So I am studying elliptic curves over finite fields and I am a little confused about something. In some texts I see a "long" Weierstrass equation and in some I see a "short" Weierstrass equation, what is the difference between the two? Are they equivalent?

"normal" form Weierstrass equation: $$y^2=x^3+ax+b$$ where $$a,b ∈Z$$

Also since I am considering finite fields should $a,b∈F_q$?

long Weierstrass equation: $$E:y^2+a_1 xy+a_3 y=x^3+a_2 x^2+a_4 x+a_6$$ where $$a_1,a_2,a_3,a_4,a_5,a_6∈F_q$$

Thanks for the clarification!

  • $\begingroup$ You can go from the "long" form to the short form by atransformation $\endgroup$ Apr 30, 2015 at 17:53
  • $\begingroup$ Sorry, what transformation? $\endgroup$
    – Math Major
    Apr 30, 2015 at 17:54
  • $\begingroup$ @MathMajor you should revisit this question. The answer marked as "correct" is not correct. $\endgroup$ May 4, 2015 at 2:19

2 Answers 2


The correct transformations are the following (assuming the characteristic of the field of definition is not $2$ or $3$). First change $y\longrightarrow y-(a_1x+a_3)/2$, so the new equation has the form $$y^2=x^3+Ax^2+Bx+C.$$ And now change $x\longrightarrow x-A/3$, so that the new equation has the form $$y^2=x^3+ax+b.$$ Clearly, both changes of variables are invertible, so their composition is also invertible. This means that the rational points on both equations are in bijective correspondence.

Be warned that if your field of definition is $\mathbb{F}_q$ and $q$ is divisible by $2$ or $3$, then it may not be possible to find a short Weierstrass form for your elliptic curve. Clearly, the first change of variables above may not work in char 2, and the second one may not work in char 3.


You can go from the "long" form to the short form by the transformation $$z=y+\frac12 a_1 x$$

Since you will be interested in rational solutions, and since $a_1$ is rational, studying the short form tells you what you need to know about the long form as well.

  • $\begingroup$ That's not correct. This transformation will only take care of the xy term, but will not take care of the y term, or the x^2 term. $\endgroup$ Apr 30, 2015 at 18:43

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