I'm struggling with the proof of 2.21 of Saito's "Fermat's Last Theorem".

Let $\omega$ be a primitive 3rd root of unity, $X(3) = \mathbb{P}^1_{\mathbb{Q}(\omega)}$, and $E = \{ X^3 + Y^3 + Z^3 - 3 \mu XYZ \} \subseteq \mathbb{P}^2_{X(3)}.$ (where $\mu$ is an inhomogeneous coordinate of $X(3)$.)

Let $O = [ 0:1:-1], P= [0:\omega:-1], Q = [1:0:-1].$
In order to show that $X(3)$ is the fine moduli scheme (over $\mathbb{Q}$) of full level 3 structure, I want to show that $E$ has a structure of a generalized elliptic curve with the $0$-section $[0 : 1 : -1]$, such that $P, Q$ is a basis of the $3$-torsion points.

Here is what I tried:

Let $Y(3) = X(3) - \{ 1, \omega, \omega^2, \infty\}$. Then $E$ is smooth over $Y$.
And for $\mu \in X - Y$, the fibre of $E$ at $\mu$ is

$$E_\infty = \{ XYZ = 0 \}, \\ E_1 = \{ (X + Y + Z)(X + \omega Y + \omega^2Z)(X + \omega^2 Y + \omega Z) = 0 \}, \\ E_\omega = \{ (X + Y + \omega Z)(X + \omega Y + Z)(X + \omega^2 Y + \omega^2 Z) = 0 \}, \\ E_{\omega^2} = \{ (X + Y + \omega^2 Z)(X + \omega^2 Y + Z)(X + \omega Y + \omega Z) = 0 \}.$$

These fibres are Neron 3-gon, so we can define generalized elliptic curve structures on them, such that the fibres of $P, Q$ are bases of their 3-torsion points.

How can I define the generalized structure on $E$ globally?


Seeing a comment of Will Sawin, I have 2 idea of defining a generalized elliptic curve law.:

  1. Using that $E$ is a plain cubic curve, define $E^\text{sm} \times E \to E$ by Bézout's theorem.
    First writing down the explicit addition law for $E \times_{X(3)} Y(3)$, then define $E \times E \to E$ by that explicit formula. And check the group action for $E^\text{sm}$ and the group action axiom for $E^\text{sm} \times E \to E$.
    But the addition law is too complicated to write down explicitly...

  2. Extend the morphism $E \times Y(3) \to \operatorname{Isom}(E, E) \times Y(3) : P \mapsto (Q \mapsto P + Q)$ to the morphism $\varphi : E^\text{sm} \to \operatorname{Isom}(E, E)$.
    If for each closed point $s$ of $X(3)$ and for generic point $x$ of $E_s$ the morphism $\varphi$ is defined at $x$, then by a general theorem (see proposition 1.3 of Artin's Neron models), we have that $\varphi$ is defined everywhere. (Now since $E$ is in the projective space, the Hilbert scheme exits, hence so does the scheme of isomorphisms.)

To define $\varphi$ for a such point $x$, it sufficies to see that $\operatorname{Isom}$ is proper over $X(3)$. (for example, see 4.1.16 of Liu's Algebraic geometry and arithmetic curves.) But I think this is false, since for a Neron n-gon $C$ over a field, $\operatorname{Aut}(C)$ is not proper. (see II 1.8 of Deligne-Rapoport)

Thank you very much!

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    $\begingroup$ There is a fairly detailed discussion of a slightly different model here: arxiv.org/pdf/1803.09962.pdf $\endgroup$ Apr 7, 2020 at 22:14
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    $\begingroup$ Yes, you can define the structure using Bezout's theorem. $\endgroup$
    – Will Sawin
    Apr 7, 2020 at 22:21
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    $\begingroup$ What's the largest open set on which you can figure out how to define the map? What is the closed complement? Can you use the behavior near the closed complement to prove that it extends? $\endgroup$
    – Will Sawin
    Apr 7, 2020 at 23:27
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    $\begingroup$ @k.j. There are at least two ways to do it. One way, that will certainly work, is to write down the explicit formula and compute precisely what happens at the singular curves. The other is a more abstract argument. I know the abstract argument must exist because there are general theorems about objects of this form that can't be proven by working with every example by hand. (A third way would be to look up one of these general theorems but that seems like cheating...) $\endgroup$
    – Will Sawin
    Apr 8, 2020 at 2:44
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    $\begingroup$ For a more abstract argument, you could start with what you wrote about the Picard variety, which should give you a morphism of schemes on the open set of smooth fibers, then try to give an argument that it extends to generic points on the singular fiber, then special points on the singular fiber. I'm not sure exactly what is the best way to go about it. $\endgroup$
    – Will Sawin
    Apr 8, 2020 at 2:46

1 Answer 1


I have understood.

  1. $E$ is normal connected scheme.

