Lecture 19: J_0(N) mod N
The purpose of this lecture is to show that $J_0(N)$, and any quotient of it, has completely toric reduction at $N$. We do this in three steps: (1) We analyze the minimal regular model of $X_0(N)$ and show that its special fiber is a nodal curve whose irreducible components are $\bP^1$'s; (2) We recall a theorem of Raynaud, relating the Néron model of the Jacobian of a curve to the Picard scheme of its minimal model; (3) Combining the first two results shows that the special fiber of the Néron model of $J_0(N)$ is the Picard scheme of a nodal curve whose irreducible components are $\bP^1$'s. We explicitly compute this and find that it is a torus.
This lecture's goal
Let $N \gt 7$ be a prime. Thanks to the previous lecture, our task is to find an abelian variety $A/\bQ$ and a map $f \colon X_0(N) \to A$ satisfying the following four hypotheses:

$A$ has good reduction away from $N$.

$A$ has completely toric reduction at $N$.

The JordanHolder constituents of $A[p](\ol{\bQ})$ are trivial or cyclotomic.

$f(0) \ne f(\infty)$.
We know an example of an abelian variety to which $X_0(N)$ maps: the Jacobian $J_0(N)$ of $X_0(N)$. In fact, this is the universal such abelian variety. Thus to find $A$ we need only consider quotients of $J_0(N)$.
As we have seen, the first condition above ("$A$ has good reduction away from $N$") comes for free for quotients of $J_0(N)$, thanks to the following results:

$X_0(N)$ admits a smooth model away from $N$.

The Jacobian of such a curve has good reduction away from $N$.

Any quotient of an abelian variety having good reduction has good reduction.
Today, we're going to show that the second condition ("$A$ has completely multiplicative reduction at $N$") also comes for free. The proof is similar to the above, but a bit more complicated. We proceed as follows:

The special fiber of the minimal regular model of $X_0(N)$ at $N$ is a nodal curve whose irreducible components are $\bP^1$.

The Jacobian of such a curve has completely toric reduction.

Any quotient of an abelian variety having completely toric reduction has completely toric reduction.
The first step will consume most of our time. For the second, we appeal to a theorem of Raynaud relating Néron and minimal regular models. The third step follows easily from the theory of Néron models.
Completely toric reduction for quotients
Proposition. Let $\cO$ be a DVR with fraction field $K$ and residue field $k$. Let $A/K$ be an abelian variety and let $B$ be a quotient of $A$. Suppose $A$ has completely toric reduction. Then the same is true for $B$.
Proof. Let $f \colon A \to B$ be the quotient map. Since the isogeny category is semisimple, there exists a map $g \colon B \to A$ such that $fg=[n]$ for some positive integer $n$. The maps $f$ and $g$ extend to maps of the Néron models $\cA$ and $\cB$ by the Néron mapping property. These extended maps still satisfy $fg=[n]$, since it holds generically. It follows that $f$ and $g$ induce maps between $\cA_k$ and $\cB_k$ satisfying $fg=[n]$. In particular, $f \colon \cA_k \to \cB_k$ is surjective, which shows that $\cB_k$ is a torus. ◾
Raynaud's theorem on the relative Picard functor
Let $f \colon X \to S$ be a proper flat map. We define the relative Picard functor, denoted $\Pic_{X/S}$, to be the sheafification of the functor $S' \mapsto \Pic(X_{S'})$ on the big fppf site of $S$. A lot is known about this functor, but we'll only mention the few results we need. We refer to Raynaud's article (MR0282993) as a reference.
To begin with, we have the following result of Murre:
Theorem. If $S$ is a field then $\Pic_{X/S}$ is representable by a group scheme. (Note: nothing about the singularities of $X$ is assumed.)
Thus, when $S$ is a field, we have a component group $\Pic^0_{X/S}$, which we can think of as a subsheaf of $\Pic_{X/S}$. For a general base $S$, we define $\Pic^0_{X/S}$ to be the subsheaf of $\Pic_{X/S}$ consisting of those sections that restrict into $\Pic^0_{X_s/s}$ for every geometric point $s \to S$.
Suppose now that $S=\Spec(\cO)$ where $\cO$ is a DVR. Let $K$ be the fraction field of $\cO$ and $k$ the residue field. Suppose also that $X$ is a curve (i.e., its fibers are pure of dimension 1). Let $\{X_i\}$ be the irreducible components of the special fiber $X_k$. The local ring of $X_i$ at its generic point is artinnian; let $d_i$ be its length. This is the multiplicity of $X_i$ in $X$.
Theorem. Suppose that $X_K$ is smooth over $K$, $X$ is regular, and the gcd of the $d_i$ is 1. Let $\cJ$ be the Néron model of $\Jac(X_K)$ over $\cO$ and let $\cJ^0$ be its identity component. Then $\Pic^0_{X/S}$ is representable by smooth group scheme over $\cO$, and coincides with $\cJ^0$. In particular, $\cJ^0_k$ is isomorphic to $\Pic^0_{X_k/k}$.
Remark. The functor $\Pic_{X/S}$ is not necessarily representable by a scheme, but it is represented by an algebraic space. Let $E$ be the schemetheoretic closure of the identity section of $\Pic_{X/S}$. If $\Pic_{X/S}$ were separated, this would simply be the identity section, but $\Pic_{X/S}$ can fail to be separated. The quotient sheaf $\Pic_{X/S}/E$ is representable by a separated and smooth group scheme over $\cO$. It admits a degree function $\Pic_{X/S}/E \to \bZ$, the kernel of which is the full Néron model $\cJ$ of $\Jac(X_K)$.
The minimal regular model of X_0(N)
Given Raynaud's theorem, to understand the Néron model of $J_0(N)$ we should first understand the minimal regular model of $X_0(N)$. One might guess that $\ol{M}_0(N)$, the coarse space of $\ol{\fM}_0(N)$, would be the minimal regular model. This is almost the case, but not quite: the automorphisms groups in $\ol{\fM}_0(N)$ cause its coarse space to be nonregular. However, the singularities are very mild and easy to resolve.
To study $\fM_0(N)$ and its coarse space, we first pass to a finite étale Galois cover which is a scheme, see what goes on there, and then take the quotient to obtain the coarse space.
The covering space and its structure
We now change notation and use $p$ in place of $N$. We assume that $p$ is a prime $\gt 3$. Let $\ell$ be a prime satisfying the following: (1) $\ell \ne p$; (2) $\ell \gt 2$; and (3) $\ell \ne \pm 1$ modulo $p$. Let $G=\GL_2(\bF_{\ell})$. The order of $G$ is $(\ell^21)(\ell^2\ell)=\ell (\ell1)^2 (\ell+1)$, and is therefore prime to $p$. We work over $\bZ[1/6\ell]$ in this section (we really only care about what goes on at $p$).
We will be concerned with the following moduli spaces:

