# Counting points of intersections of linear and quadratic projective varieties over finite fields

Some computation I was doing reduced to counting the number of points in a algebraic subvariety of the 34-dimensional projective space over $\mathbb{F}_q$ (expressed as a polynomial in $q$). Googling a little for methods to solve this type of problems I found that this is the subject of the notorious Weil conjectures, whose proof I do not expect to understand in my lifetime.

However, my particular variety is really user-friendly, being the intersection of the zero sets of polynomials that are all of degree either 1 or 2. I would intuitively expect that in this case the counting problem is much simpler and a method of doing it has been known already around the time dr. Weil was born. Is that true and if so, can someone point me to a place in the literature where I can learn the method of solving the counting problem for these projective varieties of very low degree? Any help is appreciated.

• You don't need to understand the proofs of the Weil conjectures to use them! Also, it's known that every projective variety is the intersection of zero sets of quadratics. One quadratic is easy but two quadratics can already get you elliptic curves. (You can eliminate the linear equations immediately by substitution; those aren't the hard part.) So if there's an easy answer you'll need to use some special feature of your polynomials. – Qiaochu Yuan Oct 20 '17 at 20:17
• Wait, can you elaborate on how to get rid of the linear equations? This could be useful beyond this particular problem – Vincent Oct 20 '17 at 21:00
• If one of the linear equations is $\sum a_i x_i = 0$ then pick some $x_i$ such that $a_i \neq 0$ and rewrite the equation as $x_i = \frac{\sum_{j \neq i} a_j x_j}{a_j}$, then substitute for all of the appearances of $x_i$ in your other equations. – Qiaochu Yuan Oct 20 '17 at 21:08
• Great! I will see what things look like after these substitutions. Also how is it possible that every subvariety is an intersection of quadrics? Isn't that really weird? – Vincent Oct 20 '17 at 21:13
• I'm not saying that every subvariety of $\mathbb{P}^n$ is an intersection of quadrics, that's definitely false, I'm saying that every projective variety is isomorphic to a subvariety of some $\mathbb{P}^n$ which is an intersection of quadrics. So with no further hypotheses the assumption that the defining equations have degree $1$ or $2$ does not actually reduce the generality of the problem. – Qiaochu Yuan Oct 20 '17 at 22:34