## IV.4: Elliptic Curves $\star$

:::{.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:
\[
E: y^2 = x^3-x, \qquad \lambda=-1,\, j(E) = 2^6 \cdot 3^3 = 1728
.\]

![](Reading%20Notes/4_Hartshorne/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"}
\[
\size \Aut(X, p) =
\begin{cases}
2 & j(E) \neq 0,1728\\
4 & j(E) = 1728, \characteristic k \neq 3 \\
6 & j(E) = 0,\characteristic k \neq 3 \\ 
12 & j(E) = 0,1728, \characteristic k = 3
\end{cases}
.\]
:::

:::{.remark title="Proof idea"}
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, $\abs{\Jac(X)}\cong \abs{\Pic^0(X)}$ 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

Everywhere $X$ is an elliptic curve over $\CC$.

:::{.remark}
On classifying elliptic curves using elliptic functions:

- $f: \CC\to \CC$ is an **elliptic function** with respect to a lattice $\Lambda = \gens{1, \tau}$ is a meromorphic function where $f(z + \lambda) = f(z)$ for all \( \lambda\in \Lambda \), sometimes called **doubly periodic functions**.
  - An example: the Weierstrass $\wp\dash$function,
  \[
  \wp(z)=\frac{1}{z^2}+\sum_{\omega \in \Lambda^{\prime}}\left(\frac{1}{(z-\omega)^2}-\frac{1}{\omega^2}\right), \qquad \Lambda' \da \Lambda\smz
  .\]
  Converges on $\CC\sm \Lambda$ and has poles of order 2 at points of \( \Lambda \).
  - Another example: its derivative
  \[
  \wp^{\prime}(z)=\sum_{\omega \in \Lambda} \frac{-2}{(z-\omega)^3}
  .\]

- The set of elliptic functions form a field $F = \gens{ \wp, \wp'}$ which satisfy
  \[
  \left(\wp^{\prime}\right)^2=4 \wp^3-g_2 \wp-g_3,
  \]
  where
  \[
  g_2=60 \sum_{\omega \in \Lambda^{\prime}} \frac{1}{\omega^4} \quad \text { and } \quad g_3=140 \sum_{\omega \in A^{\prime}} \frac{1}{\omega^6} \text {. }
  .\]

- So there is a holomorphic map
\[
\phi: \CC &\to \PP^2\slice \CC \\
z &\mapsto (\wp(z), \wp'(z))
,\]
where $\im(\phi) \subseteq X \da V(y^2 = 4x^3 - g_2 x - g_3)$ (a curve).
  - This descends to a bijection $\phi: \CC/ \Lambda \iso X$.
  - Under $\phi$, $F$ corresponds to $\CC(X)$, the field of rational functions on $X$.
- Any elliptic function $f$ has a divisor $(f) = \sum n_i (a_i)$ with $a_i\in \CC/ \Lambda$.
  - For any set of distinct points $\ts{a_1,\cdots, a_\ell}\in \CC/ \Lambda$ and any sequence $\ts{n_1,\cdots, n_\ell} \subseteq \ZZ$, there exists an elliptic function $f$ with $(f) = \sum n_i a_i \iff \sum n_i = 0$ and $\sum n_i a_i = 0$ in the group $\CC/ \Lambda$.
  - Thus $a_1 + a_2 \equiv b \mod \Lambda \iff$ there is an elliptic function $f$ with zeros at $a_1, a_2$ and poles at $b, 0$.
  - Since $f\in \CC(X)$, one has $\phi(a_1) + \phi(a_2) \sim \phi(b) + \phi(0)\in \Div(X)$.
  - Set $P_0 \da \phi(0)$ (the point $i\infty$ on the $y\dash$axis), give $X$ a group structure with identity $P_0$ to get a group law $\phi(a_1) + \phi(a_2) = \phi(b)$ which induces a group isomorphism $\CC/ \Lambda\cong X$.

- Given $c_2, c_3\in \CC$ with $\Delta(c_2, c_3)\da c_2^3 - 27c_3^2 \neq 0$, there is a $\tau\in \CC\sm\RR$ and $\alpha\in \CC\smz$ such that the lattice \( \Lambda\da \gens{1, \tau}_\ZZ \) yields $g_2 =\alpha^4 c_2$ and $g_3 = \alpha^6 c_3$ as above.

