Elliptic Curves

Conic

\(\textbf{Definition: }\)A conic is the solution set to a quadratic euqation in two variables, such as \(x^2+y^2=1\).

There is no garantee that there is a cubic is rational.

\(\textbf{Example. }\)\(x^3+y^3=1\) has no rational points. (Not obvious)

We will focus on the ratioanl points on the unit circle \(x^2+y^2=1\).

Give a unit circle \(x^2+y^2=1\), we can find a rational point on the circle, \(P = (-1, 0)\).

If \(Q\) is a rational coordinate on the unit circle, then the line \(L\) passing through \(P\) and \(Q\) has rational slope.

If a line through \(P\) has rational slope, then the other intersection with unit circle is a rational point.

Activity: Find the rational points on the unit circle. The line \(y = t(x+1)\) intersects the unit circle at \(P=(-1, 0)\).

\[ x^2+y^2=1, \] \[ x^2+t^2(x+1)^2=1, \] \[ (1+t^2)x^2+2t^2x+(t^2-1)=0, \] \[ x^2 +\frac{2t^2}{1+t^2}x+\frac{t^2-1}{1+t^2}=0, \] \[ x^2 +\frac{2t^2}{1+t^2}x+\frac{t^4}{(1+t^2)^2}=-\frac{t^2-1}{1+t^2}+\frac{t^4}{(1+t^2)^2}, \]

Irreducibility

\(\textbf{Definition: }\)Let \(C: f(x, y)=0\) be a conic. \(C\) is irreducible if \(f(x, y)\) cannot be factored (Over \(\mathbb{C}\)).

\(\textbf{Example. }\) \(x^2+y^2=1\) is irreducible. Since \(f(x, y) = x^2+y^2-1\) cannot be factored.

\(\textbf{Example. }\) \(x^2+y^2=0\) is reducible. Since \(f(x, y) = x^2+y^2\) can be factored as \(f(x, y) = (x+iy)(x-iy)\).

\(\textbf{Theorem 1. }\) Let \(C: f(x, y)=0\) be an irreducible conic with a rational point. Then there exist rational functions \(x=a(t), y=b(t)\) such that \[ \{(a(t), b(t))\;|\;t\in\mathbb{Q}\} \] is the set of all rational points of \(C\) except for one point.

Example 1

\(\textbf{Example. }\) \(x^2+y^2=7\) has no rational points.

Legendre's Theorem

\(\textbf{Theorem 2. }\) Let \(A, B, C\) be square-free integers. \[ Ax^2+By^2+CZ^2=0 \] has a non-trivial solution in \(\mathbb{Q}\) if and only if

• \(A, B, C\) don't all have the same sign,
• \(-AB\) is a square module \(|C|\)
• \(-AC\) is a square module \(|B|\)
• \(-BC\) is a square module \(|A|\)

Affine Space

\(\textbf{Definition: }\)Affine space over \(\mathbb{Q}\), denoted \(\mathbb{A}^n(\mathbb{Q})\), is defined as \(\mathbb{A}^n(\mathbb{Q}) = \{(x_1, \ldots, x_n) \mid x_i \in \mathbb{Q}\}\).

\(\textbf{Definition: }\)Projective space over \(\mathbb{Q}\), denoted \(\mathbb{P}^n(\mathbb{Q})\), is defined as \(\mathbb{P}^n(\mathbb{Q}) = \{(x_0, \ldots, x_n) \mid x_i \in \mathbb{Q}\}\).

