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By conic we understand a conic on the projective plane $\mathbb{P}_2=\mathbb{P}(V)$, where $V$ is $3$-dimensional. I'd like to ask how to find the number of points in the intersection of two given smooth conics $Q_1$ and $Q_2$ (given by $q_1,q_2\in S^2V^*$ respectively). So, we are to find the set of all $(x_0,x_1,x_2)\in V$ such that $q_1(x_0,x_1,x_2)=q_2(x_0,x_1,x_2)=0$. First, is this number necessarily finite? And second, how to describe such intersection of quadrics (i.e. 'conics' in $\mathbb{P}^n$) in higher dimensions? I'm sorry for the lack of my 'research', I'm really stuck on this problem.

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  • $\begingroup$ Search for "Bezout's theorem". Short answer for distinct smooth conics in the plane: the number is finite --- indeed, at most 4. $\endgroup$ – user64687 Jan 8 '14 at 13:04
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First, is this number necessarily finite?

No: a degenerate conic factors into two lines. Two such degenerate conics might have a line in common. This line would be a continuum of intersections. Barring such a common component, the number of intersections is limited to four.

And second, how to describe such intersection of quadrics (i.e. 'conics' in $\mathbb P^n$) in higher dimensions?

If you want to count them, then Bézout's Theorem, as mentioned in a comment, is indeed the way to go. Wikipedia includes a generalization to higher dimensoons. Applied to quadrics, you get the result that barring shared components, $n$ quadrics intersect in up to $2^n$ points. In exactly $2^n$ if you include complex points of intersection and take multiplicities into account.

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    $\begingroup$ Nitpick of your first answer: the OP did say two smooth conics. $\endgroup$ – user64687 Jan 9 '14 at 12:26

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