Weierstrass models and nonsingular cubics
Identify the smooth projective curve behind a cubic equation and distinguish an elliptic curve from a singular drawing.
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The question for today
What you’ll understand
- State the genus-one and rational-base-point definition of an elliptic curve.
- Check a short Weierstrass model and its projective point at infinity under the correct characteristic restrictions.
- Distinguish a singular model from an invariant of its smooth curve.
A cubic equation can conceal several different objects: its affine solutions, a projective closure, a singular scheme, or a smooth curve equipped with an identity point. Elliptic-curve arithmetic uses the last of these. The missing point at infinity is not a cosmetic addition; it will become the group identity and the location of the poles of the coordinate functions. Begin by checking smoothness and the field of definition before applying any addition formula or counting algorithm.
The curve includes a chosen rational point
An elliptic curve over a field k is a smooth projective geometrically integral curve of genus one together with a specified k-rational point O. Geometric integrality concerns the curve after extending to an algebraic closure, while rationality asks whether coordinates lie in k. A genus-one curve without a k-point is not an elliptic curve over k with an available identity. It may instead be a torsor under an elliptic curve, a distinction that later becomes central to descent.
For fields of characteristic different from two and three, every elliptic curve admits a short Weierstrass presentation. Homogenizing the cubic gives a plane projective equation with a unique point on the line at infinity, namely O. The affine chart omits precisely that point. A smooth plane cubic has genus one by the plane-curve genus formula. Thus, in this setting, checking the projective equation and its nonzero discriminant supplies the geometric hypotheses as well as a distinguished rational point.
Smoothness is a condition on all partial derivatives
On the affine chart, write F=y²−x³−Ax−B. A singular point must satisfy F=0, F_x=−3x²−A=0, and F_y=2y=0. Since two is invertible, it has y=0 and corresponds to a repeated root of the cubic in x. Eliminating that common root yields the discriminant condition. At infinity, the derivative of the homogeneous equation with respect to Z is nonzero at O, so that point is smooth as well.
A zero derivative in one coordinate does not make the curve singular: both affine partials must vanish. The point (0,0) on y²=x³−x has a vertical tangent, but F_x there is one. By contrast, (0,0) on y²=x³ has both partials zero and is a cusp. The cuspidal curve is not an elliptic curve despite having a cubic equation and an evident point. Chord constructions on singular cubics lead to different algebraic groups and require their own interpretation.
A model is a presentation, not every arithmetic invariant
Changing coordinates can alter coefficients and discriminant while preserving the elliptic curve over the chosen field. The j-invariant is unchanged, but equality of j-invariants guarantees isomorphism over an algebraic closure, not necessarily over k. Twists can have the same geometric curve and different rational points. Specifying the field and the identity-preserving isomorphism therefore matters whenever a computation is transported from one equation to another.
In characteristics two and three the divisions used to obtain a short equation may fail. The general Weierstrass equation includes xy, y, and x² terms and has its own discriminant formulas. One must not reuse the displayed short-model test as a universal classification in those characteristics. Later reduction arguments will also distinguish an integral equation with a singular special fiber from genuinely bad reduction of the curve, which is determined after appropriate local changes of model.
Work it through
Check the standard cubic and its missing point
Over Q, inspect y²=x³−x. Compute its discriminant and j-invariant, and check the projective point at infinity.
The coefficients are A=−1 and B=0. Thus 4A³+27B²=−4, giving discriminant 64, which is nonzero.
The j-invariant is 1728·(−4)/(−4)=1728. This records geometric isomorphism information rather than a complete classification over Q.
On Z=0 the projective equation forces X=0, leaving O=[0:1:0]. For F=Y²Z−X³+XZ², F_Z(O)=1, so this point is smooth.
The projective cubic with O is an elliptic curve over Q. Its affine vertical tangent at (0,0) does not contradict smoothness.
The living notebook
Explore the model
Interactive lab · matrix
Keep coordinate changes invertible
Pause and reset, then set both controls to diagonal scale a = 1 and shear b = 0. Predict determinant one and equal singular values. For a fresh comparison, pause and reset, then set a = 0.25 and b = 0; predict determinant one quarter, singular values one and one quarter, and condition number four. Both maps remain invertible. This real matrix instrument illustrates the requirement on coordinate changes; it does not plot a cubic or calculate its discriminant.
- Area scale · det A
- 1.4
- Singular values · σmax, σmin
- 1.597, 0.876
- Condition number · κ₂(A)
- 1.823
The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. 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 forAn invertible change can distort coordinates while preserving the underlying geometric object.
Pause & reason
Check your understanding
Your turn
Try it yourself
For y²=x³+2x+2, calculate the discriminant over Q. Reduce the same equation modulo five and show directly that (1,0) is singular. Explain why these conclusions concern different fields.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Compute 4A³+27B² first, then evaluate the equation and both affine partial derivatives modulo five.
Compare with the solution
The inner expression is 32+108=140, hence the discriminant is −2240 and the rational curve is smooth. Modulo five, (1,0) satisfies the equation because 1+2+2=0. Its partials are −3·1²−2=0 and 2·0=0 modulo five. The reduced model is therefore singular. Nonzero over Q does not mean invertible modulo every prime: the integer discriminant is divisible by five. Reduction and local minimality must be checked separately when deciding the curve's reduction type.
Keep the idea
Take it with you
- An elliptic curve includes smooth projective genus-one geometry and a rational identity point.
- The short Weierstrass discriminant test requires the stated characteristic restrictions.
- Coordinate equations and their discriminants can change while the curve remains isomorphic.
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