$\fM_0(p)$ and its coarse space $M_0(p)$.

The moduli space $\fM_0(p; \ell)$ of elliptic curves with $\Gamma_0(p)$ and $\Gamma(\ell)$structure. This is a scheme since $\ell \gt 2$, so we denote it by $M_0(p; \ell)$.

The moduli space $M(\ell)$ of elliptic curves with $\Gamma(\ell)$structure, which is smooth over $\bZ[1/\ell]$.
We have a natural map $M_0(p; \ell) \to \fM_0(p)$ which is finite étale and Galois with group $G$. We have a natural identification $\fM_0(p)=[M_0(p; \ell)/G]$ and $M_0(p)=M_0(p; \ell)/G$. Note that $M_0(p; \ell)$ is affine. If we let $A$ be its coordinate ring then $M_0(p)=\Spec(A^G)$.
We need the following result, a proof of which can be found in KatzMazur:
Theorem. The scheme $M_0(p; \ell)$ is regular and flat over $\bZ$.
We now have the following result:
Proposition. The scheme $M_0(p; \ell)_{\bF_p}$ is CohenMacaulay and reduced. It is smooth away from the supersingular points. Each supersingular point is an ordinary node (i.e., the strict complete local ring is $k \lbb u, v \rbb/(uv)$).
Proof. Write $M_0(p; \ell)=\Spec(A)$, where $A$ is regular and flat over $\bZ[1/\ell]$ is a regular ring. Then $M_0(p; \ell)_{\bF_p}=\Spec(B)$ with $B=A/pA$. The ring $B$ is CohenMacauly since it is the quotient of a regular ring by a nonzerodivisor.
We have the usual maps $i,j \colon M(\ell) \to M_0(p; \ell)$ (which are closed immersions) and $f,g \colon M_0(p; \ell) \to M(\ell)$. In particular, the ordinary locus of $M_0(p; \ell)$ is isomorphic to two copies of the ordinary locus of $M(\ell)$, and thus smooth. It follows that $M_0(p; \ell)$ is reduced, as it is onedimensional, CohenMacauly, and generically reduced.
Let $M(\ell)=\Spec(C)$. Let $x \in M(\ell)$ be a supersingular point and $y=i(x)=j(x)$. Consider the map $a \colon A_y \to C_x \times C_x$ on complete local rings given by $(i^*, j^*)$. This map is injective since $A_y$ is reduced and the map hits each component. Let $t \in C_x$ be a uniformizer. Let $u=f^*(t)g^*(t^p)$ and $v=g^*(t)f^*(t^p)$. Then $a(u)=(tt^{p^2}, 0)$ while $a(v)=(0, tt^{p^2})$. It follows that any element of $\fm_{C,x} \times \fm_{C,x}$ can be expressed as a power series in $a(u)$ and $a(v)$. Thus if $z \in \fm_{A,y}$ then $a(z)=F(a(u),a(v))$ for some $F$, and so $a(zF(u,v))=0$, and so $z=F(u,v)$. It follows that the map $k \lbb u, v \rbb \to A_y$ is surjective, where $k$ is the residue field. Since $a(uv)=0$, we have $uv=0$, and so we get a surjection $k \lbb u,v \rbb/(uv) \to A_y$. This map must be injective, for otherwise we'd lose a component. ◾
Proposition. The scheme $M_0(p; \ell)$ is smooth over $\bZ[1/\ell]$ away from the supersingular points in characteristic $p$. The strict complete local ring at such a point is of the form $W \lbb u,v \rbb/(uvp)$, where $W=\bZ_p^{\un}$.
