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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

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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).

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  • $\begingroup$ 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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