Projective space over \(\mathbb{Q}\), denoted \(\mathbb{P}^n(\mathbb{Q})\), is defined as \[ \mathbb{P}^n(\mathbb{Q}) = (\mathbb{A}^{n+1}(\mathbb{Q}) - \{(0, \ldots, 0)\}) / \sim, \] where \((x_0, \ldots, x_n) \sim (y_0, \ldots, y_n)\) if and only if there is some \(\lambda \in \mathbb{Q}\) such that \(x_i = \lambda y_i\) for all \(i\), \(0 \leq i \leq n\). Projective space points are written with brackets: \[ [x_0, \ldots, x_n] \in \mathbb{P}^n(\mathbb{Q}). \] This gives homogeneous coordinates for the point. Most important for us: - \(\mathbb{A}^1\) (affine line (usual line)) - \(\mathbb{P}^1\) (projective line) - \(\mathbb{A}^2\) (affine plane (usual plane)) - \(\mathbb{P}^2\) (projective plane)

Projective Curve

\(\textbf{Definition. }\)The projective curve defined by \(f(x,y,z) = 0\) is the set of points \((x,y,z)\in\mathbb{P}^2\) such that \(f(x,y,z) = 0\).

\(\textbf{Example. }\)The projective curve defined by \(y^2z-x^3-z^3 = 0\) is a cubiic projective curve.

Relationship between Projective Curve and Affine Curve

Fix \(\mathbb{P}^2 = \mathbb{A}^2\cup \mathbb{P}^1\) where \[ \mathbb{P}^1 = \{(x,y,0):\;x,y\in\mathbb{Q}\}\qquad \text{and}\qquad \mathbb{A}^2 = \{(x,y,1):\;x,y\in\mathbb{Q}\}. \] The curve \[ D_z = \{(x,y,1)\in\mathbb{A}^2:\;f(x,y,1) = 0\} \] is the affine part of porjective curve \(D\) where \[ D = \{(x,y,z)\in\mathbb{P}^2:\;f(x,y,z) = 0\}. \] The points on \(D\) in \(\mathbb{P}^1=\{[x, y, 0]\}\) are called the points at infinity of \(D\). THe curve \(D_Z:=D\cap\mathbb{A}^2\ with equation f(x,y,1)=0\) is called the affine part or dehomogenization of \(D\).

Homogenization

\(\textbf{Definition. }\)If \(C : f(x, y) = 0\) is an affine curve, the projective closure or homogenization of \(C\) is the projective curve \(D : F(X,Y,Z) = 0\) with just enough \(Z\)'s to make \(f\) homogeneous.

\(\textbf{Example. }\)\(f(x, y) = x^3+xy+y+1\) and \(F(X, Y, Z) = X^3+XYZ+YZ^2+Z^3\).

Decomposition of Projective Curve

Let \(D: F(X, Y, Z)=0\) be a projective curve.

\(\textbf{Example. }\)Let \(D\) be \[ D: Y^2Z - X^3 - XZ^2 = 0. \] Let \(D_Z\) be the affine part of \(D\). Then \[ D_Z: y^2 - x^3 - x = 0. \] \[ \frac{\partial f}{\partial x} = -3x^2 - 1, \qquad \frac{\partial f}{\partial y} = 2y. \] For singular points, we have \[ \frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = 0, \] which implies that \(x = \pm \sqrt{\frac{1}{3}}\) and \(y = 0\). None of two is on \(D_Z\), so \(D_Z\) has no singular points. For points at infinity, we have \[ F(X, Y, 0) = -X^3 = 0, \] which implies that \(X = 0\). Hence, the points at infinity are \([0, 1, 0]\). Thus, we can get a new line \(D_Y\) \[ D_Y: Z - X^3 - XZ^2 = 0. \] We can get \[ \]

\(\textbf{Lemma. }\)If \(F(X, Y, Z)\) is a homogeneous polynomial, then \[ X\frac{\partial F}{\partial X} + Y\frac{\partial F}{\partial Y} + Z\frac{\partial F}{\partial Z} = \deg F\cdot F(X, Y, Z). \] where \(\deg F\) is the degree of \(F\).

