is a prime of good reduction, then $E[\ell^m]$, although not ramified, will have some nice properties (it comes from the generic fibre of a finite flat group scheme). As a consequence of the proof of Serre's conjecture, we have:

Theorem Consider a finite flat group scheme $A$ over $\mathrm{Spec}(\Z)$ of type $(\ell^m,\ell^m)$ for some prime $\ell > 2$ such that the corresponding representation has cyclotomic determinant. Then $A \simeq \Z/\ell^m \oplus \mu_{\ell^m}$.

So we will be able to get a contradiction to Serre's open image theorem (and Mazur's theorem) simply by ensuring that $E[\ell^m]$ is unramified at all primes above $N$. For each prime $p$ dividing $N$, to say that $D_p$ (and all its conjugates) acts trivially is equivalent to asking that $v_p(j_E)$ is divisible by $\ell^n$. So we get the desired contradiction by choosing a semi-stable elliptic curve such that $v_p(j_E) = - v_p(\Delta_E)$ is divisible by $\ell^n$ for all primes $p$. To write down such an elliptic curve, simply write

$$y^2 = x (x - a^{\ell^n})(x + b^{\ell^n}).$$

And now you get a contradiction to Serre's theorem as long as $a^{\ell^n} + b^{\ell^n} = c^{\ell^n}$ for some $c$ with $(abc,\ell) = 1$, and one can take $n$ sufficiently large. Even better, Using Mazur's theorem, one gets a contradiction in mathematics even for $n = 1$ as long as $\ell > 11$. So simply write down a solution to $a^{\ell} + b^{\ell} = c^{\ell}$ for integers $(abc,\ell) = 1$ and $\ell > 11$ and we will have found a contradiction in mathematics and can all go home and eat some pie. Good luck!

Your cubic polynomial example helped me to realize that I made one basic and crucial error in assuming that $\cap K_p = \Q$ ... That confusion seems to be the source of most of the other issues you point out.

Hmm. I'm going to push back a little here. First, I think there are a number of different errors, and the main issue is not considering that the choice of $D_p$ is very non-canonical. The questions (*) and (**) depend completely on the choice of embeddings, and it is an error to think of them as sensible hypotheses. And also ... the statement above happens to be true! (Edit: I was assuming that $K_p$ to be the fixed field of a decomposition group $D_p$, not the inertia group $I_p$.) If you take a finite extension with Galois group $G$, then the subgroup $H$ generated by $D_p$ even for all sufficiently large primes will be $G$. This is because the subgroup $H$, by Ceboratev, will contain an element from each conjugacy class of $G$, and such subgroups have to be the entire group by a theorem of Jordan. However, it is false to say that the subgroup $H$ generated by the decomposition groups of the ramified primes generates $G$; what you can deduce is that if $N$ is the normal closure of $H$ in $G$ then the extension corresponding to $G/H$ is unramified everywhere.

  • $\begingroup$ Thank you for your thorough response. Your cubic polynomial example helped me to realize that I made one basic and crucial error in assuming that $\bigcap k_{p} = \mathbb{Q}$ (whence my assumption that $\bigcap k_{p}(\mu_{\ell}) = \mathbb{Q}(\mu_{\ell})$). Instead, I suppose $\bigcap k_{p}$ is some weird non-Galois extension of $\mathbb{Q}$ where over each prime p there lies some prime which is unramified, although its largest Galois subextension is $\mathbb{Q}$. That confusion seems to be the source of most of the other issues you point out. More points to follow. $\endgroup$ Nov 28, 2015 at 15:13
  • $\begingroup$ (Also, just to be clear in case my original question gave the wrong impression, my goal was never to "find a contradiction in mathematics", but rather to pinpoint whether or not my assumption (**) must be false in general, or whether there was a misconception buried somewhere in my argument. Guess it was mainly the latter.) $\endgroup$ Nov 28, 2015 at 15:19
  • $\begingroup$ "If you take a finite extension with Galois group G, then the subgroup H generated by Dp even for all sufficiently large primes will be G..." Sure, but I don't understand why you're considering decomposition groups here when the Galois group over each $k_{p}$ is an inertia group (note each $k_{p}$ is strictly Henselian). And it seems that your cubic polynomial example shows that the Galois groups above the $k_{p}$'s don't generate all of $G$, yes? $\endgroup$ Nov 29, 2015 at 0:46
  • $\begingroup$ I was thinking $K_p$ was the fixed field of (some) $D_p$. Never in my wildest dreams did I think that one could sink to such levels of non-canonical depravity to take $K_p$ to be the fixed field of (some) $I_p$. $\endgroup$ Nov 29, 2015 at 18:21
  • $\begingroup$ Assumptions * and ** are not sensible questions in general, because they depend on the choice of $D_p$ (or $I_p$), and the answer may (and does) change depending on that choice. This, I think, was the main misconception. $\endgroup$ Nov 29, 2015 at 18:21

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