- So every elliptic curve arises as $\CC/ \Lambda$ for some \( \Lambda \)
  - If $X$ is elliptic, embed into $\PP^2$ via an equation of the form $y^2 = x(x-1)(x- \lambda)$ with \( \lambda\neq 0, 1 \).
  - Take a linear change of variables to get
  \[
  y^2 &= 4x^3 -c_2 x - c_3 \\
  c_2 &= {1\over 3} 4^{1\over 3}( \lambda^2 - \lambda+ 1) \\
  c_3 &= {1\over 27}( 2 \lambda^2 - 5 \lambda + 2) \\
  \Delta&= \lambda^2( \lambda- 1)^2\neq 0 \text{ since } \lambda\neq 0, 1
  .\]
  - The curve determined by $\Lambda$ is now exactly $X$ by making the change of variables 
  \[
  y' &\da \alpha^3 y \\
  x' &\da \alpha^2 x
  .\]

- Define
\[
J(\tau) \da {g_2^3 \over \Delta} \quad \implies j(\tau) = 1728 J( \tau)
;\]
$J$ classifies $X$ up to isomorphism: $J( \tau) = J( \tau') \iff$ they differ by an element of $\SL_2(\ZZ)$, i.e. there exist $a,b,c,d\in \ZZ$ with $ad-bc= \pm 1$ and 
\[
\tau' = {a\tau + b\over c\tau + d}
.\]
Furthermore for any $\tau'$ there is a unique $\tau$ with $J(\tau) = J(\tau')$ where $\tau$ is in the usual fundamental region $\Re(\tau) \in [-1/2, 1/2)$ where $\abs\tau \geq 1$ for $\Re(\tau) \leq 0$ and $\abs\tau > 1$ for $\Re(\tau) > 0$:

![](figures/2023-01-03_16-43-55.png)

:::

:::{.remark}
Consequences of the theory:

- $X\isoas{\Grp} (\RR/\ZZ)\cartpower{2}$ and $X[n] \cong (\ZZ/n\ZZ)\cartpower{2}$.
  - The $n\dash$torsion points correspond to ${a\over n} + {b\over n}\tau$ for $a,b \in \ts{0,1,\cdots, n-1}$.
  - The infinite order points correspond to any $w\in \CC\smts \gens{1, \tau}_\QQ$.

- Multiplication by $n$ is a finite morphism $n_X: X\to X$ of degree $n^2$
  - Why: $n_X$ is separable and a group morphism, so $\deg n_X = \size(\ker n_X) = n^2$.
- An important invariant: the ring of endomorphisms $R \da \Endo_{\Grp}(X, P_0)$, particularly endomorphisms determined by elliptic functions with periods $1,\tau$.
- There is an injection $R\injects \CC$: endomorphisms $f\in R$ correspond to \( \alpha\in \CC \) such that \( \alpha \Lambda \subseteq \Lambda \).
  - One direction of this: given such an $\alpha$, one gets a holomorphic automorphism of $\CC/ \Lambda$, which is an morphism of curves by GAGA.
- Under this correspondence, $n_X$ corresponds to literal multiplication by $n$ in $\CC$.
  Say $X$ has **complex multiplication (CM)** if $R\supsetneq \ZZ$, so the endomorphism ring is larger than expected.
  - If $X$ has CM, then $\tau \in K\da \QQ(i\sqrt{d})$ for some squarefree $d\in \ZZ_{> 0}$ and $\ZZ < R \leq \ZZ_K$ (its ring of integer) is a subring properly containing $\ZZ$.
    Conversely, if $\tau\in r + s\sqrt{-d}\in \gens{1, \sqrt{-d}}_\QQ$ then $X$ has CM and $R = \ts{a + b \tau \st a,b\in \ZZ\,\, 2br,\, b(r^2+ ds^2)\in \ZZ}$.
  - There are only countably many $j\in \CC$ such that the corresponding $X$ has CM, since there are only countably many elements in all quadratic extensions of $\QQ$.

