4.1 Elliptic curves

def: Elliptic curve

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &E \in \mathfrak E \\ &E = \cancel S(E) \\ &\text{gen}(E) = 1 \\ &O \in E \\ \hline \\ &(E,O) \in \mathcal E \\ &(E,O) - \text{elliptic curve} \end{align*}$$

note

We designate a special point $O$ to be able to define a group over elliptic curve later in this section.

Weierstrass form

Further we're going to use affine polynomials to describe projective curves. That means that we're actually describing a curve on a projective open set $U_z=\{[x,y,1], x,y \in \overline F \}$ and then there are some points of infinity that we can analyze using homogeneization.

For example affine equation $y-x^2=1$ describes a projective equation $yz-x^2=z^2$.

def: Weierstrass curve

$$\begin{align*} &\sphericalangle \\ &\mathbb P^2 \\ &f(x,y) = y^2+a_1xy+a_3y-x^3-a_2x^2-a_4x-a_6 \\ &a_i \in F \\ &C: I(C)=\lang f \rang_I \\ \hline \\ &C \in \mathcal W/F \\ &C - \text{is a curve in Weierstrass form} \\ &\text{eq}_W(C):=f - \text{Weierstrass curve equation} \end{align*}$$

Proposition 4.1.1: Elliptic curve is a Weierstrass curve

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ \hline \\ &\begin{align*} &(E,O) \in \mathcal E/F \implies \exists f, g \in F(E)_F, C \in \mathcal W / F: \\ &\phi: E \to \mathbb P^2, P \mapsto [f(P),g(P),1], O \mapsto [0,1,0], \\ &E \cong_{\phi}C \hspace{0.5cm} \tag{a}\\ &E \in \mathcal E/F, e_1 = \text{eq}_W(E), e_2 = \text{eq}_W(E)\implies \\ &\exists u \in F^*, r,s,t \in F: \\ &e_1(x,y)=e_2(u^2x+r, u^3y+su^2x+t) \hspace{0.5cm} \tag{b}\\ &C \in \mathcal W/F, C = \cancel S(C) \implies (C, [0,1,0]) \in \mathcal E \hspace{0.5cm} \tag{c}\\ \end{align*} \end{align*}$$

Proof

a.

The plan for the proof will be

1. Find a mapping to a curve in Weierstrass form
2. Prove that it's a morphism
3. Prove that it's surjective
4. Prove that it has degree 1
5. Hence it's isomopshism
6. Prove that image is smooth

From $(3.4.13)$ we have $\dim L(n(O))=n$. Note that $L(n(O))$ contains functions that don't have a pole more than of order $n$. So $\dim L(2(O))=2$ and it will have basis $\{1,f\}$ and $\dim L(3(O))=3$ and it will have a basis $\{1,f,g\}$ for some functions $f, g \in F(E)_F$. Note that $f$ has a pole of order 2 and $g$ has a pole of order 3. Thus $\{1,f, f^2, f^3, g, g^2, fg \}\subseteq L(6(O))$ But we know that $\dim L(6(O))=6$ so there's a linear relations between these fuctions:

$$A_1 + A_2f+A_3g+A_4f^2+A_5fg+A_6g^2+A_7f^3=0, \\A_i \in F, \exists A_i \ne 0\\$$

Note that $A_6A_7 \ne 0$, otherwise each other term would have a pole of different order so all $A_i$ must be $0$ in this case. Now assume $f=-A_6A_7f, g=A_6A_7^2g$ then dividing by $A^3_6A^4_7$ we'll get:

$$g^2+a_1fg+a_3g=f^3+a_2f^2+a_4f+a_6$$

So if we consider a map:

$$\phi: E \to \mathbb P^2, P \mapsto [f(P),g(P),1]$$

Then it's image will be a curve $C$ subject to Weierstrass equation. Since $E$ is smooth by $(3.4.3)$ it's a morphism. And since it's non-constant it's surjective by $(3.4.4)$. Finally note that $\text{ord}_Og = -3, \text{ord}_Of = -2$ so if we assume $g=g_1/g_2, f=f_1/f_2$ we can make $\phi(O)=[g_2(O)\cdot f_1(O)/f_2(O),g_1(O), g_2(O)] = [0,1,0]$