\(\textbf{Theorem. }\)Let \(D: F(X, Y, Z) = 0\) be a projective curve with non-singular points \(P\), then the tangent line is given by \[ \frac{\partial F}{\partial X}(P)X + \frac{\partial F}{\partial Y}(P)Y + \frac{\partial F}{\partial Z}(P)Z = 0. \]

\(\textbf{Proof. }\)Without loss of generality, we can assume that \(P = [x_0, y_0, 1]\). Then

Bezout's Theorem

\(\textbf{Theorem. }\)Let \(D_1, D_2\) be two projective curves of degree \(d_1, d_2\) respectively. If \(D_1\) and \(D_2\) have no common components, then \(D_1\) and \(D_2\) intersect in \(d_1d_2\) points.

\(\textbf{Note. }\)Not true in general that a line with rational slope drawn through a rational point on a cubic intersects the cubic in another rational point.

Let \(D\) be a projecitive non-singular cubic curve with rational coefficients. Define the operation \[ \oplus: D(\mathbb{Q})\times D(\mathbb{Q})\to D(\mathbb{Q}) \] as follow:

• If \(P\neq Q\), then \(P\oplus Q\) is the third point of intersection of the line through \(P\) and \(Q\) with \(D\).
• If \(P = Q\), then \(P\oplus Q\) is the tangent line to \(D\) at \(P\) intersected with \(D\) at a third point.

Another version of operation:

Let \(D\) be a projective nonsingular cubic curve with rational coefficients and a fixed rational point \(O\). Define the chord-tangent law \[ + : D(\mathbb{Q})\times D(\mathbb{Q})\to D(\mathbb{Q}) \] as follows: given \(P, Q\in D(\mathbb{Q})\),

• Calculate \(R = P\oplus Q\).
• Then \(P+Q = R\oplus O\)

Weierstrass Form

\(\textbf{Definition. }\)A cubic curve \(D\) is in Weierstrass form if it is of the form \[ D: y^2 = x^3 + Ax^2 + Bx + C. \]

• Weierstrass form is essetially the form for irreducible cubics with a flex at infinity with tangent line \(Z = 0\).
• Every irreducible cubic is birationally equivalent to a cubic in Weierstrass form.
• There is no always a bijection between the sets of rational points, but missing points can be easily identified.

\(\textbf{Example. }\) \[ y^2 = x^3 -4x+4. \] \[ y^2 = x^3 \]

Elliptic Curve

\(\textbf{Definition. }\)An elliptic curve \(E over \mathbb{Q}\)is a projective nonsingular cubic curve with with a chosen rational point \(O\in E(\mathbb{Q})\).

All elliptic durves have at least one rational point.

Let \(E\) be the elliptic curve \[ E: y^2 = x^3 - 4x + 4. \] Let's add \(P = (1, 1)\) and \(Q = (2, 2)\).

The equation of the line joining \(P\) and \(Q\) is \[ y = x. \] Subsititute this line formula in for y in the formula for \(E\), we get \[ x^2 = x^3 - 4x + 4. \] \[ x^3 - x^2 - 4x + 4 = 0. \] \[ (x-1)(x^2-4) = 0. \] \[ x = 1, 2, -2. \] \[ y = 1, 2, -2. \] \[ R = (1, 1), (2, 2), (-2, -2). \] Hence, we can get the third rational point. Reflect \(R\) in the x-axis, we get \[ P + Q = (-2, 2). \]

\(\textbf{Addition formulas. }\)Let \( P = (x_1, y_1) \) and \( Q = (x_2, y_2) \) be (affine) points on \( E: y^2 = x^3 + ax^2 + bx + c \).

• Say \( P \neq -Q \). Let the line joining \( P \) and \( Q \) have equation \( y = \lambda x + \nu \). Then \( P + Q = (x_3, y_3) \) where

• Say \( P \neq -P \). The x-coordinate of \( P + P \) is

(This is called the duplication formula.)

