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I am trying to understand the method of finding the points at infinity of elliptic curves. There are good questions that deal with the general theory of doing this, but I couldn't find any good examples.

I understand the following question:

$$C: x^22 + xy − 2y^2 + x − 5y +7=0$$

Homogenizing we find $$C: X^2 + XY - 2Y^2 + XZ - 5YZ + 7Z^2 = 0.$$ To find the points at infinity on $C$, we substitute $Z=0$ and solve: $$X^2 + XY - 2Y^2 = 0$$ Factoring, we have $$(X + 2Y)(X - Y) = 0 \text {, so } X = -2Y \text { or } X = Y \text {.}$$ Thus, there are two points at infinity on $C,[-2,1,0]$ and $[1,1,0]$. These correspond to the "directions" in the affine plane given by the lines $x=-2 y$ and $x=y$.

I'm confused on how to use this method to find the points at infinity of an even simpler equation. Take

$$C_0: 3x - 7y + 5 = 0$$

Homogenizing and taking $Z=0$ just yields $0=0$. What am I doing wrong?

Clyde Kertzer
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1 Answers1

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Homogenizing we find $$3X - 7Y + 5Z = 0.$$ Taking $Z=0$ we have \begin{align*} 3X - 7Y &=0 \\ 3X &= 7 \end{align*} so $X = \frac{7}{3}Y$ which yields the point at infinity $\left[\frac{7}{3}, 1, 0\right]$.

Viktor Vaughn
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Clyde Kertzer
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  • The points $\left[\frac{7}{3} Y : Y : 0\right] = \left[\frac{7}{3} : 1 : 0\right]$ are equal for all $Y \neq 0$; this is part of the definition of projective space. There is only one point on this line at infinity. This should be intuitive: the line at infinity and any other line intersect in exactly one point. – Viktor Vaughn Apr 25 '23 at 13:15