Next, we want to show that $\deg \phi = 1$. By definition this means that $[F(E):\phi^*F(C)]=1$. Let's recall how we define pullback of a function $h \in F(C)$:

$$h(x,y,z)=h_1(x,y,z)/h_2(x,y,z) \\ \phi^*h(P) = h_1(f(P),g(P),1)/h_2(f(P),g(P),1)$$

This means that $\phi^*F(C)$ will have some rational functions of $f$ and $g$, in other words $\phi^*F(C)=F(f,g)$. Additional note for clarity of notation: $F(C)$ means a functional field of a curve $C$ so here $F$ stands for functional, while $F(f,g)$ is a field extension of the base field $F$, so $F$ here is a field.

Now let's consider a map: $\phi_f: E \to \mathbb P^1, P \mapsto [f(P),1]$. In this case $\phi_x^{-1}([1,0])=O$ (since $f$ has the only pole in $O$) and $e_{\phi_f}(O)=\text{ord}_O\phi_f^*t_{[1,0]}=-\text{ord}_Of=2$. So by $(3.4.5)$ we have $\deg \phi_f=2$. Knowing that $\phi_f^*F(\mathbb P^1)=F(f)$ implies $[F(E):F(f)]=2$.

Similarly $[F(E):F(g)]=3$. But since we have tower of extensions $F(E)/F(f,g)/F(f)$ and $F(E)/F(f,g)/F(g)$ we know that:

$$2=[F(E):F(f)]=[F(E):F(f,g)][F(f,g):F(f)] \\ 3=[F(E):F(g)]=[F(E):F(f,g)][F(f,g):F(g)] \\$$

This implies that $[F(E):F(f,g)]=1$ and thus $\deg \phi = 1$. So by $(3.4.4)$ we have $E \cong_\phi C$.

Now let's prove $C$ is smooth. Assume otherwise - there exist some singular point. By linear change of variables we can assume it's $(0,0)$. Then using partial derivatives criterion we can check that the equation in this case is:

$$y^2+a_1xy=x^3+a_2x$$

Then we can make a rational map $\psi: C \to \mathbb P^1, (x,y) \mapsto [x,y]$. Then $\deg \psi = 1$ since it has an inverse $[1,t] \mapsto (t^2+a_1t-a_2, t^3+a_1t^2-a_2t)$

So we built a rational map of degree 1 $\phi \circ \psi: E \to \mathbb P^1$. So by $(3.4.4)$ it's an isomopshism. But it cannot be the case since $\text{gen}(E)=1$ and by $(3.4.13)$ $\text{gen}(\mathbb P^1)=0$. So we arrived at contradiction and thus $C$ is smooth

b.

As in $(a)$ we consider two maps $\phi=[f,g,1], \phi'=[f',g',1]$ where both $f,g$ and $f',g'$ are the bases for $L(2(O))$ and $L(3(O))$ respectively. Since they're the bases of the same spaces we can write:

$$f = u_1f'+r, g=u_2g'+s_2f'+t$$

Since in Weierstrass form $f^3$ and $g^2$ has coefficents $1$ it follows $u_1^3=u_2^2$. Letting $u=u_2/u_1$ and $s=s_2/u^2$ we have:

$$f=u^2f'+r, g=u^3g'+su^2f'+t$$

Thus defining $x = f(P), y=g(P), P \in E$ we finish the proof.

c.

Consider some point $P_0=(x_0,y_0)$. Denote $F(x,y):=\text{eq}(C)$. Since $F(x,y)\equiv 0$ we know that $0=dF(x,y)=F_x(x,y)dx+F_y(x,y)dy$ so we can define

$$\omega:=\frac{d(x-x_0)}{F_y(x,y)}=-\frac{d(y-y_0)}{F_x(x,y)}$$

Note that $\omega$ cannot have pole in $P$, otherwise $F_x(x_0,y_0) = F_y(x_0,y_0)=0$ and $P$ is singular. Consider a map:

$$\phi: E \to \mathbb P^1, [x,y,1] \to [x,1]$$

$F(E)=F(x,y), \phi^*F(\mathbb P^1)=F(x)$. Since in Weierstrass form $y^2$ is a polynomial of $x$ we'll have $\deg \phi=[F(x,y):F(x)]=2$. The other way to think about it is that the preimage of any point in $\mathbb P^1$ is two points (because we solve for $y^2$) or 1 points in which case it's ramified with index 2.

