First of all everything i'm asking about comes from the beginning of Katz and Mazur's book : Arithmetic moduli of elliptic curves (which you can find here). I'm considering an elliptic curve $f : E \to S$ where $S$ is a scheme (i.e. $E$ is a proper smooth $S$-group scheme of relative dimension $1$, with a chosen section $0 : S \to E$).

I'm a trying to understand what KM are doing on page 68/69. Since $E \to S$ is separated we know that $D := 0(S)$ is a closed subscheme, it is in fact proven in KM that $D$ is a relative effective cartier divisor i.e. $D \to S$ is flat and $I(D)$ is an invertible sheaf. We write $\omega_{E/S} = 0^*\Omega^1_{E/S} = f_* \Omega^1_{E/S}$. This is a line bundle so Zariski locally on $S$ it is generated by a global section $\omega$.

Now they say

1) Zariski locally on $S$, the formal completion of $E$ along $0$ i.e. $Spf(\varprojlim \mathcal{O}_E/I(D)^n)$ is isomorphic to $Spf(R[[T]])$ where $S = Spec(R)$.

This in turn, implies (they say) that

2) Zariski locally on $S$, $\omega$ can be written in the form $f(T)dT$ where $f(T) \in R[[T]]$.

How does one prove that first fact (I get that for open affine $U = Spec(A)$ containing $0(D)$ we have $A = R \oplus I(D)$ where $I(D)$ is the free ideal of rank $1$ corresponding to $D$, and that by iterating this we get a morphism $A \to R[[T]]$ but i don't know how to go from there).

For the second I don't even understand what they mean by the formal expansion of $\omega$.

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    $\begingroup$ My outlook is unredeemably old-fashioned. Over a field $k$, at least, the differentials are a $1$-dimensional vector space over the function field of your elliptic curve. Those that are invariant are a one-dimensional vector space over $k$. Take a local uniformizing parameter $T$ at the neutral point; then there’s a unique invariant differential writable in the form $(1+a_1T+a_2T^2+\cdots)dT$, where what’s in parentheses is the formal expansion of a suitable element of the function field. Somehow, I fear that this won’t be a bit of help to you. $\endgroup$ – Lubin Sep 27 '15 at 2:34
  • $\begingroup$ First of all thank you very much for your comment. Could you maybe explain to me how one obtain the formal expansion that you speak about. I think that it might be a similar argument that the one in the more general case of a non-field base. $\endgroup$ – J. Doe Sep 27 '15 at 6:47

Here’s how to do it for an elliptic curve over $\Bbb Q$:

Start with $y^2=x^3+ax+b$, with the neutral point $\Bbb O$ up at infinity. So you homogenize and then set $y=1$ to get $z=x^3+axz^2+bz^3$. I like to think of this as a recursive schema for expanding $z$ as a series in $x$, but you can do it in various ways, to get $$ z = x^3 + ax^7 + bx^9 + 2a^2x^{11}+\cdots $$ Next, the differential: from $z=x^3+axz^2+bz^3$, you get $dz=3x^2dx+az^2dx + 2axz\,dz+3bz^2dz$. Collect the $dx$’s and the $dz$’s together, and divide suitably to get $$ \frac{dz}{3x^2+az^2}=\frac{dx}{1-2axz-3bz^2}\,. $$ You can persuade yourself that this differential has no zeros (nor poles, of course): it’s the invariant differential I referred to in my comment. Now use your expansion of $z$ as a series in $x$, to get $$ \omega=1 + 2ax^4 + 3bx^6 + 6a^2x^8 + 20abx^{10} + (15b^2 + 20a^3)x^{12}+\cdots $$ Notice that if you take $x$ to have weight $-1$ and $z$ weight $-3$, every monomial has weight zero here, as it ought to.

Finally, since I am who I am, I have to observe that if you integrate this thing formally, you get the logarithm of the formal group of your elliptic curve.


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