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.

) and (*) are not compiling, even though they did above. Does anyone know how to fix this? $\endgroup$ – Jeff Yelton Nov 26 '15 at 23:43any 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 $q$. $\endgroup$ – Horace the grumpy bear Nov 30 '15 at 18:00up to conjugationin $\mathrm{GL}_2(\mathbf{Z}_{\ell})$ by the power of $\ell$ dividing $v_p(j_E)$. $\endgroup$ – Horace the grumpy bear Nov 30 '15 at 18:03