## IV.4: Elliptic Curves :::{.remark} Curves $E$ with $g(E) = 1$; we'll assume $\characteristic k \neq 2$ throughout. Outline: - Define the $j\dash$invariant, classifies isomorphism classes of elliptic curves. - Group structure on the curve. - $E = \Jac(E)$. - Results about elliptic functions over $\CC$. - The Hasse invariant of $E/\FF_q$ in characteristic $p$. - $E(\QQ)$. ::: ### The $j\dash$invariant :::{.remark} The $j\dash$invariant: - $j(E) \in k$, so $\AA^1\slice k$ is a coarse moduli space for elliptic curves over $K$. - Defining $j(E)$: - Let $p_0\in X$, consider the linear system $L\da \abs{2p_0}$. - Nonspecial, so RR shows $\dim(L) = 1$. - BPF, otherwise $E$ is rational. - Defines a morphism $\phi_L: E\to \PP^1\slice k$ with $\deg \phi_L = 2$. - Up to change of coordinates, $f(p_0) = \infty$. - By Hurwitz, $f$ is ramified at 4 branch points $a,b,c,p_0$. - Move $a\mapsto 0, b\mapsto 1$ by a Mobius transformation fixing $\infty$, so $f$ is branched over $0,1,\lambda,\infty$ where $\lambda \in k\smts{0,1}$. - Use $\lambda$ to define the invariant: \[ j(E) = j( \lambda) = 2^8\qty{(\lambda^2 - \lambda+ 1)^3 \over \lambda^2 (\lambda- 1)^2} .\] - Theorem 4.1: - $j$ depends only on the curve $E$ and not $\lambda$. - $E\cong E'\iff j(E) = j(E')$. - Every element of $k$ occurs as $j(E)$ for some $E$. - So this yields a bijection \[ \correspond{ \text{Elliptic curves over }k }\modiso &\mapstofrom \AA^1\slice k \\ E &\mapsto j(E) .\] - Some facts that go into proving this: - $\forall p,q\in X\,\,\exists \sigma\in \Aut(X)$ such that $\sigma^2=1, \sigma(p) = q$, for any $r\in X$, one has $r + \sigma(r) \sim p + q$. - $\Aut(X)\actson X$ transitively. - Any two degree two maps $f_1,f_2: X\to \PP^1$ fit into a commuting square. - Under $S_3\actson \AA^1\slice k\smts{0, 1}$, the orbit of $\lambda$ is \[ S_3 . \lambda= \ts{ \lambda, \lambda\inv, s_1 = 1- \lambda, s_1\inv = (1- \lambda)\inv, s_2 = \lambda(\lambda-1)\inv, s_3 = \lambda\inv (\lambda- 1)} .\] - Fixing $p\in X$, there is a closed immersion $X\to \PP^2$ whose image is $y^2=x(x-1)(x- \lambda)$ where $p\mapsto \infty = \tv{0:1:0}$ and this $\lambda$ is either the $\lambda$ from above or one of $s_1^{\pm 1}, s_2^{\pm 1}$. - Idea of proof: embed $X\injects \PP^2$ by $L\da \abs{3p}$, use RR to compute $h^0(\OO(np)) = n$ so $h^0(\OO(6p)) = 6$. - So $\ts{1,x,y,x^2,xy,y^2,x^3}$ has a linear dependence where $x^3,y^2$ have nonzero coefficients since they have poles at $p$. - Rescale $x^3, y^2$ to coefficient 1 to get \[ y^2+a_1 x y+a_3 y=x^3+a_2 x^2+a_4 x+a_6 .\] - Do a change of variable to put in the desired form: complete the square on the LHS, factor as $y^2=(x-a)(x-b)(x-c)$, send $a\to 0, b\to 1$ by a Mobius transformation. - Note that one can project from $p$ to the $x\dash$axis to get a finite degree 2 morphism ramified at $0,1, \lambda, \infty$. ::: :::{.example title="?"} An elliptic curve that is smooth over every field of non-2 characteristic: \[ E: y^2 = x^3-x, \qquad \lambda=-1,\, j(E) = 2^6 \cdot 3^3 = 1728 .\] ![](figures/2022-12-03_23-36-23.png) One that is smooth over every $k$ with $\characteristic k \neq 3$: the Fermat curve \[ E: x^3 + y^3 = z^3,\qquad \lambda = \pm \zeta_3^{k},\, j(E) = 0 .\] ::: :::{.theorem title="Orders of automorphism groups of elliptic curves"} \[ \size \Aut(X, p) = \begin{cases} 2 & j(E) \neq 0,1728\\ 4 & j(E) = 1728, \characteristic k \neq 3 \\ 6 & j(E) = 0,\characteristic k \neq 3 \\ 12 & j(E) = 0,1728, \characteristic k = 3 \end{cases} .\] ::: :::{.proof title="?"} Idea: take the degree 2 morphism $f:X\to \PP^1$ with $f(p) = \infty$ branched over $\ts{0,1, \lambda, \infty}$. Produce two elements in $G$: for $\sigma\in G$, find $\tau\in \Aut(\PP^1)$ so $f\sigma = \tau f$; then either $\tau \neq \id$, so $\ts{\sigma, \tau} \subseteq G$, or $\tau = id$ and either $\sigma=\id$ or $\sigma$ exchanges the sheets of $f$. If $\tau\neq \id$, it permutes $\ts{0, 1, \lambda}$ and sends \( \lambda\mapsto \lambda\inv, s_1^{\pm 1}, s_2^{\pm 1} \) from above. Cases: 1. $j=1728:$ If \( \lambda= -1, 1/2, 2, \characteristic k \neq 3 \), then $\lambda$ coincides with *one* other element of $S_3. \lambda$, so $\size G = 4$. 