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I am reading Katz and Mazur's book "Artihmetic moduli of elliptic curves" (available here) and I have some questions about the construction of the generalized Weierstraß equation for a family of elliptic curves (see section 2.2, pages 67 to 69 in the book, 39-40 in the pdf). My question will be a little long and I aplogize for it. Any partial help or clarification is welcome.

I am aware of the establishment of such an equation in the case of an elliptic curve over an algebraically closed field, as done in Silverman's book "The arithmetic of elliptic curves". Katz & Mazur's generalization is somewhat similar, but I do not understand the details of it.

I will not copy all the details of the construction because it would be pointless in my opinion (see the above mentioned pages). Here are however the key points:

Let $S$ be an arbitrary scheme and $E$ an elliptic curve over $S$ (that is a proper smooth curve over $S$ with geometrically connected fibers all of genus $1$ and given with a section $"0"$). We denote by $f$ the morphism $E\rightarrow S$.

  1. Serre-Grothendieck duality shows that $\underline{\omega}_{E/S}:=f_{\star}\Omega^1_{E/S}$ is an invertible sheaf on $S$, of formation compatible with base change.
  2. Zariski locally on $S$, one may choose an $\mathcal{O}_S$-basis $\omega$ of $\underline{\omega}_{E/S}$, that is a nowhere vanishing form on $E/S$, ie an isomorphism $\mathcal{O}\rightarrow \Omega^1_{E/S}$. Such an $\omega$ is translation invariant.
  3. Zariski locally on $S$, the formal completion $\hat{E}$ of $E$ along its zero-section $"0"$ is of the form $$\hat{E}=\operatorname{Spf}(A[[T]]),\quad S=\operatorname{Spec}(A)$$ Such an isomorphism is called a parametrization at zero, noted simply $T$. We may choose it so that the formal expansion of $\omega$ has the form $$(1+\text{higher terms})dT$$
  4. For any integer $n\geq 1$, the invertible sheaf $I(0)$ (defining the closed embedding $S\rightarrow E$ defined by the section $"0"$) satisfies that $f_{\star}(I^{-n}(0))$ is locally free of rank $n$ on $S$. Locally on $S$ with a fixed formal parametrization adapted to $\omega$, we see that $f_{\star}(I^{-2}(0))$ is free on $1,x$ where $$x=\frac{1}{T^2}(1+\text{higher terms})$$ and $f_{\star}(I^{-3}(0))$ is free on $1,x,y$ where $$y=\frac{1}{T^3}(1+\text{higher terms})$$ Then, looking at $f_{\star}(I^{-6}(0))$, we see that $y^2-x^3$ actually lies in $f_{\star}(I^{-5}(0))$, which is free on $1,x,y,x^2,xy$. From here, we obtain the desired Weierstraß equation.

Here are my questions about the above construction.

First, in point 2, how can one see a basis for $\underline{\omega}_{E/S}$ as an isomorphism $\mathcal{O} \rightarrow \Omega^1_{E/S}$? Actually, what should be the subscript of the $\mathcal{O}$ here? I am afraid I do not understand the nature of $\omega$ correctly.

Second, how do we know that the formal completion of $E$ along its zero section is locally given by the formal scheme associated to power series with one indeterminate?

Third, I do not understand clearly the meaning of an equality of the type $\omega = (1+\text{higher terms})dT$. How do we obtain such a writing adapted to $\omega$?

Fourth, why is it important to have a choice of a formal parameter $T$ that is adapted to $\omega$? To what extent does it change the discussion of point 4, when writing down the formal expansions of the functions $x$ and $y$?

Five, why is there no term $\frac{1}{T}$ in the formal extensions of $x$ and of $y$?

Sixth, how can one justify that $f_{\star}(I^{-n}(0))$ is locally free of rank $n$ on $S$?

With all of this, as you can guess, I am particularly confused by the role of $\Omega^1_{E/S}$ and of formal series expansions in this construction. These do not appear at all in the rather straightforward arguments used in Silverman's book.

Any contribution/clarification/reference for one of these points would be gladly appreciated. It thank you very much for your help.

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    $\begingroup$ For your $6$-th question one can use Riemann Roch and cohomology and base change (i.e. you prove it over an algebraically closed field and then it follows from general nonsense). For your second question this is probably because $E$ is smooth of relative dimenson one. I must admit I dont quite understand the arguments with the formal parameter $T$ here, I am not sure why it it necessary. $\endgroup$ – user45878 Jul 16 '18 at 7:05

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