Suppose we have some elliptic curve (just suppose we are working over $\mathbb{R}$) defined by the equation $$Y^2Z + a_1XYZ + a_3YZ^2 = X^3 + a_2X^2Z + a_4XZ^2 + a_6Z^3 $$ with a point at infinity $[0,1,0]$. Now we can go to its affine part $$y^2 + a_1xy + a_3y = x^3 + a_2x^2 + a_4x + a_6 $$ Now I can visualize this curve well, and I can visualize the point at infinity as being where all vertical lines intersect.

Now the typical thing I see (since we aren't in characteristic 2) is complete the square, letting $y \mapsto (y-a_1x - a_3)/2$ our equation becomes $$y^2 = 4x^3 + b_2x^2 + 2b_4x + b_6 $$ where we have $b_2$, $b_4$, $b_6$ defined in the appropriate way.

Now my question is, how is this transformation affecting our point at infinity? What does this transformation look like in terms of the projective equation, not just the affine equation? Im struggling to see how this works in terms of the projective curve, and why we are allowed to do this and why it stays an elliptic curve (why it still looks like the initial equation in some $X,Y,Z$).


This change of coordintes is linear, so it extends to a change of coordinates on $\Bbb P^2$ by $X\mapsto X$, $Y\mapsto (Y-a_1X-a_3Z)/2$, $Z\mapsto Z$. You can see directly that this preserves the point at infinity $[0:1:0]$, and that it gives you the projective equation $Y^2Z=4X^3+b_2X^2Z+2b_4XZ^2+b_6Z^3$.

  • $\begingroup$ Ahhh okay yes this makes sense. So now my question is; why do we know this is an elliptic curve? My working definition is a smooth projective curve satisfying the first equation I gave. This has that annoying 4 in front of the X^3, is there a way to get rid of that? $\endgroup$
    – Bagggggs
    Jan 23 '21 at 23:58
  • $\begingroup$ Yes: send $Y\mapsto 2Y$, divide the whole equation by $4$. You could have instead made the substitution $Y\mapsto Y-(a_1X-a_3Z)/2$ at the start. $\endgroup$
    – KReiser
    Jan 24 '21 at 0:03
  • $\begingroup$ okay yes yes yes. this makes sense to me. Thank you! $\endgroup$
    – Bagggggs
    Jan 24 '21 at 0:05

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