Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In page 22-23 of Rational Points on Elliptic Curves by Silverman and Tate, authors explain why is it possible to put every cubic curve into Weierstrass Normal Form. Here are relevant pages: (My question is at the end; I have put a red line across the point I am interested in)

enter image description here

enter image description here

Could someone explain to me how to "work out" the algebra part?

This seems pretty important derivation actually. I am afraid I don't even know how to start the algebra. I would appreciate any suggestions/hints. Thank you :)

Related. There is a close question I found here on MSE. However, that particular question asks more about geometrical insight. My question is how does one explicitly go from the general cubic to the equation $xy^2+(ax+b)y=cx^2+dx+e$. By the way, the equation for the general cubic is (just so the coefficients are consistent with the above formula): $ax^3+bx^2y+cxy^2+dy^3+ex^2+fxy+gy^2+hx+iy+j=0$

share|cite|improve this question
up vote 6 down vote accepted

The calculation is straightforward using the Riemann-Roch theorem, but not so much without it. Let's look at the poles and zeroes of $x$ and $y$. Let $P$ be the third point of intersection of $Z$ and $C$, and $Q$ the third point of intersection of $X$ and $C$, and $A$ and $B$ be the two other points of intersection of $Y$ with $C$. Remark that:

  1. $Z$ has a double zero at $\mathcal O$ and a simple zero at $P$;
  2. $X$ has a double zero at $P$ and a simple zero at $Q$
  3. $Y$ has a simple zero at $\mathcal O$ and zeroes at $A$ and $B$.

From (1) and (2), $x=X/Z$ has a double pole at $\mathcal O$ (and simple zeros at $P$ and $Q$);

From (1) and (3), $y=Y/Z$ has simple poles at $\mathcal O$ and at $P$ (and simple zeros at $A$ and $B$).

It follows that the function $xy$ has a triple pole at $\mathcal O$ and no other poles. Thus, the functions $x$ and $y':=xy$ only have poles at $\mathcal O$ (of orders $2$ and $3$ respectively). Therefore, the 7 functions

$$1, x, x^2, x^3, y', (y')^2, xy'$$

all have poles of order $\leq 6$ at $\mathcal O$ and no other poles. But the vector space of such functions has dimension $6$ by the Riemann-Roch theorem. Therefore, we know that there is a linear combination of them which is $0$, i.e. a Weierstrass equation for $C$.

In order to find the transformation explicitly, you can express the $7$ functions above as formal Laurent series around $\mathcal O$, with coefficients in $k(a,b,c,d,e,f,g,h,i,j)$, and find a combination which cancels the principal parts. Riemann-Roch is only required to prove that this will work, but if you want to avoid Riemann-Roch and you are extremely patient, it is possible to prove it directly. At some point in the calculation, you will have to divide by the discriminant $D(a,b,c,d,e,f,g,h,i,j)$... good luck with that! :)

share|cite|improve this answer
Thanks for this beautiful answer :) I must definitely learn more about Riemann-Roch Theorem. It is quite powerful. – Prism Sep 10 '13 at 21:13
@Prism My pleasure! – Bruno Joyal Sep 10 '13 at 21:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.