2. $j=0$: If \( \lambda= -\zeta_3, -\zeta_3^2, \characteristic k \neq 3 \) then $\lambda$ coincides with *two* elements in $S_3 . \lambda$ so $\size G = 6$. 3. $j=0=1728$: If \( \lambda= -1, \characteristic k = 3 \) then $S_3 . \lambda= \ts{ \lambda}$ and $\size G = 12$. ::: ### The group structure :::{.remark} The group structure: - Fixing $p_o\in E$, the map $p\mapsto \mcl(p-p_0)$ induces a bijection $E \iso \Pic^0(E)$, so the group structure on $E$ is the pullback along this with $p_0 = \id$ and $p+q=r\iff p+q \sim r+p_0 \in \Div(E)$. - Under the embedding of $\abs{3p_0}$, points $p,q,r$ are collinear iff $p+q+r \sim 3p_0$, so $p+q+r=0$ in the group structure. - $E$ is a group variety, since $p\mapsto -p$ and $(p, q)\mapsto p+q$ are morphisms. Thus there is a morphism $[n]: E\to E$, multiplication by $n$, which is a finite morphism of degree $n^2$ with kernel $\ker [n] = C_n^2$ if $(n, \characteristic k) = 1$.and $\ker [n] = C_p, 0$ if $n=\characteristic k$, depending on the Hasse invariant. - If $f:E_1 \to E_2$ is a morphism of curves with $f(p_1) = p_2$ then $f$ induces a group morphism. - $\Endo(E, p_0)$ forms a ring under $f+g = \mu\circ (f\times g)$ and $f\cdot g \da f\circ g$. - The map $n \mapsto ([n]: E\to E)$ defines a finite ring morphism $\ZZ\to \Endo(E, p_0)$ for $n\neq 0$. - $R \da \Endo(E, p_0)\units = \Aut(E)$, and if $j=0,1728$ then $R$ contains $\ts{\pm 1}$ and is thus bigger than $\ZZ$. ::: :::{.remark} The Jacobian: a variety that generalizes to make sense for any curve, a moduli space of degree zero divisor classes. - For $X/k$ a curve and $T\in\Sch\slice k$, define \[ \Pic^0(X\times T) \da \ts{\mcf \in \Pic(X\times T) \st \deg \ro{\mcf}{X_t} = 0 \, \forall t\in T },\qquad \Pic(X/T) \da \Pic^0(X\times T)/ p^* \Pic(T) \] where $p:X\times T\to T$ is the second projection. Regard this as *families of sheaves of degree 0 on $X$ parameterized by $T$*. - The Jacobian variety of a curve $X$: $\Jac(X) \in \Sch^\ft\slice k$ along with $\mcl \in \Pic^0(X/\Jac(X))$ such that for any $T\in\Sch^\ft\slice k$ and any $\mcm\in \Pic^0(X/T)$, $\exists ! \, f: T\to \Jac(X)$ such that $f^* \mcl = \mcm$. Thus $J$ represents the functor $\Pic^0(X/\wait)$. - For $E$ elliptic, $E = \Jac(E)$. In general, $\abs{\Jac(X)}\cong \abs{\Pix^0(X)}$ on points, since points of $\Jac(X)$ are morphisms $\spec k\to \Jac(X)$, which correspond to elements in $\Pic^0(X/k) = \Pic^0(X)$. - $\Jac(X) \in \Grp\Sch\slice k$ where $e: \spec k\to \Jac(X)$ corresponds to $0\in \Pic^0(X/k)$, $\rho: \Jac(X) \to \Jac(X)$ is $\mcl \mapsto \mcl\inv\in\Pic^0(X/\Jac(X))$, and $\mu: \Jac(X)\cartpower{2}\to \Jac(X)$ is $\mcl \mapsto p_1^* \mcl \tensor p_2^*\mcl \in\Pic^0(X/\Pic(X)\cartpower 2)$. - $\T_0 \Jac(X) \cong H^1(X; \OO_X)$: giving an element of $\T_p X$ is the same as a morphism $T\da \spec k[\eps]/\eps^2\to X$ sending $\spec k \to p$. So $\T_0 \Jac(X)$, this means giving $\mcm\in \Pic^0(X/T)$ whose restriction to $\Pic^0(X/k)$ is zero. Use the SES $H^1(X;\OO_X)\injects \Pic X[\eps] \to \Pic(X)$. - $\Jac(X)$ is proper over $k$ by the valuative criterion. Just show that an invertible sheaf $\mcm$ on $X\times \spec K$ lifts unique to $\tilde \mcm$ on $X\times \spec R$, but $X\times \spec R$ is regular, so apply $\rm{II}.6.5$. - For any $n$ there is a morphism \[ \phi^n: X\cartpower{n} &\to \Jac(X) \\ (p_1,\cdots, p_n) &\mapsto \mcl(\sum p_i - np_0) .\] This is surjective for $n\geq g(X)$ by RR since every divisor class of degree $d\geq g$ has an effective representative. The fibers of $\phi^n$ are all tuples $(p_1,\cdots, p_n)$ such that $D = \sum p_i$ forms a complete linear system. - Most fibers are finite, so $\Jac(X)$ is irreducible of dimension $g$. - Smoothness: $\dim \T_0 \Jac(X) = \dim H^1(X;\OO_X) = g$, so smooth at zero, and group schemes are homogeneous so smooth everywhere. ::: ### Elliptic functions > Stopped at elliptic functions.