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Now I'm trying (5.3.2) of Katz-Mazur's Arithmetic moduli of elliptic curves.

Let $k$ be an algebraically closed field of characteristic $p > 0$, $E_0$ a supersingular elliptic curve over $k$, and $E/W[[T]]$ the universal deformation of $E$. (Where $W$ is the Witt vector ring with the residue field $k$.)
Let $A$ be a finite $W[[T]]$-algebra which is local, $\mathfrak{E} = E \times_{W[[T]]} A$, $P, Q \in \mathfrak{E}(A)$ be a Drinfeld $p^n$-basis (i.e., a pair of points such that $\mathfrak{E}[p^n] = \sum_{(a,b) \in (\mathbb{Z}/p^n)^2} [aP + bQ]$ as Cartier divisors), and $X$ be a parameter of the formal group of $\mathfrak{E}$.
Then $P, Q$ both lie in the formal group of $\mathfrak{E}$.
So let $f = X(P), g = X(Q) \in \mathfrak{m}_A$. Then for an artinian local ring $R$ with the residue field $k$ and for a map $A \to R$ with $f, g \mapsto 0$, we have $P \times_A R = Q \times_A R = 0$.

Does the word "$P$ lies in the formal group" mean that $P : \operatorname{Spec}A \to \mathfrak{E}$ induces $\operatorname{Spf}A \to \hat{\mathfrak{E}}$? And if so, how can I show it?

I know the Grothendieck existence theorems. (In particular $\operatorname{Hom}(X, Y) \cong \operatorname{Hom}(\hat{X}, \hat{Y})$ over a complete noetherian local ring.)
But $\hat{\mathfrak{E}}$ is the formal completion along the $0$-section, not along $\mathfrak{m}_A \mathscr{O}_{\mathfrak{E}}$. So we cannot apply it.

And I know that since $E_0$ is supersingular, $P \times k = Q \times k = 0 \in E_0(k)$.

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  • $\begingroup$ Where is the quote taken from? What is a Drinfeld $p^n$-basis? $\endgroup$ Commented May 24, 2020 at 14:54
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    $\begingroup$ The formal completion on the zero section should equal the formal completion on the maximal ideal since the base ring $A$ is already complete. $\endgroup$
    – Will Sawin
    Commented May 24, 2020 at 17:24
  • $\begingroup$ @WillSawin Thank you very much! Would you explain it in a little more detail? I think that in order to show that the completion along the 0 section and one along $\mathfrak{m}_A$ are the same, we need these two closed subsets are the same. And now this fails. $\endgroup$
    – k.j.
    Commented May 24, 2020 at 18:13
  • $\begingroup$ It's not true that to show two completions are the same, you need the two closed subsets to be the same. $\endgroup$
    – Will Sawin
    Commented May 24, 2020 at 18:16
  • $\begingroup$ @WillSawin For a nice scheme $X$ and its closed subset $Z$, the fomal completion along $Z$ is the space $Z$ with some sheaf of rings. So I think we need it. $\endgroup$
    – k.j.
    Commented May 25, 2020 at 4:32

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