@David, here I really mean isomorphic. Equal would be "easy" since one may compute the relation ideal for $B$ and $C$ and then test for equality.

@Yves, yes indeed you are right. So may be you (or Yiftach) could write the proof. Is there a natural generalization of this result? It seems to me that the notion of profinite $p$-group is really peculiar and does not seem to generalize easily...

@Yiftach, I think I understand how to do it when the groups are finite. So let $K=<x_1,\ldots,x_r>$ and let $\{n_1,\ldots,n_s\}$ be the generators of the $\mathbf{Z}_p[[K]]$-module. Then by what you said one has that the normal closure of $<x_i*n_j: i,j>$ which I denote by $NC<x_i*n_j: i,j>$ generated $N$. Now if $<x_i*n_j: i,j>$ was contained in a maximal subgroup $M$ of $N$ then because $N$ is a $p$-group we know that $M$ is normal and thus we would have $NC<x_i*n_j: i,j>\subseteq M$ which is a contradiction.

Hi @Chandan, I was wondering if you had a reference for the following statement that you made: If $\Delta(\Omega/K')$ is a finitely generated $Z_p[[\Gamma(K'/K)]]$-module then you may find finitely many elements of $\Gamma(\Omega/K')$ such that the normal closure of these elements generate it.

Yes of course you are right!

Ok I was being stupid, ma section is not well defined.

@Kevin, I agree that the terminology "universal elliptic curve" (away from $0$ and $1728$) confusing but I could not come up with a better terminology.

@Misha, so what you told me in the comments look good, I like this idea of using an explicit positive line bundle. So concerning your addendum keep in mind that not all finite index subgroups are congruence subgroups. So I don't quite understand why your $X_d(N)$ is projective...

@Donu, note that my non-trivial section $s:P^1(\mathbf{C})\rightarrow Y$ for the Hopf surface $Y=(C^2-(0,0))/<(2,2)>$ is given by $s([x_0,x_1])\mapsto [2x_0,2x_1]$.

Well I don't think you can always extend it in the analytic category when the base has a non-trivial fundamental group.

@Donu, I'm a bit puzzle by what you said about the existence of a section. Take $Y=(\mathbf{C}^2-(0,0))/<(2,2)>$, then it seems to me that the map $s:\mathbf{P}^1\rightarrow Y$ given by $s(x)=(2,2)$ gives you section of your elliptic fibration. In this case, all fibers are isomorphic to the elliptic curve $\mathbf{C}^{\times}/<2>$.