- Examples:
  - $\tau = i$ yields $R = \ZZ[i] = \ZZ_K$ for $K = \QQ(i)$.
  Note $R\units = \ts{\pm 1, \pm i} \cong C_4$, so $\size \Aut(X) = 4$ and $j(\tau) = 1728$ and $J(\tau) = 1$
  Another way to see $j(i) = 1$: write $\Lambda = \gens{1, i}_\ZZ$, then multiplication by $i$ is an automorphism so
  \[
  g_3=140 \sum_{\omega \in \Lambda^{\prime}} \omega^{-6}=140 \sum_{\omega \in \Lambda^{\prime}} i^{-6} \omega^{-6}=-g_3 \implies g_3 = 0 \implies J(\tau) = 1
  .\]
  The equation of $X$ is $y^2 = x^3 - Ax$ for some $A$.
  - $\tau = \zeta_3$ yields $R = \ZZ[\zeta_3] = \ZZ_K$ for $K = \QQ(\zeta_3)$ and $R\units = \ts{\pm 1, \pm\zeta_3, \pm\zeta_3^2}\cong C_6$ and $j=0$, so $X$ has equation $y^2= x^3-B$ for some $B$.
  - If $\tau = 2i$ then $R= \ZZ[2i] < \ZZ_K$ is a proper subring for $K=\QQ(i)$ with conductor 2.

- CM is easy to detect in terms of $\tau$, but it's hard to relate $\tau$ and $j$ in general -- e.g. given an equation of a curve in $\PP^2$, it's hard to tell if it has CM.
  One approach goes through class field theory.
  - Some major results: for $X$ with CM and $R\da \Endo(X, P_0)$, let $K \da \QQ(i\sqrt{d})$ be the quotient field of $R$.
  Then $j$ is an algebraic integer, $K(j)/K$ is an abelian extension of degree $\size \Pic(R)$, and $j\in \ZZ \iff \size \Pic(R) = 1$, and there are exactly 13 such $j\dash$values.

:::

:::{.remark}
Positive characteristic stuff.
Let $X\slice k$ with $\characteristic k = p > 0$ with Frobenius $F:X\to X$.

- $F$ induces a map on homology $F^*: H^1(\OO_X)\selfmap$ which is not linear but is $p\dash$linear, i.e. $F^*(\lambda a) = \lambda^p F(a)$ for all \( \lambda\in k \).
- Since $X$ is elliptic, $h^1(\OO_X) = 1$, and since $k$ is perfect $F^*$ is either zero or bijective.
- Definition: if $F^* = 0$ say $X$ has **Hasse invariant 0** and $X$ is **supersingular**, otherwise if $F^*$ is bijective $X$ has **Hasse invariant 1**.
- The Hasse invariant criterion: if $X\embeds \PP^2$ as a cubic curve with homogeneous equation $f(x,y,z) = 0$, then $X$ has Hasse invariant zero iff the coefficient of $(xyz)^{p-1}$ in $f^{p-1}$ is zero.
  - Check that the ideal sheaf is $\OO_{\PP^2}(-3)$, use the SES $\OO_{\PP^2}(-3)\injectsvia{f}\OO_{\PP^2}\surjects \OO_X$.
  - Check $h^1(\OO_{\PP^2}) = h^2(\OO_{\PP^2}) = 0$, so the LES yields an isomorphism $H^1(\OO_X)\isovia{\delta} H^2(\OO_{\PP^2}(-3))$; the latter is 1-dimensional with basis ${1\over xyz}$.
  - Use this to compute $F^*$: let $F_1$ be Frobenius on $\PP^2$, then $F_1^*$ sends $\OO_X\to \OO_{X^p}$ where $X^p = V(f^p) \subseteq \PP^2$.
  Since $X\leq X^p$ is a closed subscheme, get a diagram:
  ![](figures/2023-01-03_18-12-50.png)
  and thus 
  ![](figures/2023-01-03_18-14-58.png)
  
  - Note ${1\over xyz}\mapsto \qty{1\over xyz}^p$ by $F_1^*$ and its image in $H^2(\OO(-3))$ is ${f^{p-1}\over (xyz)^p}$.
  - OTOH $H^2(\OO(-3)) = \gens{(xyz)\inv}$ and any monomial with any exponent $n\geq 0$ on $x,y,z$ must have coefficient zero.
  - Thus the image is $(xyz)\inv$ times the coefficient of $(xyz)^{p-1}$ in $f^{p-1}$, and the Hasse invariant is determined by whether or not this coefficient is zero.
- For $p\neq 2$ and $X: y^2 = x(x-1)(x- \lambda)$ with \( \lambda\neq 0, 1 \) the Hasse invariant of $X$ is zero $\iff h_p( \lambda) = 0$ where
\[
h_p( \lambda)\da \sum_{i=0}^k {k\choose i}^2 \lambda^i, \qquad k = {p-1\over 2}
.\]
- As a corollary, for any fixed $p$ there are at most $[p/12] + 2$ elliptic curves up to isomorphism over $k$ having Hasse invariant zero.