Proof. Let $R$ be the strict complete local ring at a supersingular point. Then $R$ is regular and flat over $W$ of dimension 2. Furthermore, $R/p$ is of the form $k \lbb u,v \rbb/(uv)$. It follows that $R$ is a quotient of $W \lbb u, v \rbb$. Now, $uv \in pR$, and so we have a relation of the form $uv=pw$ for some $w \in R$. Let $R'=W \lbb u,v \rbb/(uvpw)$. One easily see that $R'$ is flat. Furthermore, the surjection $R' \to R$ induces an isomorphism mod $p$. It follows that this map must be an isomorphism: killing any nonzero element of $R'$ would either introduce $p$torsion or kill something in $R'/p$. We thus find $R=W \lbb u, v \rbb/(uvpw)$.
Now, consider the cotangent space to $R$, the quotient of $\fm=(u,v,p)$ by its square. This is the $k$vector space spanned by $u$, $v$, and $p$ modulo $pw$. If $w \in \fm$ then $pw \in \fm^2$, and so $\fm/\fm^2$ would have dimension 3. This contradicts the regularity of $R$. Thus $w$ is a unit, and so, replacing $u$ with $u/w$, we may assume $w=1$. ◾
The structure of M_0(p)
Recall that $M_0(p)=M_0(p; \ell)/G$, and that, if $M_0(p; \ell)=\Spec(A)$ then $M_0(p)=\Spec(A^G)$. Since $G$ is prime to $p$, formation of $G$ invariants commutes with reduction mod $p$. In particular, formation of the coarse space of $\fM_0(p)$ commutes with reduction mod $p$.
The element $1 \in G$ acts trivially on $M_0(p; \ell)$. Let $\ol{G}=G/\{\pm 1\}$. Then $M_0(p)=M_0(p; \ell)/\ol{G}$. Let $x$ be a point in $M_0(p)$ in characteristic $p$ with automorphism group $H$. The group $\ol{G}$ transitively permutes the points of $M_0(p; \ell)$ above $x$, and the stabilizer of any point is a subgroup $\ol{G}$ isomorphic to $\ol{H}=H/\{\pm 1\}$. It follows that the strict completion $R$ of the local ring at $x$ is isomorphic to the $\ol{H}$invariants of the strict completion of the local ring $S$ at any point $y$ over $x$. We have $S=W \lbb u, v \rbb/(xyp)$ by the previous section, where $W$ is the Witt ring of $\ol{\bF}_p$. Now, since $p \gt 3$, we have the following:

If $j(x) \ne 0, 1728$ then $\ol{H}$ is trivial.

If $j(x) = 1728$ then $\ol{H}=\bZ/2\bZ$.

If $j(x) = 0$ then $\ol{H}=\bZ/3\bZ$.
Thus if $j(x) \ne 0, 1728$ then $R=S$ and $x$ is a regular point. Note that if $j(x)$ is 0 or 1728 then, since $\ol{H}$ does not fix a point in a neighborhood of $x$, it acts nontrivially on $S$. It follows that, for an appropriate choice of $u$, $v$, the generator of $\ol{H}$ acts by $u \mapsto \zeta u$, $v \mapsto \zeta^{1} v$, where $\zeta$ is a primitive $k$th, where $k=\# \ol{H}$. It follows that $R=S^{\ol{H}}$ is generated by $U=u^k$, $V=v^k$, and $uv=p$. We have $UV=(uv)^k=p^k$, and so $R=W \lbb U, V \rbb/(UVp^k)$. We have thus shown that following:
Theorem. Let $x$ be a characteristic $p$ point of $M_0(p)$ and let $R$ be the strict complete local ring at $x$.

