dimension of the intersection of two quadrics The answer to this question seems obvious to me but I don't know how to prove it or where I can find a reference for it (assuming it is true). If I have two quadrics in R^3, I want to say that they intersect either in a point or a curve. Namely that they can't intersect in a surface unless they're identical. Any help would be greatly appreciated.
 A: If at least one of the two quadrics is irreducible then they can't intersect on a surface, here a visual argument:
Let $Q_1$, $Q_2$ be the quadrics, $Q_1$ irreducible (i.e. different from the union of two planes), and let $Z$ be $Q_1\cap Q_2$ the intersection, $Z\neq Q_1$. Let's suppose $Z$ be a surface.
Let $p$ be a smooth point of $Q_1$, $p\notin Z$ (if $Q_1$ is not a cone than all points of $Q_1$ are smooth, otherwise we need just to consider a point different from the center of the cone).
Let's consider $T_p$, the tangent plane in $p$ to $Q_1$, then $T_p\cap Q_1$ contains at most two lines. Let $Z'=Z-(T_p\cap Q_1)$, then $Z'$ is a surface too. 
Since $Z'$ is a surface, we can consider a plane $\Pi$ such that $p\in \Pi$ and $\Pi \cap Z'$ is a curve (in particular we ask $\Pi\neq T_p$). Now let $C_i=Q_i\cap\Pi$, $W=Z\cap\Pi$.
Then $C_1$ is an irreducible conic (since $\Pi$ is not tangent in $p$ to $Q_1$), and $W$ is equal to $C_2 \cap C_1$, but $W$ is a curve different from $C_1$ since $p \in C_1-W$. So we have an irreducible conic $C_1$ intersecting with a conic $C_2$ in a curve different from $C_1$, which is the initial statement but in dimension two.
In dimension two the assert is easy to prove since a conic can defined from five points, so we conclude.