- A very interesting problem: fix the curve $X$ and vary $p$.

  - How to make sense of this: let $X = V(f) \subseteq \PP^2\slice \ZZ$ be a cubic curve where $f\in \ZZ[x,y,z]$ where $X\slice \CC$ is smooth.
    Then for almost all primes, the curve $X_{(p)} \subseteq \PP^2\slice{\FF_p}$ (where $X\to X_{(p)}$ corresponds to $f\mapsto f\mod p$) will be smooth over $\bar{\FF_p}$.
  - So one can consider
  \[
  B \da \ts{p \in \spec \ZZ \st X_{(p)} \text{ is smooth over } \bar{\FF_p},\, X_{(p)} \text{ has Hasse invariant } 0 }
  ,\]
  and it is a theorem that $B$ has **density** $1/2$, i.e.
  \[
  \lim_{x\to \infty} {\size\ts{p\in B\st p\leq x } \over \size\ts{p\in \spec \ZZ \st p\leq x} } = {1\over 2}
  .\]
  - More is true: if $X_{(p)}$ is smooth at $p$, then $X$ has Hasse invariant 0 $\iff p$ is ramified *or* $p$ remains prime in the imaginary quadratic field containing the ring of CM of $X$.
  - If $X$ does not have CM, $B$ has density zero but is still infinite.
  - Conjecture: $\size{p\in B\st p\leq x}\sim {\sqrt x\over \log x}$.

- An example: $X: y^2 = x^3-x$ has $j(X) = 1728$ and CM by $\tau = i$.
  - By the Hasse invariant criterion, for $p\neq 2$, one sets $k = {p-1\over 2}$ and looks at the coefficient of $x^k$ in $(x^2-1)^k$.
  It's zero when $k$ is odd, and if $k=2m$ it is $(-1)^m {k\choose m}$.
  - So the Hasse invariant is 1 when $p\equiv 1\mod 4$ and zero when $p\equiv 3 \mod 4$ (so Hasse invariant 0 iff $p$ is prime in $\ZZ[i]$).
  - Note $B = \ts{p\in \spec \ZZ \st p\equiv 3\mod 4}$. 
    By Dirichlet's theorem on primes in APs, this has density $1/2$, and there are infinitely many.

- An example: $X: y^2 = x(x-1)(x-2)$ has $j(X) = {2^6 \cdot 7^3\over 3^2}\in \QQ\sm\ZZ$.
  - Check $X_{(p)}$ is smooth at $p\neq 2,3$ and that the only $p\leq 73$ or so with Hasse invariant zero is $p=23$, so one might conjecture $B$ has density zero.
  - True because $j$ is not an integer, so $X$ doesn't have CM.

:::

:::{.remark}
On rational points:

- For $X\slice k$ with $k = \kbar$ and $P_0\in X$ a fixed point, embed $X\embeds \PP^2\slice k$ by $\abs{3P_0}$.
- If $X = V(f)$ where $f$ can be defined in a smaller field $k_0 \subseteq k$, we say $(X, P_0)$ is **defined over $k_0$**.
- In this case, $X(k_0)\leq X(k)$ is a subgroup. If $k=\CC$ and $k_0 = \QQ$, one can clear denominators to get a Diophantine equation.
- Mordell's theorem: $X(\QQ) \in \Ab\Grp^\fg$. Some examples:
  - $X: x^3 + y^3 = z^3$; since Fermat is true for $n=3$,
  \[
  X(\QQ) = \ts{\tv{1:-1:0}, \tv{1:0:1}, \tv{0:1:1}} \cong C_3
  ,\]
  and these are the three inflection points.
  - $X: y^2 + y = x^3-x$; taking $P_0 = \tv{0:1:0}$ for the identity, $X(\QQ) = \gens{P}\cong \ZZ$ where $P = \tv{0, 0}$ in affine coordinates.

:::