# Confusion: homogeneous Weierstraß equation for elliptic curves

I am learning about the basics of elliptic curves, using Vakil's notes on algebraic geometry (that can be found here, chapter 19, "genus one curves"). Here, he establishes that an elliptic curve over a field $k$ (with $\operatorname{char}(k) \not = 2,3$) may be described as a cubic regular curve in $\mathbb{P}^2_k$ defined by an homogeneous equation of the form $$\begin{equation}\tag{E} y^2z=x^3+ax^2z+bz^3 \end{equation}$$ where $a$ and $b$ are parameters taken in $k$ (Weiestraß normal form).

Now, in many sources, a Weierstraß equation is rather defined as $$\begin{equation} \tag{E'}y^2=x^3+ax+b \end{equation}$$ which is supposedly obtained from $(\operatorname{E})$ by looking at the intersection with the distinguished open affine subset of $\mathbb{P}^2_k$ defined by $z\not = 0$.

This correspondance is even stated in the answer of this question.

But when taking $z=1$ in $(\operatorname{E})$, the equation we actually obtain is $y^2=x^3+ax^2+b$, right? What happened to the $x^2$ term then?

I could guess two possible explanations for this, and I am confused about which one is correct.

First explanation, there is a typo, and the actual Weierstraß normal form would rather be $y^2z=x^3+axz^2+bz^3$. This would be consistent with the fact that in Vakil's note, page 516, he actually states that we make the coefficient of $x^2z$ equal $0$ by translating the variable $x$ with a multiple of $z$.

Second explanation, after taking $z=1$ in $(\operatorname{E})$ and getting the equation $y^2=x^3+ax^2+b$, we translate the variable $x\rightarrow x-\frac{a}{3}$ to eliminate the $x^2$ term. Then, when we associate the equations $(\operatorname{E})$ and $(\operatorname{E'})$, it is implicitly meant that the coefficients $a$ and $b$ are not necessarily the same (namely after translating the variable $x$, we obtain coefficients $a'=-\frac{a^2}{3}$ and $b'=b+\frac{2a^3}{27}$ for $(\operatorname{E'})$).

Is any of these two explanations the correct one?
I thank you very much for your clarification.

• I'm sure your first explanation is the correct one. – Lord Shark the Unknown Jul 5 '18 at 6:14
• Thank you very much @LordSharktheUnknown for clarifying ! Then, I shall always use the the equation $y^2z=x^3+axz^2+bz^3$ instead when dealing with elliptic curves. – Suzet Jul 5 '18 at 23:50

Both your explanations are correct. Surely, there is a typo in the notes you are reading, as the standard short Weierstrass model is of the form $y^2=x^3+ax+b$. However, as you note, it is also true that every elliptic curve over a field of characteristic $\neq 2,3$ has a model of the form $y^2=x^3+ax^2+b$ (and one can go back and forth using the change of variables you note). I would suggest you also read a more standard source along with the notes you are reading, such as Silverman's "The Arithmetic of Elliptic Curves" which contain more details that may be useful and clarifying.