This question arose when I was trying to understand the Guth-Katz paper on Erdos distance problem, namely, related to Proof of Lemma 2.9 in http://arxiv.org/abs/1011.4105. The proof of that lemma is probably wrong (or incomplete) as written in the version I linked, although the lemma itself is of course true. The thing is I don't have any background in algebraic geometry and I'm sure the question will be trivial for experts, but please reply to it using as simple vocabulary as possible :)

What Lemma 2.9 says is basically the following. We have two families of lines in $R^3$. The first one is a ruling of a regulus (since the notion of regulus is a bit obscure, we can assume for simplicity it's just a hyperboloid), which is a 1-dimensional algebraic set of lines. The second one is a certain 2-dimensional algebraic family of lines in $R^3$ that is 'generic' in some sense (for each point in the plane we have one line from this family and the lines are pairwise skew; more precisely, our family is $\mathcal L_p'= \{L_{pq}:q\in R^2\}$ where $p\in R^2$ is fixed and the exact formula for $L_{pq}$ is given on page 8 of http://arxiv.org/PS_cache/arxiv/pdf/1011/1011.4105v3.pdf). The lemma states that if these two families share many lines (take any constant you wish) then the first one must be fully contained in the second one.

Supposing we can somehow express these sets of lines as algebraic sets in some $R^n$, then the problem boils down to showing a statement of roughly this form:

(the numbers below are chosen arbitrarily just to make the question more concrete)

Let $p$ and $q$ be two real polynomials of 10 variables, of degrees at most 20, such that

$p$ is irreducible, and let their zero sets be $P,Q\subset R^{10}$. Suppose $P$ is1-dimensionaland $Q$ is2-dimensional. Then I would like to claim the following: if $P\cap Q$ contains at least a million points, then $P$ must be a subset of $Q$.

* First question*: is the above reasoning correct and how is the theorem on which it's based called? I would like to see the simplest possible formulation of a general theorem that treats such cases (I would guess it's some Bezout-type theorem, but Bezout's theorem from http://en.wikipedia.org/wiki/B%C3%A9zout%27s_theorem doesn't quite do the job). if it's wrong then what would be a natural assumption (instead of irreducibility of $p$), that would save the statement?

EDIT: It turned out (see below the example by algori) the irreducibility condition is not the right one. So we are looking for some other additional conditions for the two sets that would make them satisfy the statement (and that are satisfied for our initial sets from Lemma).

* Second question* : we used above the terms '1- and 2-dimensional'. I cannot really find a definition of the dimension that would be completely elementary, so I wonder whether the above reasoning is correct if we assume the dimension is definied as follows:
a real algebraic set $A\subset R^n$ has dimension $k$ if it contains a subset homeomorphic to the $k$-dimensional ball and doesn't contain a subset homeomorphic to the $(k+1)$-dimensional ball?

NOTE: In fact, the initial motivation for this question (Proof of Lemma 2.9 by Guth and Katz) expired a long time ago, as I realized the above approach is not the right one (the lemma can actually be shown in a really clean way). For that reason, I voted to close this question.