The uniformizer in $\mathbb P^1$ at point $\phi(P)=[x_0,1]$ is $t_\phi(P)=x-x_0$, so $\phi^*t_{\phi(P)}=x-x_0$. Since $2=\deg \phi=\sum_{Q \in \phi^{-1}(\phi(P))}\text{ord}_Q(\phi^*t_{\phi(Q)})$ and one of $Q$ is $P$ itself, we have $\text{ord}_{P}(x-x_0) \le 2$. It also follows that we either have one point in pre-image and then $\text{ord}_{P}(x-x_0)=2$ or two points and then $\text{ord}_{P}(x-x_0)=1$. In the first case the polynomial $F(x_0,y)$ has double root and so $F_y(x_0,y_0)=0$, in the second case two distincts roots and $F_y(x_0,y_0) \ne 0$. In any case:

$$\text{ord}_P(\omega) = \text{ord}_P(x-x_0)-\text{ord}_P(F_y)-1=0$$

Note that in case $F_y(x_0,y_0)=0$ we know that $\text{ord}_P(F_y) \ge 1$ but it cannot be strictly greater because that order of $\omega$ would be strictly less than $0$ and we establised above that in cannot be the case.

So we proved that $\omega$ has zero order in any $P$ in affine plane. One last thing is we need to consider its order in $O$. We know that $\text{ord}_Ox=-2$ and $\text{ord}_Oy=-3$ so $x=ut_O^{-2}, y = vt_O^{-3}$ where $u(O),v(O)\ne 0, \infty$ (units in local ring at $O$) and:

$$\omega = \frac{dx}{2y+a_1x+a_3}=\frac{-2ut^{-3}+u't^{-2}}{2vt^{-3}+a_1ut^{-2}+a_3}dt=\\ =\frac{-2u+u't}{2v+a_1ut+a_3t^3}dt$$

Since $u$ is regular, by $(3.4.11.f)$ $u'$ is regular as well, $t(O)=0$ meaning the neither numerator nor denominator goes to $0$ at $O$ and thus $\text{ord}_O(\omega)=0$. So we proved:

$$\text{div}(\omega) = 0$$

Finally by $(3.4.13.b)$ we have

$$2\text{gen}(C)-2=\deg \text{div} \omega = 0 \implies \\ \text{gen}(C)=1$$

and taking $O = [0,1,0]$ we finish the proof

$\square$.

note

$(b)$ show that having this form of linear transformation is necessary condition for the curve to be the same

Short Weierstrass form

Consider a curve in Weierstrass form:

$$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$$

We can simplify the equation using linear transformations. First, let's map $y \mapsto \frac{1}{2}(y-a_1x-a_3)$. Note that transformation matrix is $\begin{pmatrix}-a_1 & 1 \\ 1 & 0\end{pmatrix}$ with $\det A=1$ so it's invertible and thus isomopshism. Then the equation becomes:

$$y^2 = 4x^3+b_2x^2+2b_4x+b_6, \\ b_2=a_1^2+4a_4 \\ b_4=2a_4+a_1a_3 \\ b_6=a_3^2+4a_6$$

We define some auxiliary quantities for future reference:

$$b_8=(b_2b_6-b_4^2)/4 \\ c_4=b_2^2-24b_4 \\ c_6=-b_2^3+36b_2b_4-216b_6 \\ \Delta=(c_4^3-c_6^2)/1728 \\ j=c_4^3/\Delta \\ \omega = \frac{dx}{2y+a_1x+a_3}=\frac{dy}{3x^2+2a_2x+a_4-a_1y}$$

