Good evening,

I have some questions about methods of proving using power series of the invariant differential in Silverman's book Arithmetic of Elliptic Curves (direct link below). On page 118 (Chap. IV) we are told that the invariant differential $\omega(z) = \frac{dx(z)}{2y(z) + a_1 x(z) +a_3} $ has the power series $$ \omega(z) = (1 + a_1z + ({a_1}^2 + a_2)z^2 + ({a_1}^3 + 2a_1a_2 + 2a_3)z^3 + \cdots) dz \, . $$ This is not clear to me.

  1. How can the denominator $ 2y +a_1x +a_3 $ be successively canceled?

The setting is introduced at the page 115: We analyse the Weierstrass equation around $O$ using the variable transformation $(x,y) \mapsto (z,w)$ so that $z = \frac{-x}{y}$, $w = \frac{-1}{y}$ and show that $w$ can be expressed as a formal series depending on $z$. It's also clear to me how to get the formal series of $x(z)$ and $y(z)$: simply divide the Ws equation $w = f(z,w)$ by $z^2 w$ and solve for $z/w$ .

Second question: Subsequently it is shown that $w(z)$ has coefficients from $ \mathbb{Z}[a_1,..., a_6] $ by using the conclusion (?) that $w(z)$ has coefficients in $\mathbb{Z}[\frac{1}{2},a_1,..., a_6][[z]] $ as well as in $\mathbb{Z}[\frac{1}{3},a_1,..., a_6][[z]]$. This is shown by the fact that $w(z)$ has two representations: $$ \omega(z)/dz = \frac{dx(z) / dz}{2y + a_1 x +a_3} = \frac{-2z^{-3} + \cdots}{-2z^{-3} + \cdots} \in \mathbb{Z}[{1/2},a_1,\ldots, a_6][[z]] \, , $$ as well as $$ \omega(z)/dz = \frac{dx(z) / dz}{3x^2 + 2a_2 x +a_4 -a_1y} = \frac{-3z^{-4} + \cdots}{-3z^{-4} + \cdots} \in \mathbb{Z}[{1/3},a_1, \ldots, a_6][[z]] \, . $$ Then the author compares the numerators and denominators. My question is:

  1. Why $ \omega(z)/dz $ is in $\mathbb{Z}[\frac{1}{2},a_1,..., a_6][[z]]$ respectively $\mathbb{Z}[\frac{1}{2},a_1,..., a_6][[z]]$ included, and why does this show that any denominator is simultaneously a power of $2$ and $3$?

Images of the relevant pages:

The both problem places...




It has to do partly with what the expansions of $x$ and $y$ look like, as Laurent series in $z$, the local uniformizing parameter at the point $\Bbb O$ at infinity. the situation is that $$ x=z^{-2}+z^{-1}\cdot(\text{power series in $z$ with coeffs involving the $a$’s})\,, $$ so that similarly $$ dx=\bigl[-2z^{-3}+z^{-2}\cdot(\text{another such power series in $z$})\bigr]dz\,. $$ Now, $2y+a_1x+a_3$ also has a Laurent-series expansion in $z$ starting out $-2z^3$, again with the other coefficients being polynomials in the $a$’s. Once you form the quotient, you may multiply top and bottom of the fraction by $-z^3/2$, to get something that looks like $$ \omega=\frac{1+zP(z)}{1+zQ(z)}dz\,, $$ where $P$ and $Q$ are power series in $z$ with coefficients in $\Bbb Z[a_1,\cdots,a_6,\frac12]$. Since the bottom of the fraction has $1$ for its constant term, it has a reciprocal in $\bigl(Z[a_1,\cdots,a_6,\frac12]\bigr)[[z]]$, and there you are.

I hope this has helped.


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