The chord–tangent group law
Compute sums on a smooth cubic while separating rational formulas from the proof of associativity.
On this page
The question for today
What you’ll understand
- Compute secant and tangent addition with all exceptional cases stated.
- Explain the reflection and multiplicity in the geometric construction.
- Relate associativity to degree-zero divisor classes rather than a drawing.
A line usually meets a cubic in three points counted with multiplicity. If two intersections are known, the third is algebraically determined, and reflection produces their sum. This turns a geometric construction into rational arithmetic. The construction also contains traps: a tangent counts one intersection twice, a vertical line passes through infinity, and a persuasive picture does not prove associativity. A complete group law must handle all those cases and extend regularly across the apparent denominators.
Intersect first, then reflect
Work with a smooth short Weierstrass curve over a field of characteristic different from two and three. For distinct affine points with different x-coordinates, the joining line has a well-defined slope. Substituting its equation into the cubic produces a monic cubic in x. Two roots are the known coordinates, so the sum of roots determines the third intersection coordinate. The group sum is the negative of that intersection, obtained by reversing its y-coordinate.
The minus sign is forced by the identity convention. A vertical line through P also passes through its reflection and O, so P plus its reflection must equal O. For this short model, negation is therefore (x,y) mapped to (x,−y). O acts as identity, and three collinear intersections sum to O when multiplicities are included. These statements concern projective intersections; ignoring O would leave the most basic cancellation operation undefined.
Tangency and vertical lines complete the recipe
For doubling, the tangent slope follows from implicit differentiation. Its denominator is nonzero when y is nonzero. The same coordinate formulas then apply with the two initial points equal. The tangent intersection carries multiplicity at least two; at a flex it carries multiplicity three, yielding three times that point equal to O. Over a general field these are algebraic intersection multiplicities, not statements about how a real graph happens to look.
When y=0, the point equals its own negative and its double is O. When two affine points share an x-coordinate, they are either the same point or opposite points in characteristic different from two; the vertical case is handled before dividing. Addition involving O is handled by the identity rule. A software implementation that applies one slope fraction indiscriminately will divide by zero precisely at legitimate group elements, rather than discovering a defect in the elliptic curve.
Associativity needs a structural argument
Commutativity is visible in the symmetric construction, but associativity compares several intersecting lines and is much less transparent. A reliable justification identifies a point P with the degree-zero divisor class of P minus O. Riemann–Roch in genus one makes this identification bijective over an algebraic closure, and addition of divisor classes is already associative. A line section has divisor of zeros equal to its three intersections and has pole divisor three O after taking a suitable ratio.
Consequently, the line construction matches addition in the divisor class group and inherits its group axioms. The formulas are defined over the coefficient field, so rational points are closed under the operation. This is stronger than observing that a few rational computations work. It also explains why changing the rational base point changes the displayed identity while leaving an isomorphic genus-one group structure. The next lesson examines the functions and divisors that make this proof precise.
Work it through
Double a rational point with a nonintegral answer
On E: y²=x³−2 over Q, double P=(3,5), retaining exact fractions.
The point lies on E because 25=27−2. Its tangent slope is 3·3²/(2·5)=27/10.
The new x-coordinate is (27/10)²−6=729/100−600/100=129/100.
The reflected y-coordinate is (27/10)(3−129/100)−5=4617/1000−5000/1000=−383/1000. Substitution gives 383²/10⁶=129³/10⁶−2, verifying the curve equation.
Twice P is (129/100,−383/1000). Rational closure does not imply integral coordinates remain integral.
The living notebook
Explore the model
Interactive lab · function
Inspect what a tangent slope measures
Pause and reset, then set both controls to coefficient a = 0.5 and position x = 1. Predict y = 0.5 and tangent slope one. For a fresh comparison, pause and reset, then set a = 0.5 and x = −1; predict the same height and slope minus one. The graph is the fixed parabola ax², not an elliptic curve. Use it to separate the tangent's local slope from the global third-intersection step, which must be calculated on the cubic itself.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forA derivative supplies a tangent line; a separate intersection calculation supplies the group sum.
Pause & reason
Check your understanding
Your turn
Try it yourself
On y²=x³−x over Q, add P=(0,0) and Q=(1,0). Then compute P+P and P+(P+Q) using the appropriate branches.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
The secant between different x-coordinates is horizontal. Each of the three roots of x³−x supplies a point with y=0.
Compare with the solution
The first slope is zero, so P+Q has x=−1 and y=0. Each y-zero point is its own inverse, giving P+P=O. Adding P to (−1,0) uses another horizontal secant: its sum has x=1 and y=0, hence P+(P+Q)=Q. This checks a particular associative identity, while the divisor-class argument supplies the general proof. The four points O,(0,0),(1,0),(−1,0) form a subgroup isomorphic to two copies of Z/2Z, rather than a cyclic group of order four.
Keep the idea
Take it with you
- The sum is the reflected third intersection, with intersection multiplicity counted.
- Vertical lines and y-zero doubling require explicit exceptional branches.
- Divisor classes justify associativity and rational closure of the geometric construction.
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