• Let \( A \) be an abelian group. An element \( x \in A \) has order \( n \), denoted \( \text{{ord}}(x) = n \), if \( n \) is the smallest positive integer such that \( nx = 0 \).
• Define \( A[n] = \{ x \in A \,|\, nx = 0 \} \).
• Any element of \( A \) of finite order is called a torsion element. The set of torsion elements of \( A \) is a subgroup of \( A \), denoted \( A_{\text{{tors}}} \).
• Let \( E \) be an elliptic curve (over \( \mathbb{Q} \)).
• Questions for us:

• What is \( E(\mathbb{Q})_{\text{{tors}}} \)? Is it finite? Can we find all the points? What is the group structure?
• Today: Points of order 2 and 3

• A point \( P \) on the elliptic curve \( E: y^2 = f(x) \) has order 2 if and only if \( P = (x_0, 0) \). In particular, \( E(\mathbb{Q})[2] \) has 1, 2, or 4 elements, corresponding to whether \( f(x) \) has 0, 1, or 3 rational roots, respectively.
• What about order 3?

The polynomial \( \psi_3(x) = 3x^4 + 4ax^3 + 6bx^2 + 12cx + 4ac - b^2 \) is the 3-division polynomial for \( E \).

Theorem:

1. The nonidentity point \( P \in E(\mathbb{Q}) \) has order 3 if and only if \( \psi_3(x(P)) = 0 \).
2. \( \psi_3 \) has no repeated roots.
3. E has exactly eight points of order 3 with complex coordinates.

Discriminant

\(\textbf{Definition. }\)Let \(E\) be an elliptic curve in Weierstrass: \[ f(x) = x^3 + Ax^2 + Bx + C. \] The discriminant of \(E\) is \[ \Delta = -4A^3C + A^2B^2 + 18ABC - 4B^3 - 27C^2. \]

\(\textbf{Theorem. }\)Set \[ f(x) = x^3 + Ax^2 + Bx + C, \] with \(A, B, C\in\mathbb{Z}\), set \[ E: y^2 = f(x), \] and let \(D\) be the discriminant of \(f(x)\). Suppose \(P = (x_0, y_0)\in E(\mathbb{Q})\) has finite order. Then

• \(x_0, y_0\in\mathbb{Z})
• Either \(y_0 = 0\) or \(y_0\;|\;D\)

\(\textbf{Corollary: }\)\(E(\mathbb{Q})\) is finite.

\(\textbf{Example. }\)Find all points of finite order on \[ E: y^2 = x^3 + x^2 +3x - 1. \] We first calculate the discriminant of \(E\): \[ D = -171. \] Are there points with \(y_0 = 0\)? No.

Mazur's Theorem

\(\textbf{Theorem. }\)Let \(E\) be an elliptic curve over \(\mathbb{Q}\). Then \(E(\mathbb{Q})_{\text{{tors}}}\) is isomorphic to one of the following fifteen groups: \[ \text{Trivial }(E(\mathbb{Q})_{\text{{tors}}}\cong\{O\}), \] \[ \mathbb{Z}/N\mathbb{Z}\text{ for some }\mathbb{N}\text{ with }2\leq N \leq 12\text{ except }N = 11. \] \[ \mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2N\mathbb{Z}\text{ for some }N\text{ with }1\leq N \leq 4. \]

\(p\)-adic

\(\textbf{Definition. }\)Let \(p\) be a prime number. The \( p \)-adic valuation is the function \[ \text{ord}_p : \mathbb{Q} \rightarrow \mathbb{Z} \cup \{\infty\} \] defined as follows:

• \(\text{ord}_p(0) = \infty\)
• For nonzero \(x\in\mathbb{Q}\) \[ x = \frac{m}{n}p^\nu \] with \(m, n \in \mathbb{Z}\) and \(p \nmid m, n\). Then \[ \text{ord}_p(x) = \nu. \]
• A number with a large negative \( p \)-adic valuation is \( p \)-adically "big" (lots of \( p \)'s in the denominator).
• A number with a large positive \( p \)-adic valuation is \( p \)-adically "small".
• \(x \in \mathbb{Q}\) is an integer if and only if \(\text{ord}_p(x) \geq 0 \)for all \(p\).

References

• First Initial. Last name, "Book name", Publications, Edition, Year.