Next, we can eliminate the $x^2$ term by applying the substitution $(x, y) \mapsto \left(\frac{x-3b_2}{36}, \frac{y}{108}\right)$. Again the transformation matrix is $\begin{pmatrix}1/36 & 0 \\ 0 & 1/108\end{pmatrix}$ so this is isomopshism. Then if $\text{char} F \ne 2,3$ we get:

$$y^2 = x^3 + Ax + B, \\ A=-27c_4, B=-54c_6$$

This form of the elliptic curve is called a Short Weiestrass form that we'll denote $\text{eq}_{Ws}(E)$. Note that when we transform the curve from Weierstrass curve to Short Weierstrass curve we actually change the curve so it's not the same curve anymore. But we do it via $\mathbb P^2$ automorphism so we can actually take some points on the original curve, transform these points via automorphism to Short Weierstrass curve, do some operations there and then come back to original curve via the inverse automorphism.

Thus essentially we can consider only Short Weiestrass curves. If it happens that the curve is not Short Weierstrass then we just use the algorithm above and thus we need to learn to deal with only the Short Weierstrass curves.

Singularity and invariants

Consider some transformation in the form of $(4.1.1.b)$:

$$\begin{align*} x' &= u^2 x + r, \\ y' &= u^3 y + u^2 s x + t \end{align*}$$

In can be checked that in this case we have the following transformations of the coefficents:

$$\begin{align*} u a'_1 &= a_1 + 2s, \\ u^2 a'_2 &= a_2 - s a_1 + 3r - s^2, \\ u^3 a'_3 &= a_3 + r a_1 + 2t, \\ u^4 a'_4 &= a_4 - s a_3 + 2r a_2 - (t + rs)a_1 + 3r^2 - 2st, \\ u^6 a'_6 &= a_6 + r a_4 + r^2 a_2 + r^3 - t a_3 - t^2 - r t a_1, \\ u^2 b'_2 &= b_2 + 12r, \\ u^4 b'_4 &= b_4 + r b_2 + 6r^2, \\ u^6 b'_6 &= b_6 + 2r b_4 + r^2 b_2 + 4r^3, \\ u^8 b'_8 &= b_8 + 3r b_6 + 3r^2 b_4 + r^3 b_2 + 3r^4, \\ u^4 c'_4 &= c_4, \\ u^6 c'_6 &= c_6, \\ u^{12} \Delta' &= \Delta, \\ j' &= j, \\ u^{-1} \omega' &= \omega. \end{align*}$$

For the Short Weierstrass curve we have the following values for $\Delta, j$:

$$\Delta = -16(4A^3+27B^2) \\ j=-1728\frac{(4A)^3}{\Delta}$$

Proposition 4.1.2: Singularity criterion

$$\begin{align*} &\sphericalangle \\ &(E,O) \in \mathcal E \\ \hline \\ &E = \cancel S(E) \iff \Delta \ne 0 \end{align*}$$

Proof

$\implies$

Note that in the table of transformation above we see that at this particular transformation $\Delta$ just scales by $u^{12}$. It can be shown that the transformation $y \mapsto \frac{1}{2}(y-a_1x-a_3)$ does the same thing. So it doesn't change $\Delta$ being zero or non-zero. Thus we can consider the equation in the form:

$$y^2 = 4x^3+b_2x^2+2b_4x+b_6 \\$$

The point $(x_0, y_0)$ is singular iff partial derivatives are zero, that is

$$2y_0=12x_0^2+2b_2x_0+2b_4=0$$

So singular point has a form $(x_0, 0)$, moreover polynomial $p(x):=4x^3+b_2x^2+2b_4x+b_6$ has $p(x_0)=0$, $p'_x(x_0)=0$ so $x_0$ should be a root of order at least 2 and thus discriminant of the cubic should go to zero. But it's discriminant is $16\Delta$ so we finished the proof. As a side note since cubic can have only one root of order 2, there could be only one singular point in Weiestrass.

$\impliedby$

Consider an elliptic curve in the Long Weierstrass form:

$$f(x,y)=y^2+a_1xy+a_3y-x^3-a_2x^2-a_4x-a_6$$

And its projective form:

$$F(x,y,z)=y^2z+a_1xyz+a_3yz^2-x^3-a_2x^2z-a_4xz^2-a_6z^3$$

At point $O$ it has $\partial F/\partial z(O)=1$, so $O \in \cancel S(E)$.

