They ask me to calculate the points of intersection of the projective curves $ C_1 = Z (x^2 + y^2-z^2) $ and $ C_2 = Z (x^2 + y^2 -2z^2) $

What I have done:

I tried to solve the system

$ x^2 + y^2 = z^2 $

$ x^2 + y^2 = 2z^2 $

equaling $ z^2 = x^2 + y^2 = 2z^2 $ we have $ z^2 = 2z^2 $, which translates to $ z = 0 $, that is, points $ [x: y: 0] $ such that $ x^2 + y^2 = 0 $, which is true if and only if $ x = y = 0 $, that is, its intersection is only the point $ [0: 0: 0] \not \in \mathbb{P}^2 $

I do not know what the error is in what I did, on the other hand, we have by Bezout's Theorem that the curves have to intersect in 4 points (counting multiplicity)

Could someone help me with this?

Thank you


Bezout's Theorem depends on the field you are working over. Roughly, the Theorem states that if you have curves of degree $d$ and $e$, then the curves will intersect in exactly $de$ points in the Algebraic Closure Of Your Field. It seems like you are viewing your curves over $\mathbb{R}$ which is not algebraically closed. Try doing your calculations in $\mathbb{C}$ and see what you get (for instance, $[i:1:0]$ should be a point of intersection).

  • $\begingroup$ In this case, the field is algebraically closed, thanks for the comment $\endgroup$ – Erick David Luna Núñez Jun 4 at 17:17

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