# Number of solutions to polynomials over finite fields

Let $m = p_{1}p_{2}\cdots p_{s}$. Let $N_{f}(n)$ denote the number of solutions to $f \equiv 0 \pmod{n}$. If $f = f(x)$, a polynomial in the single variable $x$, then from the Chinese Remainder Theorem, we have $N_{f(x)}(m) = \prod_{i = 1}^{s}N_{f(x)}(p_{i})$. Now suppose $f = f(x, y)$, a polynomial in the variables $x$ and $y$. Is $N_{f(x, y)}(m) = \prod_{i = 1}^{s}N_{f(x, y)}(p_{i})$? Can I reduce this down to the single variable case?

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What sort of form do you want the answer to be in? And also, what makes you suspect such a reduction can be made? It seems quite unnatural that the number of roots of a multivariate polynomial is related to the roots of certain single variable polynomials in a nontrivial way. – dinoboy Nov 28 '12 at 6:58
There's no need to reduce to the single-variable case. If the $p_i$ are distinct, it still follows from CRT. – Qiaochu Yuan Nov 28 '12 at 6:59
You ask two questions, but I think you answered the first, because the argument with CRT you gave doesn't use the number of variables. For the second question, I think the object $N_{f(x)}(p)$ is much much simpler than $N_{f(x,y)}(p)$, so it is hard for me to believe there will be a way to understand the latter based on the former (which is well understood). – Lior B-S Nov 28 '12 at 7:01