Next, consider some singular point $P \notin \cancel S(E)$ and $P=(x_0,y_0)$. We can consider transformation $x'=x+x_0, y'=y+y_0$. Under this transformation we have $\Delta$ not changing (see above for $u=1, s = 0, r=x_0, t=y_0$). So we may assume that the singular point is $(0,0)$. Since the curve is singular at $(0,0)$ it means that $f'_x(0,0)=0$ and $f'_y(0,0)=0$. Since $(0,0) \in E$ it also means that $f(0,0)=0$. But we know that

$$a_6=-f(0,0)=0 \\ a_4=-f_x(0,0)=0 \\ a_3=f_y(0,0)=0 \\$$

So the equation becomes:

$$f(x,y)=y^2+a_1xy-x^3-a_2x^2$$

In this case we can calculate:

$$b_2=a_1^2+4a_4=a_1^2 \\ b_4=2a_4+a_1a_3=0 \\ b_6=a_3^2+4a_6=0 \\ b_8=(b_2b_6-b_4^2)/4=0 \\ c_4=b_2^2-24b_4=a_1^4 \\ c_6=-b_2^3+36b_2b_4-216b_6=-a_1^6 \\ \Delta=(c_4^3-c_6^2)/1728=(a_1^{12}-a_1^{12})/1728=0 \\$$

So we proved $\exists P \notin \cancel S(E) \implies \Delta = 0$

$\square$

Proposition 4.1.3: j-invariant

$$\begin{align*} &\sphericalangle \\ &(E,O), (E',O') \in \mathcal E \\ \hline \\ &\begin{align*} & E \cong E' \iff j(E)=j(E') \hspace{0.5cm} \tag{a} \\ & \forall j \in \overline F: \exists (E,O) \in \mathcal E: j(E)=j \hspace{0.5cm} \tag{b} \\ \end{align*} \end{align*}$$

Proof

a.

$\implies$

If we take two Weierstrass forms of $E_1$ and $E_2$ they will be isomorphic. Thus by $(4.1.1.b)$ they're related by transformation that doesn't change the j-invariant.

$\impliedby$

We will prove for $\text{char } F\ge 5$ and curves in short Weierstrass form:

$$E: y^2=x^3+Ax+B \\ E': y^2=x^3+A'x+B' \\$$

Then

$$\frac{(4A)^3}{4A^3 + 27B^2} = \frac{(4A')^3}{4A'^3 + 27B'^2} \implies \\ A^3B'^2=A'^3B^2$$

We want to build isomopshism in the form:

$$x \mapsto u^2x', y \mapsto u^3y'$$

Consider 3 cases:

1. $A = 0\, (j=0)$ then since $\Delta \ne 0$ it follows that $B\ne 0$ so $A'=0$ so we can make an isomopshism using $u = (B/B')^{1/6}$

2. $B = 0\, (j=1728)$. Then likewise $B'=0$ so we can take $u = (A/A')^{1/4}$

3. $A,B \ne 0\, (j\ne 0, 1728)$. Then $A', B' \ne 0$ because if one of them is $0$ then another one is $0$ as well and $\Delta = 0$ which is a contradiction of smoothness of $E$. So we can take $u = (A/A')^{1/4}=(B/B')^{1/6}$

b.

If $j_0 \ne 0, 1728$ then consider equation:

$$E: y^2+xy=x^3-\frac{36}{j_0-1728}x-\frac{1}{j_0-1728}$$

In this case $j(E)=j_0, \Delta=\frac{j_0^3}{(j_0-1728)^3}$. The following curves work for other values of $j$:

$$E: y^2+y=x^3, \Delta = -27, j=0 \\ E: y^2=x^3+x, \Delta = -64, j=1728 \\$$

$\square$

Proposition 4.1.4: Invariant differential

$$\begin{align*} &\sphericalangle \\ &(E,O) \in \mathcal E \\ &\omega = \frac{dx}{2y+a_1x+a_3}=\frac{dy}{3x^2+2a_2x+a_4-a_1y} \\ \hline \\ &\text{div}\omega = 0 \end{align*}$$

Proof

Was proved in $(4.1.1.c)$

$\square$