First, since $X$ is connected and $E \to X$ is proper flat of finitely presentation, $E$ is connected.
Next, since $E^\text{sm} \to X$ is smooth and since $X$ is regular, $E^\text{sm}$ is regular. Now since $E \to X$ is local complete intersection, in particular Cohen-Macaulay, and is 2-dimensional, $E$ is normal.

  1. The rational map $\varphi : E^\text{sm} \to \operatorname{Isom}(E, E), P \mapsto (Q \mapsto P+Q)$, a priori defined on $E \times_X Y$, is defined everywhere.

By (1.3) of Artin's Neron Model (in Cornel-Silverman's Arithmetic Geometry), it suffices to see that this $\varphi$ is defined at $x \in E^\text{sm}$, which is a generic point of $E^\text{sm}_s$. (Where $s$ is the image of $x$ under $E \to X$. Now if $s$ is the generic point, then since $\varphi$ is defined at $x$, it suffices to check this for closed $s$.)
Since $E^\text{sm} \to X$ is flat, such a point is exactly a point which is codimension 1 in $E^\text{sm}$. Thus it suffices to see that the morphism $\operatorname{Spec}k(E^\text{sm}) \to E^\text{sm} \to \operatorname{Isom}$ is extended to $\operatorname{Spec}\mathscr{O}_{E^\text{sm}, x} \to \operatorname{Isom}$. (Where $k(E)$ is the function field of the scheme $E$.)

So we show:

Let $R$ be a DVR with the prime element $\pi$, the residue field $k$, and the fractional field $K$ of charactericstic $\neq 3$. Let $E$ be a curve over $R$, defined by $$X^3 + Y^3 + Z^3 - 3 \mu XYZ = 0$$ or $$\mu (X^3 + Y^3 + Z^3) - 3 XYZ = 0.$$ ($\mu \in R$). Assume that $E$ is smooth over $K$ and is singluar over $k$.
Let $f : E_K \to E_K$ be an isomorphism over $K$. Then this $f$ extends to a isomoprhism $g : E \to E$ over $R$.

First, by (4.1.16) of Liu's Algebraic Geometry and Arithmetic Curves, the morphism $f : E_K \to E_K$ extends to a rational map $g : E \to E$, defined at every codimension 1 point.
On the other hand, write $\mu = \epsilon \pi^n$. Since $E_k$ is singular, $n \ge 1$. Now blowing-up $E$ at every singular point $\lfloor n/2 \rfloor$-times, we get the regular model $\mathfrak{E}$ over $R$, of $E_K$. Seeing its special fibres, we can see that $\mathfrak{E}$ is minimal. Thus $f : E_K \to E_K$ extends to the isomorphism $\mathfrak{E} \to \mathfrak{E}$. Since $\require{AMScd}$ \begin{CD} \mathfrak{E} @>{h}>> \mathfrak{E}\\ @V{p}VV @V{p}VV\\ E @>{g}>> E \end{CD}

is commutative, this $\mathfrak{E} \to \mathfrak{E}$ induces $g : E \to E$, defined everywhere. (First, since $g$ is defined on a dense open, we can show that $h(\text{a contractible line}) = (\text{a contractible line})$. On the other hand, since $p: \mathfrak{E} \to E$ is a closed map, $E$ is some quotient of $\mathfrak{E}$ as a topological space. Thus $h$ induces $g$ as topological spaces. Finally, since $p_*\mathscr{O}_\mathfrak{E} = \mathscr{O}_E$, $h$ induces a map of schemes $g$.)
(Or, considering affine neighbourhood, we can reduce the situation into an easy ring lemma: Let $A$ be a normal noetherian local ring of dimension $2$ and let $f : \operatorname{Spec}A - \{ \mathfrak{m}\} \to \operatorname{Spec}A - \{ \mathfrak{m}\}$ be a morphism of schemes. Then this $f$ is defined everywhere, i.e., defines the ring endomorphism on $A$. Since now $A = \cap A_\mathfrak{p}$, where $\mathfrak{p}$ runs through all prime ideals of $A$ of heght $1$, this is almost trivial.)

  1. The morphism $\varphi : E^\text{sm} \to \operatorname{Isom}(E, E)$ defines a group structure on $E^\text{sm}$ and defines a group action of $E^\text{sm}$ on $E$.

Since the "group axiom diagrams" and "the group action axiom diagram" commute over $Y$, and since $E^\text{sm} \times E \times E$ is reduced and since $E$ is separated, these diagrams are commutative.

  1. This is a generalized elliptic curve.

By II.1.15 of Deligne-Rapoport, this is trivial.


  1. $P$ and $Q$ defines the full level $3$ structure.

Over $Y$, by Bezout. ($E \cap \{ \mu \omega^2 X + Y + \omega Z = 0 \} = 3P$ and $E \cap \{ \mu \omega X + Y + \omega^2 Z = 0 \} = 3Q$ as Cartier divisors.)
Over $X$, checking at each fibre, $(\mathbb{Z}/3)^2 \to E^\text{sm}[3]$ is an isomorphism fibre-by-fibre. Thus is an isomorphism globally.


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