# Intersection between projective curves

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

## 1 Answer

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).

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