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How do you compute $V(x^2+y^2-1) \cap V(x^2+y^2-2)$ in $A_{R}^{2}$ and $A_{C}^{2}$ ?

In the real affine 2-plane there are no intersection points (this is clear since these these are just two circles with different radii and do not intersect).

What happens in $A_{C}^2$? My first intuition is that there will be two solutions and they will be complex conjugates of each other. But how do you compute them?

The problem continues and asks you to take the equations into the projective plane. Then asks you to compute their intersection again, in $P_{R}^{2}$ and then in $P_{C}^{2}$, so basically solve;
$V(X^2+Y^2-Z^2) \cap V(X^2+Y^2-2Z^2)$

I learnt Bezouts theorem before attempting this problem and here both my homogeneous polynomials are of degree 2, there should be exactly 4 intersection points in the complex plane. Making both $=0$ and solving simultaneously, I guess that [1:0:0] and [0:1:0] are the two solutions in the real projective plane. (it's a guess but these points work...). but where are the two other solutions for the complex?

So my question is how do I do these computations? Also, is there anywhere where my understanding of things is wrong?

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Hm ... in $\mathbb C$ the intersection is also empty. For you cannot have a point $(x,y) \in \mathbb A_{\mathbb C}^2$ with $x^2 + y^2 = 1$ and $x^2 + y^2 = 2$ ... – martini Apr 22 '12 at 16:59
Dear Kiv, in the projective plane your guess is wrong and your points unfortunately do not work since $1\neq 0$. – Georges Elencwajg Apr 22 '12 at 17:22
For the projective case: Pluging in $[1:0:0]$ I get $1^2 + 0^2 - 0^2 = 1 \ne 0$? Over $\mathbb R$ you have no solutions either: For $Z \ne 0$ is impossible (since this would give affine solutions), so we need to have $Z = 0$, which gives $X = Y = 0$ over $\mathbb R$. Over $\mathbb C$, we also need to have $X^2 + Y^2 = 0$, so which gives $X = 1$, $Y = i$ or $X = 1$, $Y = -i$, so $[1:i:0]$ and $[1:-i:0]$ are two points. Since $X^2 + Y^2 =(X + iY)(X - iY)$ I can't see other two points? Perhaps they are double points? I just dont' know. – martini Apr 22 '12 at 17:24
makes sense ! thank you for your help :) – Kiv Efehe Apr 22 '12 at 17:53
The comments above answered your question, but you should learn a little about the resultant of two polynomials to solve equations more complicated. It is easy and very useful. I recommend Walker, R. J. Algebraic Curves, p. 23. – rla Apr 23 '12 at 22:53
up vote 2 down vote accepted

1) Since the ideals $(x^2+y^2-1), (x^2+y^2-2)$ are comaximal over any field, the corresponding varieties are disjoint in any version of algebraic geometry (classical, scheme-theoretic).

2) To use a pleasantly old-fashioned terminology, the two complex projective circles $V(X^2+Y^2-Z^2)$ and $V(X^2+Y^2-2Z^2)$ go through the cyclic points $(1:\pm i:0)\in \mathbb P^2(\mathbb C) $ and are tangent there.

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