If $x$ is not supersingular, then $M_0(p)$ is smooth at $x$.

If $x$ is supersingular and $j(x) \ne 0, 1728$, then $M_0(p)$ is regular at $x$ and $R=W \lbb x, y \rbb/(xyp)$.

If $x$ is supersingular and $j(x)=1728$, then $R=W \lbb x, y \rbb/(xyp^2)$.

If $x$ is supersingular and $j(x)=0$, then $R=W \lbb x, y \rbb/(xyp^3)$.
Remark. It is also true that the cuspidal points of $\ol{M}_0(p)$ are smooth, since they are smooth in the special fiber.
The minimal regular model
The scheme $\ol{M}_0(p)$ is a flat proper model of its generic fiber which is regular except at possibly two points. The singularities of these points can be resolved with one or two blowups. The result is that an additional $\bP^1$ is added at $j=1728$ if that point is supersingular, and two additional $\bP^1$'s are added in a chain at $j=0$ if that point is supersingular. The resulting model is minimal. We thus have:
Proposition. Let $C$ be the special fiber of the minimal regular model of $\ol{M}_0(N)$. Then $C$ is reduced curve, all of its components are $\bP^1$'s, and all of its singularities are simple nodes.
The special fiber of the Néron model
Proposition. Let $C$ be a curve over an algebraically closed field $k$ with the following properties: $C$ is reduced, all of its components are $\bP^1$'s, and all of its singularities are simple nodes. Then $\Pic^0_{C/k}$ is a torus.
Proof. Let $\Gamma$ be the graph corresponding to $C$: its vertices are the irreducible components of $C$, and there is one edge between two components at each point they touch. Given a line bundle on $C$, we get a line bundle on each component, and an identification of the fibers at the touching points. Every line bundle on $\bP^1$ is of the form $\cO(n)$. Thus if we assign to each vertex of $\Gamma$ an integer and to each edge an element of $\bG_m$ then we can build a line bundle on $C$, and all line bundles are of this form. We have thus produced a surjection from a torus to $\Pic^0_{C/k}$, which proves the proposition. ◾
Actually, we can say a bit more. Suppose we have data as above defining some line bundle. For the bundle to be trivial, it must be trivial on each component, and so the integers at each node must be 0. The nonvanishing sections of the bundle on one of the components is given by $\bG_m$. Given sections on each component (i.e., elements of $\bG_m$ at each node), they glue if and only if at each edge the quotient of their values is equal to the value of the edge. In other words, the original data defines the trivial bundle if and only if the integers are 0 and the values on the edges are a 1coboundary.
We thus see that the identity component of $\Pic_{C/k}$ is $\rH^1(\Gamma, \bG_m)$. This is a torus with character lattice $\rH_1(\Gamma, \bZ)$.
Theorem. $J_0(N)$ has completely toric reduction at $N$.
Proof. This follows from the above computation, Raynaud's theorem, and the form established for the minimal regular model of $X_0(N)$ at $N$. ◾
Injectivity of the reduction map on torsion
To end this lecture, I want to give a proof of the following theorem. This result was crucial to yesterday's lecture, and it was pointed out to me that we had not yet given a proof in all cases.
Theorem. Let $K/\bQ_p$ be a finite extension with ramification index $\lt p1$. Let $\cO$ be the ring of integers of $K$ and $k$ the residue field of $\cO$. Let $A/K$ be an abelian variety, and let $\cA/\cO$ be its Néron model. Then the reduction map $\cA(\cO)_{\tors} \to \cA(k)_{\tors}$ is injective.
Proof. Let $G_0=A(K)_{\tors}$, regarded as a closed subscheme of $A(K)$. Note that as a group scheme, $G_0$ is constant. Let $G$ be the schemetheoretic closure of $G_0$ in $\cA$. Then $G$ is a flat group scheme over $\cO$. And it is finite: every fieldvalued point of $G$ is defined over $K$, and every $K$point of $G$ extends to an $\cO$point of $G$ by the Néron mapping property. Thus $G$ is proper, and therefore finite (since we know it to be quasifinite). Since $G_0$ obviously extends to a constant group scheme over $\cO$, Raynaud's theorem implies that $G$ itself is a constant group scheme. The theorem follows, since the reduction map $G(\cO) \to G(k)$ is clearly injective for constant groups. ◾